Mapping degrees between spherical $3$-manifolds

TL;DR

This paper explicitly calculates the degrees of maps between spherical 3-manifolds, linking algebraic properties of fundamental group homomorphisms to geometric degrees, thereby determining the set of possible degrees.

Contribution

It provides explicit formulas for degrees of maps between spherical 3-manifolds based on fundamental group homomorphisms, extending previous understanding.

Findings

Explicit calculation of degree sets for surjective homomorphisms

Method to determine degrees for arbitrary homomorphisms

Complete characterization of possible degrees between spherical 3-manifolds

Abstract

Let $D(M,N)$ be the set of integers that can be realized as the degree of a map between two closed connected orientable manifolds $M$ and $N$ of the same dimension. For closed $3$-manifolds with $S^3$-geometry $M$ and $N$, every such degree $deg f\equiv \overline{deg}\psi$ $(|\pi_1(N)|)$ where $0\le \overline{deg}\psi <|\pi_1(N)|$ and $\overline{deg}\psi$ only depends on the induced homomorphism $\psi=f_{\pi}$ on the fundamental group. In this paper, we calculate explicitly the set $\{\overline{deg}\psi\}$ when $\psi$ is surjective and then we show how to determine $\overline{deg}(\psi)$ for arbitrary homomorphisms. This leads to the determination of the set $D(M,N)$.

Tables

Table 1. Table 2.1. types of fundamental groups of spherical 3 3 3 -manifolds

group	condition	presentation	normal form
${\mathbb{Z}}_n$		$⟨c\mid c^n=1⟩$	$c^s$,
$D_{4n}^{\ast }$	$2|n$	$⟨b,a\mid a^2=b^n={\left(ab\right)}^2,a^4=1⟩$	$b^s$, $b^{s}a$,
$O_{48}^{\ast }$		$⟨b,a\mid a^2=b^3={\left(ab\right)}^4,a^4=1⟩$	see Lemma 2.5
$I_{120}^{\ast }$		$⟨b,a\mid a^2=b^3={\left(ab\right)}^5,a^4=1⟩$	see Lemma 2.7
$T_{8\cdot 3^q}^{\prime }$	$q\ge 1$	$\begin{array}{cc}⟨b,a,w\mid a^2=b^2={\left(ab\right)}^2,& \\ a^4=w^{3^q}=1,& \\ wa=bw,wb=abw⟩& \end{array}$	$b^s{w}^t$, $b^{s}a{w}^t$,
$D_{n\cdot 2^q}^{\prime }$	$2|̸n,n>1,q>1$	$⟨u,w\mid u^n=w^{2^q}=1,uwu=w⟩$	$u^s{w}^t$

Table 2. Table 2.2. characteristic subgroups

group	commutators	abelianization	center
${\mathbb{Z}}_n$	$1$	$⟨c⟩\cong {\mathbb{Z}}_n$	$⟨c⟩\cong {\mathbb{Z}}_n$
$D_{4n}^{\ast }$ ($2|n$)	$⟨b^2⟩\cong {\mathbb{Z}}_n$	$⟨\overline{b},\overline{a}⟩\cong {\mathbb{Z}}_2\oplus {\mathbb{Z}}_2$	$⟨b^n⟩\cong {\mathbb{Z}}_2$
$O_{48}^{\ast }$	$T_{24}^{\ast }$	${\mathbb{Z}}_2$	$⟨a^2⟩\cong {\mathbb{Z}}_2$
$I_{120}^{\ast }$	$I_{120}^{\ast }$	$1$	$⟨a^2⟩\cong {\mathbb{Z}}_2$
$T_{8\cdot 3^q}^{\prime }$	$⟨a,b⟩\cong D_{4\cdot 2}^{\ast }$	$⟨w⟩\cong {\mathbb{Z}}_{3^q}$	$⟨b^2,w^3⟩\cong {\mathbb{Z}}_{2\cdot 3^{q-1}}$
$D_{n\cdot 2^q}^{\prime }$ ($2|̸n$)	$⟨u⟩\cong {\mathbb{Z}}_n$	$⟨w⟩\cong {\mathbb{Z}}_{2^q}$	$⟨w^2⟩\cong {\mathbb{Z}}_{2^{q-1}}$

Table 3. Table 2.3 . conjugacy classes of D 4 ​ n ∗ subscript superscript 𝐷 4 𝑛 D^{*}_{4n}

representative	order	all elements
$\left[1\right]$	$1$	$\left\{1\right\}$
$\left[b^s\right]$	$\frac{2n}{(n,s)}$	$\{b^s,b^{2n-s}\}$
$\left[b^n\right]$	$2$	$\left\{b^n\right\}$
$\left[a\right]$	$4$
$\left[ba\right]$	$4$

Table 4. Table 2.4 . subgroups and quotients of D 4 ​ n ∗ subscript superscript 𝐷 4 𝑛 D^{*}_{4n}

subgroup type	conjugacy classes	quotient
${\mathbb{Z}}_{\frac{2n}{k}}$ ($k|n$)	$\left\{⟨b^k⟩\right\}$	$D_{2k}$
${\mathbb{Z}}_{\frac{2n}{k}}$ ($k|2n$, $k|̸n$)	$\left\{⟨b^k⟩\right\}$	$D_{4\cdot \frac{k}{2}}^{\ast }$
${\mathbb{Z}}_4$		${\mathbb{Z}}_2$ if $n=2$
${\mathbb{Z}}_4$		${\mathbb{Z}}_2$ if $n=2$
$D_{4\cdot \frac{n}{k}}^{\ast }$ ($k|n$, $2|\frac{n}{k}$)		${\mathbb{Z}}_2$ if $k=2$
$D_{\frac{n}{k}\cdot 2^2}^{\prime }$ ($k|n$, $2|̸\frac{n}{k}$)		${\mathbb{Z}}_2$ if $k=2$
$D_{4\cdot \frac{n}{k}}^{\ast }$ ($k|n$, $2|\frac{n}{k}$)		${\mathbb{Z}}_2$ if $k=2$
$D_{\frac{n}{k}\cdot 2^2}^{\prime }$ ($k|n$, $2|̸\frac{n}{k}$)		${\mathbb{Z}}_2$ if $k=2$

Table 5. Table 2.5 . conjugacy classes of O 48 ∗ subscript superscript 𝑂 48 O^{*}_{48}

class	order	all elements
$\left[a\right]$	$4$	$\{a,a^{-1},ba{b}^{-1},b{a}^{-1}{b}^{-1},b^{-1}ab,b^{-1}{a}^{-1}b,aba{b}^{-1}a$,
		$aba{b}^{-1}{a}^{-1},a{b}^{-1}aba,a{b}^{-1}ab{a}^{-1},ba{b}^{-1}aba,ba{b}^{-1}ab{a}^{-1}\}$
$\left[baba\right]$	$4$	$\{baba,bab{a}^{-1},b^{-1}a{b}^{-1}a,b^{-1}a{b}^{-1}{a}^{-1},ba{b}^{-1}ab,ba{b}^{-1}{a}^{-1}b\}$
$\left[b^2\right]$	$3$	$\{b^2,b^{-2},aba,a{b}^{-1}a,ba{b}^{-1}a,b^{-1}aba,aba{b}^{-1},a{b}^{-1}ab\}$
$\left[b\right]$	$6$	$\{b,b^{-1},ab{a}^{-1},a{b}^{-1}{a}^{-1},ba{b}^{-1}{a}^{-1},b^{-1}ab{a}^{-1},ab{a}^{-1}{b}^{-1},a{b}^{-1}{a}^{-1}b\}$
$\left[ba\right]$	$8$	$\{ba,b^{-1}{a}^{-1},ab,a^{-1}{b}^{-1},bab,b^{-1}{a}^{-1}{b}^{-1}\}$
$\left[b{a}^{-1}\right]$	$8$	$\{b{a}^{-1},b^{-1}a,a{b}^{-1},a^{-1}b,b{a}^{-1}b,b^{-1}a{b}^{-1}\}$
$\left[a^2\right]$	$2$	$\left\{a^2\right\}$
$\left[1\right]$	$1$	$\left\{1\right\}$

Table 6. Table 2.6 . conjugacy classes of I 120 ∗ subscript superscript 𝐼 120 I^{*}_{120}

class	order	all elements
$\left[a\right]$	$4$	$\begin{array}{c}\{a,a^{-1},ba{b}^{-1},b{a}^{-1}{b}^{-1},b^{-1}ab,b^{-1}{a}^{-1}b,aba{b}^{-1}a,aba{b}^{-1}{a}^{-1},a{b}^{-1}aba,\\ a{b}^{-1}ab{a}^{-1},baba{b}^{-1}a{b}^{-1},baba{b}^{-1}{a}^{-1}{b}^{-1},ba{b}^{-1}aba{b}^{-1},ba{b}^{-1}ab{a}^{-1}{b}^{-1},\\ b^{-1}aba{b}^{-1}ab,b^{-1}aba{b}^{-1}{a}^{-1}b,b^{-1}a{b}^{-1}abab,b^{-1}a{b}^{-1}ab{a}^{-1}b,baba{b}^{-1}aba,\\ baba{b}^{-1}ab{a}^{-1},b^{-1}a{b}^{-1}aba{b}^{-1}a,b^{-1}a{b}^{-1}aba{b}^{-1}{a}^{-1},ba{b}^{-1}aba{b}^{-1}ab,\\ ba{b}^{-1}aba{b}^{-1}{a}^{-1}b,aba{b}^{-1}aba{b}^{-1}a,aba{b}^{-1}aba{b}^{-1}{a}^{-1},a{b}^{-1}aba{b}^{-1}aba,\\ a{b}^{-1}aba{b}^{-1}ab{a}^{-1},ba{b}^{-1}aba{b}^{-1}aba,ba{b}^{-1}aba{b}^{-1}ab{a}^{-1}\}\end{array}$
$\left[baba\right]$	$5$	$\begin{array}{c}\{baba,b^{-1}a{b}^{-1}a,abab,a{b}^{-1}a{b}^{-1},ba{b}^{-1}{a}^{-1}b,b^{-1}ab{a}^{-1}{b}^{-1},aba{b}^{-1}aba,\\ a{b}^{-1}aba{b}^{-1}a,ba{b}^{-1}aba{b}^{-1}a,b^{-1}aba{b}^{-1}aba,aba{b}^{-1}aba{b}^{-1},a{b}^{-1}aba{b}^{-1}ab\}\end{array}$
$\left[bab{a}^{-1}\right]$	$10$	$\begin{array}{c}\{bab{a}^{-1},b^{-1}a{b}^{-1}{a}^{-1},ab{a}^{-1}b,a{b}^{-1}{a}^{-1}{b}^{-1},ba{b}^{-1}ab,b^{-1}aba{b}^{-1},aba{b}^{-1}ab{a}^{-1},\\ a{b}^{-1}aba{b}^{-1}{a}^{-1},ba{b}^{-1}aba{b}^{-1}{a}^{-1},b^{-1}aba{b}^{-1}ab{a}^{-1},aba{b}^{-1}ab{a}^{-1}{b}^{-1},\\ a{b}^{-1}aba{b}^{-1}{a}^{-1}b\}\end{array}$
$\left[b^2\right]$	$3$	$\begin{array}{c}\{b^2,b^{-2},aba,a{b}^{-1}a,baba{b}^{-1},ba{b}^{-1}a{b}^{-1},b^{-1}abab,b^{-1}a{b}^{-1}ab,baba{b}^{-1}a,\\ ba{b}^{-1}aba,b^{-1}aba{b}^{-1}{a}^{-1},b^{-1}a{b}^{-1}ab{a}^{-1},aba{b}^{-1}ab,aba{b}^{-1}{a}^{-1}{b}^{-1},\\ a{b}^{-1}abab,a{b}^{-1}ab{a}^{-1}{b}^{-1},baba{b}^{-1}ab,ba{b}^{-1}abab,b^{-1}aba{b}^{-1}{a}^{-1}{b}^{-1},\\ b^{-1}a{b}^{-1}ab{a}^{-1}{b}^{-1}\}\end{array}$
$\left[b\right]$	$6$	$\begin{array}{c}\{b,b^{-1},ab{a}^{-1},a{b}^{-1}{a}^{-1},bab{a}^{-1}{b}^{-1},ba{b}^{-1}{a}^{-1}{b}^{-1},b^{-1}ab{a}^{-1}b,b^{-1}a{b}^{-1}{a}^{-1}b,\\ baba{b}^{-1}{a}^{-1},ba{b}^{-1}ab{a}^{-1},b^{-1}aba{b}^{-1}a,b^{-1}a{b}^{-1}aba,aba{b}^{-1}a{b}^{-1},aba{b}^{-1}{a}^{-1}b,\\ a{b}^{-1}aba{b}^{-1},a{b}^{-1}ab{a}^{-1}b,baba{b}^{-1}{a}^{-1}b,ba{b}^{-1}ab{a}^{-1}b,b^{-1}aba{b}^{-1}a{b}^{-1},\\ b^{-1}a{b}^{-1}aba{b}^{-1}\},\end{array}$
$\left[ba\right]$	$10$	$\begin{array}{c}\{ba,b^{-1}{a}^{-1},ab,a^{-1}{b}^{-1},bab,b^{-1}{a}^{-1}{b}^{-1},ba{b}^{-1}{a}^{-1},b^{-1}ab{a}^{-1},ab{a}^{-1}{b}^{-1},\\ a{b}^{-1}{a}^{-1}b,babab,b^{-1}a{b}^{-1}a{b}^{-1}\}\end{array}$
$\left[b{a}^{-1}\right]$	$5$	$\begin{array}{c}\{b{a}^{-1},b^{-1}a,a{b}^{-1},a^{-1}b,b{a}^{-1}b,b^{-1}a{b}^{-1},ba{b}^{-1}a,b^{-1}aba,aba{b}^{-1},a{b}^{-1}ab,\\ bab{a}^{-1}b,b^{-1}a{b}^{-1}{a}^{-1}{b}^{-1}\}\end{array}$
$\left[a^2\right]$	$2$	$\left\{a^2\right\}$
$\left[1\right]$	$1$	$\left\{1\right\}$

Table 7. Table 2.7 . conjugacy classes of T 8 ⋅ 3 q ′ subscript superscript 𝑇 ′ ⋅ 8 superscript 3 𝑞 T^{\prime}_{8\cdot 3^{q}}

	representative	order	all elements
1	$\left[w^{3t}\right]$	$\frac{3^q}{(3^q,3t)}$	$\left\{w^{3t}\right\}$
2	$\left[b{w}^{3t}\right]$	$\frac{4\cdot 3^q}{(3^q,3t)}$	$\{b{w}^{3t},b^3{w}^{3t},a{w}^{3t},ba{w}^{3t},b^{2}a{w}^{3t},b^{3}a{w}^{3t}\}$
3	$\left[b^2{w}^{3t}\right]$	$\frac{2\cdot 3^q}{(3^q,3t)}$	$\left\{b^2{w}^{3t}\right\}$
4	$\left[w^{3t+1}\right]$	$3^q$	$\{w^{3t+1},b^3{w}^{3t+1},ba{w}^{3t+1},b^{2}a{w}^{3t+1}\}$
5	$\left[b{w}^{3t+1}\right]$	$2\cdot 3^q$	$\{b{w}^{3t+1},b^2{w}^{3t+1},a{w}^{3t+1},b^{3}a{w}^{3t+1}\}$
6	$\left[w^{3t+2}\right]$	$3^q$	$\{w^{3t+2},b{w}^{3t+2},a{w}^{3t+2},b^{3}a{w}^{3t+2}\}$
7	$\left[b^2{w}^{3t+2}\right]$	$2\cdot 3^q$	$\{b^2{w}^{3t+2},b^3{w}^{3t+2},ba{w}^{3t+2},b^{2}a{w}^{3t+2}\}$

Table 8. Table 2.8 . subgroups and quotients of T 8 ⋅ 3 q ′ subscript superscript 𝑇 ′ ⋅ 8 superscript 3 𝑞 T^{\prime}_{8\cdot 3^{q}}

	subgroup type	conjugacy classes	quotient
1	${\mathbb{Z}}_{3^{q-r}}$ ( )	$\left\{⟨w^{3^r}⟩\right\}$	$T_{8\cdot 3^r}^{\prime }$
2	${\mathbb{Z}}_{4\cdot 3^{q-r}}$ ()	$\{⟨b{w}^{3^r}⟩,⟨b^3{w}^{3^r}⟩,⟨a{w}^{3^r}⟩,⟨ba{w}^{3^r}⟩,⟨b^{2}a{w}^{3^r}⟩,⟨b^{3}a{w}^{3^r}⟩\}$	-
3	${\mathbb{Z}}_{2\cdot 3^{q-r}}$ ()	$\left\{⟨b^2{w}^{3^r}⟩\right\}$	$\left({\mathbb{Z}}_2\times {\mathbb{Z}}_2\right)⋊{\mathbb{Z}}_{3^r}$
4	${\mathbb{Z}}_{3^q}$	$\{⟨w⟩,⟨b^{3}w⟩,⟨baw⟩,⟨b^{2}aw⟩\}$	-
5	${\mathbb{Z}}_{2\cdot 3^q}$	$\{⟨bw⟩,⟨aw⟩,⟨b^{2}w⟩,⟨b^{3}aw⟩\}$	-
6	$D_{4\cdot 2}^{\ast }\times {\mathbb{Z}}_{3^{q-r}}$ ()	$\left\{⟨b,a,w^{3^r}⟩\right\}$	${\mathbb{Z}}_{3^r}$

Table 9. Table 2.9 . conjugacy classes of D n ⋅ 2 q ′ subscript superscript 𝐷 ′ ⋅ 𝑛 superscript 2 𝑞 D^{\prime}_{n\cdot 2^{q}} with 2 |̸ n conditional 2 𝑛 2\not|n

representative	order	all elements
$\left[w^{2t+1}\right]$  ()	$2^q$
$\left[w^{2t}\right]$ ()	$\frac{2^q}{(2^q,2t)}$	$\left\{w^{2t}\right\}$
$\left[u^s{w}^{2t}\right]$  ()	$\frac{n}{(s,n)}\times \frac{2^q}{(2^q,2t)}$	$\{u^s{w}^{2t},u^{n-s}{w}^{2t}\}$

Table 10. Table 2.10 . subgroups and quotients of D n ⋅ 2 q ′ subscript superscript 𝐷 ′ ⋅ 𝑛 superscript 2 𝑞 D^{\prime}_{n\cdot 2^{q}} with 2 |̸ n conditional 2 𝑛 2\not|n

	subgroup type	conjugacy classes	quotient
1	${\mathbb{Z}}_{2^q}$	$\{⟨u^{s}w⟩\mid s=0,1,\dots ,n-1\}$	-
2	${\mathbb{Z}}_{2^{q-1}}$	$⟨w^2⟩$	${\mathbb{Z}}_n⋊{\mathbb{Z}}_2$
3	${\mathbb{Z}}_{2^{q-r}}$  ()	$⟨w^{2^r}⟩$	$D_{n\cdot 2^r}^{\prime }$
4	${\mathbb{Z}}_n$	$⟨u⟩$	${\mathbb{Z}}_{2^q}$
5	${\mathbb{Z}}_k$  ()	$⟨u^k⟩$	$D_{k\cdot 2^q}^{\prime }$
6	${\mathbb{Z}}_{n\cdot 2^{q-r}}$  ()	$⟨u{w}^{2^r}⟩$	${\mathbb{Z}}_{2^r}$
7	${\mathbb{Z}}_{\frac{n}{k}\cdot 2^{q-1}}$  ()	$⟨u^k{w}^2⟩$	${\mathbb{Z}}_k⋊{\mathbb{Z}}_2$
8	${\mathbb{Z}}_{\frac{n}{k}\cdot 2^{q-r}}$  ()	$⟨u^k{w}^{2^r}⟩$	$D_{k\cdot 2^r}^{\prime }$
9	$D_{\frac{n}{k}\cdot 2^q}^{\prime }$   ()	$\{⟨u^k,u^{s}w⟩\mid s=0,1,\dots ,n\}$	-

Table 11. Table 3.1 . lens spaces that cover S 3 / G superscript 𝑆 3 𝐺 S^{3}/G

$G$	$H$	$\left|H\right|$	lens space
${\mathbb{Z}}_m\times D_{4n}^{\ast }$	$⟨v,a⟩$	$4m$	$L(4m;4-m,4+m)$
${\mathbb{Z}}_m\times D_{4n}^{\ast }$	$⟨v,ba⟩$	$4m$	$L(4m;4-m,4+m)$
${\mathbb{Z}}_m\times D_{4n}^{\ast }$	$⟨v,b⟩$	$2mn$	$L(2mn;2n-m,2n+m)$
${\mathbb{Z}}_m\times O_{48}^{\ast }$	$⟨v,b⟩$	$6m$	$L(6m;6-m,6+m)$
${\mathbb{Z}}_m\times O_{48}^{\ast }$	$⟨v,ba⟩$	$8m$	$L(8m;8-m,8+m)$
${\mathbb{Z}}_m\times O_{48}^{\ast }$	$⟨v,b{a}^{-1}⟩$	$8m$	$L(8m;8-3m,8+3m)$
${\mathbb{Z}}_m\times I_{120}^{\ast }$	$⟨v,a⟩$	$4m$	$L(4m;4-m,4+m)$
${\mathbb{Z}}_m\times I_{120}^{\ast }$	$⟨v,b⟩$	$6m$	$L(6m;6-m,6+m)$
${\mathbb{Z}}_m\times I_{120}^{\ast }$	$⟨v,ba⟩$	$10m$	$L(10m;m-10,m+10)$
${\mathbb{Z}}_m\times I_{120}^{\ast }$	$⟨v,bab{a}^{-1}⟩$	$10m$	$L(10m;3m-10,3m+10)$
${\mathbb{Z}}_m\times T_{8\cdot 3^q}^{\prime }$	$⟨v,bw⟩,⟨v,b^2{w}^2⟩$	$2m{3}^q$	$L(2m{3}^q;2m+2\cdot 3^q-m{3}^{q-1},2m+2\cdot 3^q+m{3}^{q-1})$
${\mathbb{Z}}_m\times T_{8\cdot 3^q}^{\prime }$	$⟨v,a,w^3⟩$	$4m{3}^{q-1}$	$L(4m{3}^{q-1};4m+4\cdot 3^{q-1}-m{3}^{q-1},4m+4\cdot 3^{q-1}+m{3}^{q-1})$
${\mathbb{Z}}_m\times D_{n\cdot 2^q}^{\prime }$	$⟨v,w⟩$	$m{2}^q$	$L(m{2}^q;m+2^{q-1},m+2^{q-1}+m{2}^{q-1})$
${\mathbb{Z}}_m\times D_{n\cdot 2^q}^{\prime }$	$⟨v,u,w^2⟩$	$mn{2}^{q-1}$	$L(mn{2}^{q-1};n{2}^{q-1}+mn+m{2}^{q-1},n{2}^{q-1}+mn-m{2}^{q-1})$

Table 12. Table 4.1 . endomorphisms of D 4 ​ n ∗ subscript superscript 𝐷 4 𝑛 D^{*}_{4n}

$\psi$	$\ker \psi$		$\psi \left(b\right)$	$\psi \left(a\right)$	$\mathrm{Im}\psi$	$\overline{\mathrm{deg}}\left(\psi \right)$
${\psi }_1$	$D_{4n}^{\ast }$		$1$	$1$	$1$	$0$
${\psi }_2$	$⟨b^2,a⟩\cong D_{4\cdot \frac{n}{2}}^{\ast }$	($2|\frac{n}{2}$)	$b^n$	$1$	$⟨b^n⟩\cong {\mathbb{Z}}_2$	$2n$
	$⟨b^2,a⟩\cong D_{\frac{n}{2}\cdot 2^2}^{\prime }$	($2|̸\frac{n}{2}$)	$b^n$	$1$	$⟨b^n⟩\cong {\mathbb{Z}}_2$	$0$
${\psi }_3$	$⟨b⟩\cong {\mathbb{Z}}_{2n}$		$1$	$b^n$	$⟨b^n⟩\cong {\mathbb{Z}}_2$	$0$
${\psi }_{4,k}$	$⟨b^{\frac{2n}{k}}⟩\cong {\mathbb{Z}}_k$	$\left(k\right|n,2|̸k)$	$b^k$	$a$	$⟨b^k,a⟩\cong D_{4\cdot \frac{n}{k}}^{\ast }$	$k^2$
${\psi }_{5,l}$	$1$		$b^l$ $(2n,l)=1$	$a$	$D_{4n}^{\ast }$	$l^2$
${\psi }_6$	$1$		$b$	$ba$	$D_{4n}^{\ast }$	$1$
${\psi }_7$	$⟨b^4⟩\cong {\mathbb{Z}}_{\frac{n}{2}}$	($2|̸\frac{n}{2}$)	$b^{\frac{n}{2}}a$	$a$	$⟨b^{\frac{n}{2}},a⟩\cong D_{4\cdot 2}^{\ast }$	$\frac{n^2}{4}$
${\psi }_8$	$⟨b^4⟩\cong {\mathbb{Z}}_{\frac{n}{2}}$	($2|̸\frac{n}{2}$)	$a$	$b^{\frac{n}{2}}$	$⟨b^{\frac{n}{2}},a⟩\cong D_{4\cdot 2}^{\ast }$	$\frac{n^2}{4}$

Table 13. Table 4.2 . domain D 4 ​ n ∗ subscript superscript 𝐷 4 𝑛 D^{*}_{4n}

$G$	$S^3/G$	$\psi \left(b\right)$	$\psi \left(a\right)$	$\overline{\mathrm{deg}}\left(\psi \right)$
$D_{4m}^{\ast }=⟨\overline{b},\overline{a}\mid {\overline{a}}^2={\overline{b}}^n={\left(\overline{a}\overline{b}\right)}^2,{\overline{a}}^4=1⟩$	$S^3/D_{4m}^{\ast }$	$\overline{b}$	$\overline{a}$	$\frac{n}{m}$
${\mathbb{Z}}_2=⟨c\mid c^2⟩$	$L(2;1,1)$	$c$	$1$	$1+\frac{n}{2}$
${\mathbb{Z}}_2=⟨c\mid c^2⟩$	$L(2;1,1)$	$1$	$c$	$0$

Table 14. Table 4.3 . endomorphisms of T 8 ⋅ 3 q ′ subscript superscript 𝑇 ′ ⋅ 8 superscript 3 𝑞 T^{\prime}_{8\cdot 3^{q}}

$\psi$	$\ker \psi$		$\psi \left(a\right)$	$\psi \left(w\right)$	$\mathrm{Im}\psi$	$\overline{\mathrm{deg}}\left(\psi \right)$
${\psi }_1$	$1$		$b^{-1}$	$w^2$	$T_{8\cdot 3^q}^{\prime }$	$4-3^{2q+1}$
${\psi }_2$	$⟨b,a,w^{3^{q-1}}⟩\cong D_{4\cdot 2}^{\ast }\times {\mathbb{Z}}_3$		$1$	$w^3$	$⟨w^3⟩\cong {\mathbb{Z}}_{3^{q-1}}$	$9-3^{2q+2}$
${\psi }_3$	$⟨b,a⟩\cong D_{4\cdot 2}^{\ast }$		$1$	$w$	$⟨w⟩\cong {\mathbb{Z}}_{3^q}$	$1-3^{2q}$
${\psi }_4$	$T_{8\cdot 3^q}^{\prime }$		$1$	$1$	$1$	$0$

Equations

$$b(b^{s}aw)b^{-1}=b^{s+1}a(ab{)}^{-1}w=b^{s+2}w,b(b^{s}w)b^{-1}=b^{s+1}(ab{)}^{-1}w=b^{s+2}aw,$$

$$a(b^{s}aw)a^{-1}=b^{-s}{a}^2{b}^{-1}w=b^{1-s}w,a(b^{s}w)a^{-1}=a{b}^s{b}^{-1}w=b^{1-s}aw.$$

$$\begin{array}[]{llll}aw\sim b^{2}w,&baw\sim b^{3}w,&b^{2}aw\sim w,&b^{3}aw\sim bw;\\ w\sim b^{2}aw,&bw\sim b^{3}aw,&b^{2}w\sim aw,&b^{3}w\sim baw;\\ aw\sim bw,&baw\sim w,&b^{2}aw\sim b^{3}w,&b^{3}aw\sim b^{2}w;\\ w\sim baw,&bw\sim aw,&b^{2}w\sim b^{3}aw,&b^{3}w\sim b^{2}aw.\\ \end{array}$$

$$w(b^{s}aw)w^{-1}=(ab)^{s}bw=\left\{\begin{array}[]{ll}b^{s+1}w&\mbox{ if }s\equiv 0\;(2),\\ b^{s+1}aw&\mbox{ if }s\equiv 1\;(2),\end{array}\right.$$

$$w(b^{s}w)w^{-1}=(ab)^{s}w=\left\{\begin{array}[]{ll}b^{s}w&\mbox{ if }s\equiv 0\;(2),\\ b^{s-2}aw&\mbox{ if }s\equiv 1\;(2),\end{array}\right.$$

$$b(b^{s}a{w}^2)b^{-1}=b^{s+1}a{a}^{-1}{w}^2=b^{s+1}{w}^2,b(b^s{w}^2)b^{-1}=b^{s+1}{a}^{-1}{w}^2=b^{s+3}a{w}^2,$$

$$a(b^{s}a{w}^2)a^{-1}=b^{-s}{a}^2(ab{)}^{-1}{w}^2=b^{-s-1}a{w}^2,a(b^s{w}^2)a^{-1}=a{b}^s(ab{)}^{-1}{w}^2=b^{1-s}{w}^2.$$

$$\begin{array}[]{llll}aw^{2}\sim bw^{2},&baw^{2}\sim b^{2}w^{2},&b^{2}aw^{2}\sim b^{3}w^{2},&b^{3}aw^{2}\sim w^{2};\\ w^{2}\sim b^{3}aw^{2},&bw^{2}\sim aw^{2},&b^{2}w^{2}\sim baw^{2},&b^{3}w^{2}\sim b^{2}aw^{2};\\ aw^{2}\sim b^{3}aw^{2},&baw^{2}\sim b^{2}aw^{2},&b^{2}aw^{2}\sim baw^{2},&b^{3}aw^{2}\sim aw^{2};\\ w^{2}\sim bw^{2},&bw^{2}\sim w^{2},&b^{2}w^{2}\sim b^{3}w^{2},&b^{3}w^{2}\sim b^{2}w^{2}.\end{array}$$

$$b=\left(\begin{array}[]{cccc}\cos\frac{\pi}{n}&0&0&-\sin\frac{\pi}{n}\\ 0&\cos\frac{\pi}{n}&\sin\frac{\pi}{n}&0\\ 0&-\sin\frac{\pi}{n}&\cos\frac{\pi}{n}&0\\ \sin\frac{\pi}{n}&0&0&\cos\frac{\pi}{n}\end{array}\right),\ a=\left(\begin{array}[]{cc}R(\frac{\pi}{2})&0\\ 0&R(-\frac{\pi}{2})\\ \end{array}\right);$$

$$b=\frac{1}{2}\left(\begin{array}[]{cccc}1&-1&-1&-1\\ 1&1&1&-1\\ 1&-1&1&1\\ 1&1&-1&1\end{array}\right),\ a=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}0&1&1&0\\ -1&0&0&1\\ -1&0&0&-1\\ 0&-1&1&0\end{array}\right);$$

$$b=\frac{1}{2}\left(\begin{array}[]{cccc}1&-1&-1&-1\\ 1&1&1&-1\\ 1&-1&1&1\\ 1&1&-1&1\end{array}\right),\ a=\frac{1}{2}\left(\begin{array}[]{cc}R(-\frac{\pi}{2})&R(-\frac{2\pi}{5})\\ R(\frac{7\pi}{5})&R(\frac{\pi}{2})\\ \end{array}\right),$$

$$a=\left(\begin{array}[]{cc}R(-\frac{\pi}{2})&0\\ 0&R(\frac{\pi}{2})\\ \end{array}\right),\ w=\frac{1}{2}\left(\begin{array}[]{cccc}-1&-1&-1&-1\\ 1&-1&1&-1\\ 1&-1&-1&1\\ 1&1&-1&-1\end{array}\right)\left(\begin{array}[]{cc}R(\frac{2\pi}{3^{q}})&0\\ 0&R(\frac{2\pi}{3^{q}})\\ \end{array}\right);$$

$$u=\left(\begin{array}[]{cc}R(\frac{2\pi}{n})&0\\ 0&R(-\frac{2\pi}{n})\\ \end{array}\right),\ w=\left(\begin{array}[]{cc}0&R(\frac{\pi}{2^{q-1}})\\ R(\frac{\pi}{2^{q-1}})&0\\ \end{array}\right).$$

$$c=\left(\begin{array}[]{cc}R(\frac{2\pi r_{1}}{m})&0\\ 0&R(\frac{2\pi r_{2}}{m})\\ \end{array}\right).$$

$$c=\left(\begin{array}[]{cc}e^{\frac{2\pi r_{1}}{m}i}&0\\ 0&e^{\frac{2\pi r_{2}}{m}i}\\ \end{array}\right).$$

$$\mathrm{deg}(f)=l^2{r}_{21}{r}_{22}\frac{m_1}{m_2}[r_{11}{]}_{(m_1)}^{-1}[r_{12}{]}_{(m_1)}^{-1}+k{m}_2\text{for some }k\in \mathbb{Z},$$

$$\left(\begin{array}[]{c}r_{1}e^{\theta_{1}i}\\ r_{2}e^{\theta_{2}i}\end{array}\right)\stackrel{{\scriptstyle c_{1}}}{{\mapsto}}\left(\begin{array}[]{c}r_{1}e^{\theta_{1}i+\frac{2\pi r_{11}}{m_{1}}i}\\ r_{2}e^{\theta_{2}i+\frac{2\pi r_{12}}{m_{1}}i}\end{array}\right)\stackrel{{\scriptstyle\tilde{g}}}{{\mapsto}}\left(\begin{array}[]{c}r_{1}e^{k_{1}\theta_{1}i+\frac{2\pi k_{1}r_{11}}{m_{1}}i}\\ r_{2}e^{k_{2}\theta_{2}i+\frac{2\pi k_{2}r_{12}}{m_{1}}i}\end{array}\right),$$

$$\left(\begin{array}[]{c}r_{1}e^{\theta_{1}i}\\ r_{2}e^{\theta_{2}i}\end{array}\right)\stackrel{{\scriptstyle\tilde{g}}}{{\mapsto}}\left(\begin{array}[]{c}r_{1}e^{k_{1}\theta_{1}i}\\ r_{2}e^{k_{2}\theta_{2}i}\end{array}\right)\stackrel{{\scriptstyle c_{2}^{l}}}{{\mapsto}}\left(\begin{array}[]{c}r_{1}e^{k_{1}\theta_{1}i+\frac{2\pi r_{21}l}{m_{2}}i}\\ r_{2}e^{k_{2}\theta_{2}i+\frac{2\pi r_{22}l}{m_{2}}i}\\ \end{array}\right).$$

$$\mathrm{deg}(\tilde{g})=k_1{k}_2=l^2{r}_{21}{r}_{22}(\frac{m_1}{m_2}{)}^2[r_{11}{]}_{(m_1)}^{-1}[r_{12}{]}_{(m_1)}^{-1}.$$

$$\mathrm{deg}(g)=l^2{r}_{21}{r}_{22}\frac{m_1}{m_2}[r_{11}{]}_{(m_1)}^{-1}[r_{12}{]}_{(m_1)}^{-1}.$$

$$L(\frac{m_2}{(l,m_2)};r_{21},r_{22})\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\mathrm{deg}(\tilde{f})=(\frac{l}{(l,m_2)}{)}^2{r}_{21}{r}_{22}\frac{m_1}{m_2/(l,m_2)}[r_{11}{]}_{(m_1)}^{-1}[r_{12}{]}_{(m_1)}^{-1}=\frac{l^2}{(l,m_2)}{r}_{21}{r}_{22}\frac{m_1}{m_2}[r_{11}{]}_{(m_1)}^{-1}[r_{12}{]}_{(m_1)}^{-1}.$$

$$\{j^2{r}_{21}{r}_{22}\frac{m_1{m}_2}{(m_1,m_2{)}^2}[r_{11}{]}_{(m_1)}^{-1}[r_{12}{]}_{(m_1)}^{-1}+k{m}_2\mid k\in \mathbb{Z},j=0,1,\dots ,(m_1,m_2)-1\}.$$

$$\overline{\mathrm{deg}}(\psi )=\overline{\mathrm{deg}}(\xi )\overline{\mathrm{deg}}(\phi )\frac{|G_2|}{|G_{\ast }|},$$

$$\left(\begin{array}[]{cccc}1&1&1&1\\ -\cos\frac{\pi}{n}-\sin\frac{\pi}{n}&\cos\frac{\pi}{n}-\sin\frac{\pi}{n}&\cos\frac{\pi}{n}+\sin\frac{\pi}{n}&\sin\frac{\pi}{n}-\cos\frac{\pi}{n}\\ \sin\frac{\pi}{n}-\cos\frac{\pi}{n}&-\cos\frac{\pi}{n}-\sin\frac{\pi}{n}&\cos\frac{\pi}{n}-\sin\frac{\pi}{n}&\cos\frac{\pi}{n}+\sin\frac{\pi}{n}\\ -1&1&-1&1\end{array}\right)$$

$$\begin{array}[]{rcl}\overline{\deg}\psi|_{\langle w\rangle}&=&(2-3^{r-1})(2+3^{r-1})3^{q-r}[2-3^{q-1}]^{-1}_{(3^{q})}[2+3^{q-1}]^{-1}_{(3^{q})}\\ &=&3^{q-r}(2-3^{r-1})(2+3^{r-1})\frac{1-3^{q-1}}{2}\frac{1-3^{q-1}}{2}\\ &\equiv&3^{q-r}\;(3^{r}).\end{array}$$

$$\begin{array}[]{rcl}\overline{\deg}\psi&=&3^{r}(4-3^{r+1})[3^{q-r}]^{-1}_{(8)}+3^{q-r}(1-4\cdot 3^{r}+3^{2r+1})\\ &=&3^{r}(4-3^{r+1})(4-3^{q-r+1})+3^{q-r}(1-4\cdot 3^{r}+3^{2r+1})\\ &\equiv&3^{q-r}\;(8\cdot 3^{r})\end{array}$$

$$x\equiv s^2(n),x\equiv t^2(2^q).$$

$$1\to De{r}_{\alpha }({\mathbb{Z}}_{2^q},{\mathbb{Z}}_n)\to Aut({\mathbb{Z}}_n{⋊}_{\alpha }{\mathbb{Z}}_{2^q})\to \eta ({\mathbb{Z}}_n{⋊}_{\alpha }{\mathbb{Z}}_{2^q})\to 1$$

$$Aut({\mathbb{Z}}_n{⋊}_{\alpha }{\mathbb{Z}}_{2^q})\cong {\mathbb{Z}}_n⋊({\mathbb{Z}}_n^{\ast }\times {\mathbb{Z}}_{2^q}^{\ast }).$$

Full text

This work was initiated during the first and second authors’ visit to Capital Normal University February 16 - 23, 2013. The first author was supported in part by Projeto Temático Topologia Algébrica Geométrica e Diferencial 2012/24454-8. The third author was supported in part by the NSF of China (11431009).

1. Introduction

The study of mapping degree has a very long history in topology. In particular, the existence of degree one maps between manifolds is an important aspect in the classification of manifolds. Given two closed connected orientable manifolds $M$ and $N$ of the same dimension, let $D(M,N)=\{\deg f|f:M\to N\}$ denote the possible values of mapping degrees of maps from $M$ to $N$. While the determination of $D(M,N)$ has been of great interest to topologists, in general, very little has been done concerning the computation of $D(M,N)$. For example, the finiteness of the set $D(M,N)$ has been studied by various authors (see e.g. [3, 4, 6, 7, 15, 1]) and the study of $D(M):=D(M,M)$ for self-maps have also been investigated. We refer the reader to the survey paper [22] for the study of degree up to 2000, and to the papers [5, 18, 20] and [14] for mapping degrees on $3$-manifolds that are most relevant to our work. Prompted by our earlier work [13] on the fixed point theory of spherical $3$-manifolds, in this paper we determine $D(M,N)$ when $M$ and $N$ are closed $3$-manifolds with $S^{3}$ geometry.

2. Fundamental groups of spherical $3$-manifolds

A spherical $3$-manifold is an orbit space $S^{3}/G$, where $G$ is a finite group which acts freely on $S^{3}$. Hence $G$ is the fundamental group of $S^{3}/G$. Each spherical $3$-manifold is oriented by the canonical orientation of $S^{3}$.

Proposition 2.1**.**

(cf. [19]) The fundamental group of any spherical $3$-manifold is in the form of $G$ or $\mathbb{Z}_{m}\times G$, where $G$ belongs to one of the groups listed below and $(m,|G|)=1$.

The binary tetrahedral group is denoted by $T^{*}_{24}=\{a,b\mid a^{2}=b^{3}=(ab)^{3},a^{4}=1\}$, which is actually a special case of the generalized tetrahedral group $T^{\prime}_{8\cdot 3^{q}}$ with $q=1$, i.e. $T^{*}_{24}\cong T^{\prime}_{8\cdot 3^{1}}$. One such isomorphism from $T^{\prime}_{8\cdot 3^{1}}$ to $T^{*}_{24}$ is given by $b\mapsto bab^{-1},a\mapsto a,w\mapsto ba^{-1}$.

If $n$ is odd, then the generalized dihedral group $D^{*}_{4n}$ is the same as the dicyclic group $D^{\prime}_{n\cdot 2^{2}}$. If $n=1$, then $D^{\prime}_{n\cdot 2^{q}}$ is the cyclic group $\mathbb{Z}_{2^{q}}$. If $q=1$, then $D^{\prime}_{n\cdot 2^{q}}$ is not the fundamental group of a $3$-dimensional spherical manifold except for the case $n=1$. Thus, all groups in Table 2.1 are distinct.

Here are some more characteristic subgroups and quotients of these groups.

Next, we describe, for each group in Table 2.1 except for the cyclic groups, the conjugacy classes of elements, subgroups and quotients.

Lemma 2.2**.**

The set of conjugacy classes of $D^{*}_{4n}$ is given as follows.

Proof.

Note that $b(b^{k}a)b^{-1}=b^{k+2}a$, $ab^{l}a^{-1}=b^{-l}$, $a(b^{k}a)a^{-1}=b^{-k}a$. ∎

Lemma 2.3**.**

The non-trival subgroups and quotients of $D^{*}_{4n}$ are listed as follows.

Proof.

The presentations of cyclic subgroups follow from Lemma 2.2. Since $a^{2}=b^{n}$, any non-cyclic subgroup $H$ of $D^{*}_{4n}$ must have the form of either $\langle b^{k},b^{l}a\rangle$ or $\langle b^{k}a,b^{l}a\rangle$. Note that $b^{l}a$ is conjugate to either $a$ or $ba$. There are four possibilities: $\langle b^{k},a\rangle$, $\langle b^{k}a,a\rangle$, $\langle b^{k},ba\rangle$ and $\langle b^{k}a,ba\rangle$.

Clearly, $\langle b^{k}a,a\rangle=\langle b^{k},a\rangle$. Since $b$ has order $2n$, we may assume that $k|2n$. If $\frac{2n}{k}$ is even, i.e. $\frac{n}{k}$ is an integer, then we have the presentation $\langle b^{k},a\rangle=\{b^{k},a\mid a^{2}=(b^{k})^{\frac{n}{k}}=(ab^{k})^{2},a^{4}=1\}$. Thus $\langle b^{k},a\rangle$ is isomorphic to $D^{*}_{4\cdot\frac{n}{k}}$ if $\frac{n}{k}$ is even; to $D^{\prime}_{\frac{n}{k}\cdot 2^{2}}$ if $\frac{n}{k}$ is odd. In the case that $\frac{2n}{k}$ is odd, we have that $(n,k)=\frac{k}{2}$. Since $a^{2}=b^{n}$, the subgroup $\langle b^{k},a\rangle=\langle b^{k},b^{n},a\rangle=\langle b^{\frac{k}{2}},a\rangle$ is isomorphic to $D^{\prime}_{\frac{2n}{k}\cdot 2^{2}}$ by the argument above.

Suppose that $\langle b^{k},a\rangle$ is normal. Then $bab^{-1}=b^{2}a$ must lie in $\langle b^{k},a\rangle$. It follows that $b^{2}\in\langle b^{k},a\rangle$. Hence $k$ must be $1$ or $2$. Note that $\langle b,a\rangle$ is the whole group $D^{*}_{4n}$. Thus, $k=2$ is the unique non-trivial case that $\langle b^{k},a\rangle$ is normal.

The subgroup $\langle b^{k},ba\rangle$ is isomorphic to $\langle b^{k},a\rangle$, by using the automorphism given by $a\mapsto b^{-1}a,\ b\mapsto b$. This situation is the same as that of $\langle b^{k},a\rangle$.

Note that $\langle b^{k}a,ba\rangle=\langle b^{k-1},ba\rangle$. This is also the case we have already considered. ∎

Next, we consider the binary octahedral group $O^{*}_{48}$ and the binary icosahedral group $I^{*}_{120}$. The subgroups and quotients of these groups are well-known and the results can be found, for example, in [9, Appendix].

Lemma 2.4**.**

The subgroups of $O^{*}_{48}$ are $\mathbb{\mathbb{Z}}_{2},\mathbb{\mathbb{Z}}_{3},\mathbb{\mathbb{Z}}_{4},\mathbb{\mathbb{Z}}_{6},\mathbb{\mathbb{Z}}_{8}$, $Q_{8}$, $D^{\prime}_{3\cdot 2^{2}}$, $Q_{16}$ and $T^{*}_{24}$.

The normal subgroups are $\mathbb{\mathbb{Z}}_{2},Q_{8}\ and\ T^{*}_{24}$.

The only quotient which is the fundamental group of a spherical $3$-manifold is $\mathbb{\mathbb{Z}}_{2}$, which can be embedded in $O^{*}_{48}$.

Lemma 2.5**.**

The set of conjugacy classes of $O^{*}_{48}$ is given as follows.

Lemma 2.6**.**

The subgroups of $I^{*}_{120}$ are $\mathbb{\mathbb{Z}}_{2},\mathbb{\mathbb{Z}}_{3},\mathbb{\mathbb{Z}}_{4},\mathbb{\mathbb{Z}}_{5},\mathbb{\mathbb{Z}}_{6}$, $Q_{8}$, $\mathbb{\mathbb{Z}}_{10}$, $D^{\prime}_{12}$, $D^{\prime}_{20}$ and $T^{*}_{24}$.

The only normal subgroup is $\mathbb{\mathbb{Z}}_{2}$.

The only quotient is $A_{5}$ which is not the fundamental group of a spherical $3$-manifold.

Lemma 2.7**.**

The set of conjugacy classes of $I^{*}_{120}$ is given as follows.

Lemma 2.8**.**

The set of conjugacy classes of $T^{\prime}_{8\cdot 3^{q}}$ is given as follows.

Here, $t=0,1,\ldots,3^{q-1}-1$.

Proof.

Since $bwb^{-1}=b(ab)^{-1}w=a^{-1}w$ and $awa^{-1}=ab^{-1}w=baw$, two conjugate elements must lie in the same coset of $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]$. Note that $w^{3}$ commutes with all elements of $T^{\prime}_{8\cdot 3^{q}}$. It is sufficient to consider the conjugate relations in cosets $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]$, $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]w$ and $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]w^{2}$.

(1) The coset $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]$. Since $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]$ is isomorphic $D^{*}_{4\cdot 2}$, from Lemma 2.2, we have that $b\sim b^{3}$, $a\sim b^{2}a$ and $ba\sim b^{3}a$ by conjugation induced by $b$ and $a$. Note that $waw^{-1}=b$ and $wbw^{-1}=ab=b^{3}a$. Thus, the coset $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]$ contains three conjugacy classes: $[1]=\{1\}$, $[b]=\{b,b^{3},a,ba,b^{2}a,b^{3}a\}$ and $[b^{2}]=\{b^{2}\}$.

(2) The coset $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]w$. Note that

[TABLE]

[TABLE]

These relations imply that

[TABLE]

Moreover,

[TABLE]

[TABLE]

They give no new conjugacy relations. Thus, the coset $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]w$ contains two conjugacy classes: $[w]$ and $[bw]$.

(3) The coset $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]w^{2}$. Note that

[TABLE]

[TABLE]

These relations imply that

[TABLE]

Thus, the coset $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]w^{2}$ contains two conjugacy classes: $[w^{2}]$ and $[b^{2}w^{2}]$. ∎

Lemma 2.9**.**

The non-trival subgroups and quotients of $T^{\prime}_{8\cdot 3^{q}}$ are listed as follows.

Proof.

Note that $\langle w^{s}\rangle=\langle w^{(s,3^{q})}\rangle$. The first three rows come from those in Lemma 2.8, where $3^{r}=(3^{q},3t)$.

Since both $3t+1$ and $3t+2$ are coprime to $3^{q}$, we obtain that $\langle w^{3t+1}\rangle=\langle w^{3t+2}\rangle=\langle w\rangle$. Thus, all cyclic subgroups generated by the elements of order $3^{q}$ listed in Lemma 2.8 are conjugate. We obtain the row 4. This fact follows from Sylow’s theorems. Clearly, these subgroups are not normal.

Consider the elements of order $2\cdot 3^{q}$. Note that $(b^{2}w)^{6s+1}=b^{2}w^{6s+1}$, $(b^{2}w)^{6s+5}=b^{2}w^{6s+5}$. Since both $6s+1$ and $6s+5$ are relatively prime to $2\cdot 3^{q}$, we have $\langle b^{2}w\rangle=\langle b^{2}w^{6s+1}\rangle=\langle b^{2}w^{6s+5}\rangle$. Note that $(b^{2}w)^{3^{q}-(6s+2)}=b^{2}w^{-6s-3}$, $(b^{2}w)^{3^{q}-(6s+4)}=b^{2}w^{-6s-4}$. Since both $3^{q}-(6s+2)$ and $3^{q}-(6s+4)$ are relatively prime to $2\cdot 3^{q}$, we have $\langle b^{2}w\rangle=\langle b^{2}w^{-6s-2}\rangle=\langle b^{2}w^{-6s-4}\rangle$. Since $b^{2}w^{-6s-2}=(b^{2}w^{6s+2})^{-1}$ and $b^{2}w^{-6s-4}=(b^{2}w^{6s+4})^{-1}$, we obtain that $\langle b^{2}w\rangle=\langle b^{2}w^{6s+2}\rangle=\langle b^{2}w^{6s+4}\rangle$. Thus, there are four cyclic subgroups, all conjugate to each other.

Let us consider the non-cyclic subgroup $H$ of $T^{\prime}_{8\cdot 3^{q}}$. Note that the abelianization of $T^{\prime}_{8\cdot 3^{q}}=\langle w\rangle\cong\mathbb{Z}_{3^{q}}$. The abelianization $H/[H,H]$ will be a subgroup of $\langle w\rangle$. We have the following three cases:

(1) If $H/[H,H]$ is trivial then $H$ lies in the commutator $[T^{\prime}_{8\cdot 3^{q}},T^{\prime}_{8\cdot 3^{q}}]=D^{*}_{4\cdot 2}$. By Lemma 2.3, any proper subgroup of $D^{*}_{4\cdot 2}$ is cyclic, and hence $H$ must be the commutator itself. This is the row 6 for $r=q$.

(2) If $H/[H,H]=\langle w\rangle$, then by Lemma 2.8 and up to a conjugation, $H$ must contain $w$ or $b^{2}w$. If $b^{2}w\in H$, then $(b^{2}w)^{2}=w^{2}\in H$. Since $(2,3^{q})=1$, we still have that $w\in H$. Since $H$ is non-cyclic, $H$ contains an element of the form $xw^{k}$, where $x$ is a non-trivial element in the commutator. Hence $x=(xw^{k})w^{-k}$ also lies in $H$. Note that either $x$ or $x^{2}$ is $b^{2}$. If $x=b^{2}$ is the unique non-trivial element in the commutator, then $H=\langle b^{2},w\rangle$ would be cyclic. Thus, there must an element $x$ in $H$ with $x^{2}=b^{2}$. Thus, one of the six elements: $x,wxw^{-1},w^{2}xw^{-2},x^{3},x^{2}wxw^{-1},x^{2}w^{2}xw^{-2}$ must be $a$. Note that $\langle a,w\rangle=T^{\prime}_{8\cdot 3^{q}}$. Then $H$ must be the whole group.

(3) If $H/[H,H]=\langle w^{3^{r}}\rangle$ with $0<r<q$, then $H=H^{\prime}\times\langle w^{3^{r}}\rangle$ for some subgroup $H^{\prime}$ of the commutator $D^{*}_{4\cdot 2}$ since $w^{3^{r}}$ lies in the center. By Lemma 2.3, any proper subgroup of $D^{*}_{4\cdot 2}$ is cyclic, and hence $H^{\prime}$ must be the commutator itself. This is the row 6 for $0<r<q$. ∎

Lemma 2.10**.**

The set of conjugacy classes of $D^{\prime}_{n\cdot 2^{q}}$ with $2\not|n$ is given as follows.

Proof.

Note that $w(u^{k}w^{l})w^{-1}=u^{-k}w^{l}$, and note that $u(u^{k}w^{l})u^{-1}=u^{k+2}w^{l}$ if $l$ is odd; and $u(u^{k}w^{l})u^{-1}=u^{k}w^{l}$ if $l$ is even. Moreover $u^{k}w^{l}=u^{k+n^{\prime}}w^{l}$. ∎

Lemma 2.11**.**

The non-trival subgroups and quotients of $D^{\prime}_{n\cdot 2^{q}}$ with $2\not|n$ are listed as follows.

Proof.

Consider cyclic subgroups. For the conjugacy classes of the form $[w^{2t+1}]$, since $2t+1$ is coprime with the order $2^{q}$ of $w$, we have that $\langle w^{2t+1}\rangle=\langle w\rangle$. By Lemma 2.10, we obtain the first row.

For the conjugacy classes of the form $[w^{2t}]$, we have that $\langle w^{2t}\rangle=\langle u^{2^{r}}\rangle$ for some $r$ with $0<r<q$. These are the second and third row.

For the conjugacy classes of the form $[u^{s}]$, we have that $\langle u^{s}\rangle=\langle u^{k}\rangle$ for some $k$ with $k|n$, giving row $4$ and $5$.

For the conjugacy classes of the form $[u^{s}w^{2t}]$, we have that $\langle u^{s}w^{2t}\rangle=\langle u^{k}w^{2^{r})}\rangle\cong\mathbb{Z}_{\frac{n}{k}}\times\mathbb{Z}_{2^{q-r}}$ for some $k$ and $r$ with $k|n$ and $0<r<q$, which is a normal subgroup. We have the rows from $6$ to $8$, which are respectively the case $k=1$, $k>1$ and $r=1$, and $k,r>1$.

Assume that $H$ is a non-cyclic subgroup of $D^{\prime}_{n\cdot 2^{q}}$, $H$ must contains an element of the form $u^{s}w^{2t+1}$. Up to a conjugation, we may have that $w^{2t+1}\in H$. It follows that $w\in H$ because $w\in\langle w^{2t+1}\rangle$. Since $H$ is not cyclic, $H$ contain some element of the form $u^{s}$. Let $u^{k}$ be the element in $H$ such that $k$ is the minimal positive integer. Thus, $H=\langle u^{k},w\rangle$. This is the final row. If such a subgroup is normal, then $uwu^{-1}=u^{2}w$ must lies in $H$, and hence $u^{2}\in H$. Note that $k$ is odd. We have that $u^{(k,2)}=u$ lies in $H$. Thus, $H=\langle u,w\rangle$ is the whole group. ∎

2.1. Embedding $G$ in $SO(4)$

By definition of spherical $3$-manifolds, we may regard the fundamental groups of spherical $3$-manifolds as discrete subgroups of the orientation-preserving isometry group $SO(4)$ of $S^{3}$.

Proposition 2.12**.**

Let $G$ be the fundamental group of a spherical $3$-manifold. If $G$ is not cyclic, then the generators in the presentation of Proposition 2.1 can be identified as elements in $SO(4)$ as follows.

(1) if $G=\mathbb{Z}_{m}\times D^{*}_{4n}$,

[TABLE]

(2) if $G=\mathbb{Z}_{m}\times O^{*}_{48}$,

[TABLE]

(3) if $G=\mathbb{Z}_{m}\times I^{*}_{120}$,

[TABLE]

(4) if $G=\mathbb{Z}_{m}\times T^{\prime}_{8\cdot 3^{q}}$,

[TABLE]

(5) if $G=\mathbb{Z}_{m}\times D^{\prime}_{n\cdot 2^{q}}$,

[TABLE]

In all cases, the generator of $\mathbb{Z}_{m}$ is given by v=\left(\begin{array}[]{cc}R(\frac{2\pi}{m})&0\\ 0&R(\frac{2\pi}{m})\\ \end{array}\right). Here, we use $R(\theta)$ to denote the matrix (block) \left(\begin{array}[]{cc}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{array}\right).

From now on, all groups are subgroups of $SO(4)$. By Olum’s results [16, 17], every homomorphism $\psi:G_{1}\to G_{2}$ can be realized by a map $f:S^{3}/G_{1}\to S^{3}/G_{2}$ with $\psi=f_{\pi}$. Moreover, if $f$ and $g$ induce the same homomorphism $\psi$, then $\deg f\equiv\deg g$ mod $|G_{2}|$. Therefore, we have a well-defined mod $|G_{2}|$ integer associated with any homomorphism $\psi$ which we denote by $\overline{\deg}(\psi)\in\mathbb{Z}_{|G_{2}|}$.

It is known that the orbit spaces of the form $S^{3}/\mathbb{Z}_{m}$ are lens spaces. The homotopy types and homeomorphism type of $3$-dimensional lens spaces depend on the $\mathbb{Z}_{m}$ action, which can be regarded as embeddings of $\mathbb{Z}_{m}$ in $SO(4)$.

3. Lens paces covering spherical manifolds

The generalized $3$-dimensional lens space $L(m;r_{1},r_{2})$ is defined to be the orbit space $S^{3}/\mathbb{Z}_{m}$, where $\mathbb{Z}_{m}$ is the finite cyclic subgroup of $SO(4)$, which is generated by

[TABLE]

Here $r_{1}$ and $r_{2}$ are relatively prime to $m$, but $m$ is not necessarily prime. If we regard $S^{3}$ as the unit sphere of complex space $\mathbb{C}^{2}$, the element $c$ turns out to be the complex matrix

[TABLE]

By identifying the covering transformation group with the fundamental group, we call this element $c$ the standard generator of $\pi_{1}(L(m;r_{1},r_{2}))$.

It is known that $L(m;r_{1},r_{2})$ and $L(m;r^{\prime}_{1},r^{\prime}_{2})$ are homeomorphic if and only if $\{\pm\frac{r_{1}}{r_{2}},\pm\frac{r_{2}}{r_{1}}\}$ and $\{\pm\frac{r^{\prime}_{1}}{r^{\prime}_{2}},\pm\frac{r^{\prime}_{2}}{r^{\prime}_{1}}\}$ are the same subset of $\mathbb{Z}_{m}^{*}$, the multiple group of invertible elements in $\mathbb{Z}_{m}$ (see e.g. [2, Ch. V]). Usually, $L(m;1,r)$ is written as $L(m;r)$. Clearly, $L(m;r_{1},r_{2})=L(m;1,r^{\prime})$ if $r^{\prime}=r_{1}[r_{2}]^{-1}_{(m)}$. Here, $[r_{2}]^{-1}_{(m)}$ denotes the inverse of the multiplication of $\mathbb{Z}_{m}$.

The lens spaces that can cover $L(m;r_{1},r_{2})$ must be $L(k;r_{1},r_{2})$ for some $k|m$.

Example 3.1**.**

Consider the group $O^{*}_{48}$ and $H=\langle b\rangle$ the cyclic group of order $6$ generated by $b$. The eigenvalues of the matrix $b$ are $\{e^{\frac{\pi i}{3}},e^{-\frac{\pi i}{3}},e^{\frac{\pi i}{3}},e^{-\frac{\pi i}{3}}\}$, meaning either $e^{\frac{\pi i}{3}}$ or $e^{-\frac{\pi i}{3}}$ has multiplicity two. It follows that $b$ is conjugate to \left(\begin{array}[]{cc}R(\frac{\pi}{3})&0\\ 0&R(\frac{\pi}{3})\\ \end{array}\right). Thus, the corresponding lens space is $L(6;1,1)$

More generally, consider the group $\mathbb{Z}_{m}\times O^{*}_{48}$ and $H=\langle vb\rangle$, the eigenvalues of the matrix $vb$ are $\{e^{\frac{(6+m)\pi i}{3}},e^{-\frac{(6+m)\pi i}{3}},e^{\frac{(6-m)\pi i}{3}},e^{-\frac{(6-m)\pi i}{3}}\}$. Thus, $vb$ is conjugate to \left(\begin{array}[]{cc}R(\frac{(6+m)\pi}{3})&0\\ 0&R(\frac{(6-m)\pi}{3})\\ \end{array}\right).

The example above shows in general how one determines the lens space corresponding to a maximal cyclic subgroup. We summarize the results below.

Lemma 3.2**.**

Let $H$ be a maximal cyclic subgroup of some $G$, where $G$ is the fundamental group of some spherical $3$-manifold. Then the lens spaces of the form $S^{3}/H$ are listed as follows:

Proof.

By Lemma 2.1, the generator of a cyclic group can be written as a matrix in $SO(4)$. We can compute its eigenvalues for any given matrix. Note that the corresponding lens spaces are completely determined by the eigenvalues of the matrices corresponding to the generators. This proof is a straightforward computation. ∎

Lemma 3.3**.**

Let $f:L(m_{1};r_{11},r_{12})\to L(m_{2};r_{21},r_{22})$ be a map between lens spaces, inducing a homomorphism $f_{\pi}$ on fundamental group given by $f_{\pi}(c_{1})=c_{2}^{l}$, where $c_{1}$ and $c_{2}$ are respectively standard generators of $\pi_{1}(L(m_{1};r_{11},r_{12}))$ and $\pi_{1}(L(m_{2};r_{21},r_{22}))$. Then the degree of $f$ is

[TABLE]

where $[*]^{-1}_{(m_{1})}$ denotes the inverse of multiplication of $\mathbb{Z}_{m_{1}}$.

Proof.

Let us consider the case when $f_{\pi}$ is surjective. Then we have that $m_{2}|m_{1}$. We construct a map $\tilde{g}:S^{3}\to S^{3}$ by \tilde{g}\left(\begin{array}[]{c}r_{1}e^{\theta_{1}i}\\ r_{2}e^{\theta_{2}i}\end{array}\right)=\left(\begin{array}[]{c}r_{1}e^{k_{1}\theta_{1}i}\\ r_{2}e^{k_{2}\theta_{2}i}\end{array}\right). Note that

[TABLE]

[TABLE]

In order to obtain that $\tilde{g}c_{1}=c_{2}^{l}\tilde{g}$, it is sufficient to assume that $\frac{r_{11}k_{1}}{m_{1}}-\frac{r_{21}l}{m_{2}}$ and $\frac{r_{12}k_{2}}{m_{1}}-\frac{r_{22}l}{m_{2}}$ are both integers. Thus, we may pick $k_{1}=lr_{21}\frac{m_{1}}{m_{2}}[r_{11}]^{-1}_{(m_{1})}$ and $k_{2}=lr_{22}\frac{m_{1}}{m_{2}}[r_{12}]^{-1}_{(m_{1})}$. It is obvious that $\tilde{g}$ has degree

[TABLE]

The equality $\tilde{g}c_{1}=c_{2}^{l}\tilde{g}$ also implies that $\tilde{g}$ induces a map $g:L(m_{1};r_{11},r_{12})\to L(m_{2};r_{21},r_{22})$. Since $\tilde{g}$ is a lifting of $g$ with respect to the universal covering $S^{3}$ of both $L(m_{1};r_{11},r_{12})$ and $L(m_{2};r_{21},r_{22})$, we have that $\deg(\tilde{g})m_{2}=m_{1}\deg(g)$. Here $m_{1}$ and $m_{2}$ are respectively the degrees of the covering maps $S^{3}\to L(m_{1};r_{11},r_{12})$ and $S^{3}\to L(m_{2};r_{21},r_{22})$. Thus,

[TABLE]

Note that $f$ and $g$ have the same induced homomorphism on $\pi_{1}$, which is given by $c_{1}\mapsto c_{2}^{l}$. By Olum results [16, 17], we have that $\deg(f)\equiv\deg(g)\;(m_{2})$. Thus, we give the proof of $\deg(f)$ when $f_{\pi}$ is surjective.

Now, we turn to the general case. The image $\mathrm{Im}f_{\pi}$ of $f_{\pi}$ is the cyclic group generated by $c_{2}^{(l,m_{2})}$, which is isomorphic to $\mathbb{Z}_{\frac{m_{2}}{(l,m_{2})}}$. Note that the covering space of $L(m_{2},r_{21},r_{22})$ corresponding to the group $\langle c_{2}^{(l,m_{2})}\rangle$ is $L(\frac{m_{2}}{(l,m_{2})},r_{21},r_{22})$. There must be a lifting $\tilde{f}$ of $f$ satisfying the following commutative diagram:

[TABLE]

Note that $c_{2}^{l}=(c_{2}^{(l,m_{2})})^{\frac{l}{(l,m_{2})}}$. Apply our previous argument to $\tilde{f}$ which induces a surjection on $\pi_{1}$, we have that

[TABLE]

Since $\deg(f)=\deg(p)\deg(\tilde{f})=(l,m_{2})\deg(\tilde{f})$, we obtain our conclusion. ∎

It should be mentioned that the existence of the homomorphism $c_{1}\to c_{2}^{l}$ forces that $m_{1}l\equiv 0\;(m_{2})$. In other word, $\frac{lm_{1}}{m_{2}}$ must be an integer.

Theorem 3.4**.**

The set of all degrees $D(L(m_{1};r_{11},r_{12}),L(m_{2};r_{21},r_{22}))$ of maps from $L(m_{1};r_{11},r_{12})$ to $L(m_{2};r_{21},r_{22})$ is:

[TABLE]

Proof.

For any map $f$ between these two lens spaces, the homomorphism $f_{\pi}$ on fundamental groups is given by $c_{1}\to c_{2}^{l}$. Note that $m_{1}l\equiv 0\;(m_{2})$. It follows that $l=j\frac{m_{2}}{(m_{1},m_{2})}$ for $j=0,1,\ldots,(m_{1},m_{2})-1$. This theorem holds by Lemma 3.3. ∎

Remark 3.1*.*

When we write a lens space $L(m;r_{1},r_{2})$, its orientation is specified, inheriting the natural orientation of $S^{3}$. Thus, by above theorem $D(L(3;1,1),L(3;1,2))=\{3k,-1+3k\mid k\in\mathbb{Z}\}$, although $L(3;1,1)$ and $L(3;1,2)$ are homeomorphic.

4. Mapping degrees

In this section, we determine all mapping degrees between spherical $3$-manifolds. Suppose $f:M\to N$ is a map between two such manifolds with $G_{1}=\pi_{1}(M)$ and $G_{2}=\pi_{1}(N)$. If $G_{1}=\mathbb{Z}_{m}$ then the degree is already given in the last section. For the other cases of $G_{1}$, we only need to consider the case when the induced homomorphism $f_{\pi}$ is surjective.

Lemma 4.1**.**

Let $\psi:G_{1}\to G_{2}$ be a homomorphism and $G_{*}$ the quotient $\frac{G_{1}}{ker(\psi)}.$ Then

[TABLE]

where $\varphi:G_{1}\to G_{*}$ is the projection and $\xi:G_{*}\to\mathrm{Im}\psi$ is the isomorphism given by $\varphi(x)\mapsto\psi(x)$.

Proof.

Let us consider the following three maps:

i) $f:S^{2k+1}/G_{1}\to S^{2k+1}/G_{*}$ be a map such that $f_{\pi}=\varphi$

ii) $h:S^{2k+1}/G_{*}\to S^{2k+1}/\mathrm{Im}\psi$ such that $h_{\pi}=\xi$

iii) $p_{2}:S^{2k+1}/\mathrm{Im}\psi\to S^{2k+1}/G_{2}$.

It is easy to see that the homomorphism $\psi$ is $p_{2}\circ\xi\circ\varphi$, the composite of the three homomorphisms. Since the degree of the covering of $p_{2}$ is $\frac{|G_{2}|}{|G_{*}|}$ the formula follows. ∎

By using Lemma 4.1, since the degree of endomorphisms are almost listed in [5], we can compute the degree $\overline{\deg}(\psi)$, where $\psi:G_{1}\to G_{2}$ is a surjection and $G_{2}$ is a subgroup of $G_{1}$.

Lemma 4.2**.**

Each endomorphism of $D^{*}_{4n}$ is conjugate to a composition of some endomorphisms in following table.

Proof.

Let $\psi$ be an endomorphism on $D^{*}_{4n}$. By Lemma 2.2, up to a conjugation, we may assume $\psi(a)=1,a,ba,b^{k}$.

(1) $\psi(a)=1$.

(1.1) $\psi(b)=b^{s}$. Then $\psi(a^{2})=1$, $\psi(b^{n})=b^{ns}$ and $\psi((ab)^{2})=b^{2s}$. The relation $a^{2}=b^{n}=(ab)^{2}$ yields that $0\equiv ns\equiv 2s\;(2n)$. Thus, $s\equiv 0,n\;(2n)$. Both are in the table ($\psi_{1}$ and $\psi_{2}$).

(1.2) $\psi(b)=b^{s}a$. Note that $\psi((ab)^{2})=b^{s}ab^{s}a=a^{2}\neq 1=\psi(a^{2})$. Thus, this case is impossible.

(2) $\psi(a)=a$.

(2.1) $\psi(b)=b^{s}$. Then $\psi(a^{2})=a^{2}=b^{n}$, $\psi(b^{n})=b^{ns}$ and $\psi((ab)^{2})=ab^{s}ab^{s}=a^{2}$. The relation $a^{2}=b^{n}=(ab)^{2}$ yields that $ns\equiv n\;(2n)$. It follows that $s$ is odd. Note that $s=s^{\prime}s^{\prime\prime}$, where $s^{\prime}|2n$ and $(2n,s^{\prime\prime})=1$. We have that $\psi=\psi_{5,s^{\prime\prime}}\psi_{4,s^{\prime}}$.

(2.2) $\psi(b)=b^{s}a$. Then $\psi(a^{2})=a^{2}=b^{n}$, $\psi(b^{n})=(b^{s}ab^{s}a)^{\frac{n}{2}}=a^{n}=b^{n^{2}/2}$ and $\psi((ab)^{2})=ab^{s}aab^{s}a=b^{-2s}$. The relation $a^{2}=b^{n}=(ab)^{2}$ yields that $n\equiv n^{2}/2\equiv-2s\;(2n)$. It follows that $\frac{n}{2}$ is odd and $s=\frac{n}{2},\frac{3n}{2}$. Since $ab^{\frac{n}{2}}aa^{-1}=b^{-\frac{n}{2}}a=b^{\frac{3n}{2}}a$, we can consider only the case $\psi(b)=b^{\frac{n}{2}}a$, which is $\psi_{7}$.

(3) $\psi(a)=ba$. Note that there is an isomorphism $\psi_{6}$ with $b\mapsto b$ and $a\mapsto ba$ in this case. By composing with such an isomorphism, this case is the same as case (2).

(4) $\psi(a)=b^{k}$. Since $a^{4}=b^{2n}=1$, we must have that $4k\equiv 0\;(2n)$. Up to conjugation by $a$, we may assume that $k=\frac{n}{2}$ or $n$.

(4.1) $\psi(a)=b^{\frac{n}{2}}$.

(4.1.1) $\psi(b)=b^{s}$. Then $\psi(a^{2})=b^{n}$, $\psi(b^{n})=b^{ns}$ and $\psi((ab)^{2})=(b^{\frac{n}{2}}b^{s})^{2}=b^{n+2s}$. The relation $a^{2}=b^{n}=(ab)^{2}$ yields that $n\equiv ns\equiv n+2s\;(2n)$, which implies that $s$ is odd and $s\equiv 0\;(n)$. This case is impossible because $n$ is even.

(4.1.2) $\psi(b)=b^{s}a$. Then $\psi(a^{2})=b^{n}$, $\psi(b^{n})=(b^{s}ab^{s}a)^{\frac{n}{2}}=a^{n}$ and $\psi((ab)^{2})=(b^{\frac{n}{2}}b^{s}a)^{2}=a^{2}$. The relation $a^{2}=b^{n}=(ab)^{2}$ yields that $n\equiv 2\;(4)$. By Lemma 2.2, up to conjugation, there are two possibilities: either $\psi(b)=a$ or $\psi(b)=ba$. They are respectively $\psi_{8}$ and $\psi_{6}\psi_{8}$.

(4.2) $\psi(a)=b^{n}$.

(4.2.1) $\psi(b)=b^{s}$. Then $\psi(a^{2})=1$, $\psi(b^{n})=b^{ns}$ and $\psi((ab)^{2})=(b^{n}b^{s})^{2}=b^{2n+2s}$. The relation $a^{2}=b^{n}=(ab)^{2}$ yields that $0\equiv ns\equiv 2n+2s\;(2n)$, which implies that $s=0$ or $n$. They are respectively $\psi_{3}$ and $\psi_{2}\psi_{6}$.

(4.2.2) $\psi(b)=b^{s}a$. Note that $\psi((ab)^{2})=(b^{n}b^{s}a)^{2}=a^{2}\neq 1=\psi(a^{2})$. Thus, this case is also impossible.

Now we turn to the degree.

Consider the endomorphism $\psi_{6}$ with $\psi_{6}(b)=b$ and $\psi_{6}(a)=ba$. A direct computation shows that a linear map on $\mathbb{R}^{4}$ determined by the following matrix induces a self-map on $S^{3}/D^{*}_{4n}$, whose induced endomorphism on $\pi_{1}$ is exactly $\psi^{\prime}$.

[TABLE]

Since the determinant of the above matrix is $16$, the degree of the corresponding self-map must be $1$. Thus, we obtain that $\overline{\deg}(\psi_{6})=1$.

Let us write $\psi^{\prime}_{2}$ and $\psi^{\prime}_{3}$ for two homomorphisms from $D^{*}_{4n}$ to $\mathbb{Z}_{2}=\langle c\rangle$, defined by $\psi^{\prime}_{2}(b)=c$, $\psi^{\prime}_{2}(a)=1$, $\psi^{\prime}_{3}(b)=1$ and $\psi^{\prime}_{3}(a)=c$. They are two non-zero elements in $H^{1}(S^{3}/D^{*}_{4n},\mathbb{Z}_{2})=\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$. The other non-zero element is given by $\psi^{\prime}_{2}\psi_{6}$. If $2|\frac{n}{2}$, we know from [21, Theorem 2.2] that the cube of one element of these three is zero and the cube of the other two are both $1$. Since $H^{*}(S^{3}/\mathbb{Z}_{2})=\mathbb{Z}_{2}[c^{*}]$, we obtain that $\overline{\deg}(\varphi)=\varphi^{*}(c^{*})^{3}[S^{3}/D^{*}_{4n}]$ for any $\varphi:D^{*}_{4n}\to\mathbb{Z}_{2}$, where $[S^{3}/D^{*}_{4n}]\in H_{3}(S^{3}/D^{*}_{4n},\mathbb{Z}_{2})$ is the fundamental class. Note that $\overline{\deg}\psi_{6}=1\;(4n)$. It follows that $\overline{\deg}\psi^{\prime}_{2}\psi_{6}=\overline{\deg}\psi^{\prime}_{2}=1$ and hence $\overline{\deg}\psi^{\prime}_{3}=0$. These mean that $\overline{\deg}\psi_{2}=2n$ and $\overline{\deg}\psi_{3}=0$. Recall from [21, Theorem 2.2] again that the cube of any element in $H^{1}(S^{3}/D^{*}_{4n},\mathbb{Z}_{2})$ vanishes if $2\not|\frac{n}{2}$. Thus, $\overline{\deg}\psi_{2}=\overline{\deg}\psi_{3}=0$.

For $\psi_{7}$ with $b\mapsto b^{\frac{n}{2}}a$ and $a\mapsto a$. It follows that $b^{\frac{n}{2}}\mapsto(b^{\frac{n}{2}}a)^{\frac{n}{2}}$ which is $b^{\frac{n}{2}}a$ if $n\equiv 2\;(8)$, is $b^{\frac{3n}{2}}a$ if $n\equiv 6\;(8)$. Note that $\langle b^{\frac{n}{2}},a\rangle=Q_{8}$. We may identify $a$ as $i$, $b^{\frac{n}{2}}$ as $j$ in quaternion number. Thus, $\psi_{7}|_{\langle b^{\frac{n}{2}},a\rangle}(i)=i$ and $\psi_{7}|_{\langle b^{\frac{n}{2}},a\rangle}(j)=\pm k$. Hence, $\psi_{7}|_{\langle b^{\frac{n}{2}},a\rangle}$ is an isomorphism. By [14], we have that $\overline{\deg}(\psi_{7}|_{\langle b^{\frac{n}{2}},a\rangle})=1$. Clearly, $\overline{\deg}(\psi_{7}|_{\langle b^{4}\rangle})=0$. Hence, $\overline{\deg}(\psi_{7})=\frac{n^{2}}{4}$.

For $\psi_{8}$ with $b\mapsto a$ and $a\mapsto b^{\frac{n}{2}}$. Similar to the case above, $\psi_{8}|_{\langle b^{\frac{n}{2}},a\rangle}$ is an isomorphism on $Q_{8}$. By [14], we have that $\overline{\deg}(\psi_{8}|_{\langle b^{\frac{n}{2}},a\rangle})=1$. Clearly, $\overline{\deg}(\psi_{8}|_{\langle b^{4}\rangle})=0$. Hence, $\overline{\deg}(\psi_{8})=\frac{n^{2}}{4}$.

The degrees for $\psi_{4,k}$ and $\psi_{5,l}$ follow from [5, p.1248, Case II]. ∎

It should be mentioned that $\psi_{3}=\psi_{8}\psi_{2}$, if $2\not|\frac{n}{2}$. The degree for $\psi_{2},\psi_{7},\psi_{8}$ seem to be missing in [5].

Corollary 4.3**.**

If $n>2$, then the group $Out(D^{*}_{4n})\cong\mathbb{Z}_{2}\times\mathbb{Z}_{n}^{*}$ is generated by $\eta_{1},\eta_{2}$, where $\eta_{1}(b)=b$, $\eta_{1}(a)=ba$ and $\eta_{2}(b)=b^{s}$, $\eta_{2}(a)=a$. Here $s$ is an element generating $\mathbb{Z}_{n}^{*}\cong\mathbb{Z}_{\varphi(n)}$.

The group $Out(D^{*}_{4\cdot 2})\cong S_{3}$ is generated by $\eta_{3},\eta_{4}$, where $\eta_{3}(b)=ba$, $\eta_{3}(a)=a$ and $\eta_{4}(b)=a$, $\eta_{2}(a)=b$.

Theorem 4.4**.**

Let $\varphi$ be a surjective non-trivial homomorphism from $D^{*}_{4n}$ to the fundamental group $G$ of some $3$-dimensional spherical manifold $S^{3}/G$. Then $\varphi=\eta^{\prime}\psi\eta^{\prime\prime}$, where $\eta^{\prime\prime}\in Aut(D^{*}_{4n})$ and $\eta^{\prime}\in Aut(G)$. Moreover, all possible $G$ and $\psi$ are listed as follows:

Proof.

The first row comes from the degree of $\psi_{4,k}$ in Lemma 4.2, where $m=\frac{n}{k}$. The other two come from the degrees of $\psi_{2}$ and $\psi_{3}$ there. ∎

For $O^{*}_{48}$, we know that the only non-trivial proper quotient is $\mathbb{Z}_{2}$. Hence,

Theorem 4.5**.**

Let $\varphi$ be a non-trivial surjective homomorphism from $O^{*}_{48}$ to $G$ induced by a map from $S^{3}/O^{*}_{48}$ to some $3$-dimensional spherical manifold $S^{3}/G$. Then $\varphi$ is conjugate to one of the following.

[TABLE]

Proof.

The first two rows comes from [14]. Let $f:S^{3}/O*_{48}\to L(2;1,1)$ be a map, inducing a homomorphism indicated in third row. By [21, Theorem 4.13], $H^{3}(S^{3}/O^{*}_{48},\mathbb{Z}_{2})$ is generated by $\beta_{1}^{3}$ for the unique non-zero element $\beta_{1}\in H^{1}(S^{3}/O^{*}_{48},\mathbb{Z}_{2})$. It follows that the $\deg(f)\equiv 1\;(2)$ because $f^{*}(c^{3})=\beta_{1}^{3}$. We obtain the third row. ∎

We conclude that $D(O^{*}_{48})=\{m+48k\mid m=0,1,24,25,k\in\mathbb{Z}\}$. Note that the values $24+48k,k\in\mathbb{Z}$ were missing in [5]. For $I^{*}_{120}$ the only possible proper quotient is the trivial group.

Theorem 4.6**.**

Let $\varphi$ be a non-trivial surjective homomorphism from $I^{*}_{120}$ to $G$ induced by a map from $S^{3}/I^{*}_{120}$ to some $3$-dimensional spherical manifold $S^{3}/G$, then $\varphi$ is conjugate to either the identity or the automorphism $\eta$ of $I^{*}_{120}$ determined by $a\mapsto a^{-1}$, $b\mapsto b^{-1}abab^{-1}a$, where $G$ is also $I^{*}_{120}$. Here, $\overline{\deg}(\eta)\equiv 49\;(120)$.

Lemma 4.7**.**

Each endomorphism of $T^{\prime}_{8\cdot 3^{q}}$ is conjugate to a composition of some endomorphisms in the following table

Proof.

By Lemma 2.9, the non-trivial normal subgroups of $T^{\prime}_{8\cdot 3^{q}}$ are $\mathbb{Z}_{3^{q-r}}=\langle w^{3^{r}}\rangle$, $\mathbb{Z}_{2\cdot 3^{q-r}}=\langle b^{2}w^{3^{r}}\rangle$, or $D^{*}_{4\cdot 2}\times\mathbb{Z}_{3^{q-r}}=\langle b,a,w^{3^{r}}\rangle$ with $0<r\leq q$. Since neither $T^{\prime}_{8\cdot 3^{q}}/\langle w^{3^{r}}\rangle$ nor $T^{\prime}_{8\cdot 3^{q}}/\langle b^{2}w^{3^{r}}\rangle$ is a subgroup of $T^{\prime}_{8\cdot 3^{q}}$ except for the trivial case $T^{\prime}_{8\cdot 3^{q}}/\langle w^{3^{q}}\rangle=T^{\prime}_{8\cdot 3^{q}}$, any endomorphism on $T^{\prime}_{8\cdot 3^{q}}$ has kernel $1$, $D^{*}_{4\cdot 2}\times\mathbb{Z}_{3^{q-r}}$, or $T^{\prime}_{8\cdot 3^{q}}$.

(1) $\ker\psi=1$, i.e. $\psi$ is an automorphism. Note that $\psi(w)$ has order $3^{q}$. By Lemma 2.8, we may assume that $\psi(w)=w^{3t+1}$ or $w^{3t+2}$. Since $a$ has order $4$, $\psi(a)$ has also order $4$, and hence $\psi(a)\in\{b,a,ba,b^{-1},b^{2}a,b^{-1}a\}$. Note that $waw^{-1}=b$, $wbw^{-1}=ab=b^{-1}a$, $wb^{2}aw^{-1}=b^{-1}$ and $wb^{-1}w^{-1}=ba$. Up to a conjugation, we may also assume that $\psi(a)=a$ or $b^{-1}$. There four subcases:

(1.1) $\psi(w)=w^{3t+1}$ and $\psi(a)=a$. It follows that $\psi(b)=\psi(waw^{-1})=\psi(w)\psi(a)\psi(w^{-1})=b$. From number theory, we know that the set $\mathbb{Z}^{*}_{3^{q}}$ of all multiple invertible elements in $\mathbb{Z}_{3^{q}}$ is a cyclic group of order $3^{q}-3^{q-1}$, i.e. $\mathbb{Z}^{*}_{3^{q}}\cong\mathbb{Z}_{3^{q}-3^{q-1}}$. Moreover, $2$ is a generator of $\mathbb{Z}^{*}_{3^{q}}$. Thus, $\psi=\psi_{1}^{r}$, where $2^{r}\equiv 3t+1\;(3^{q})$.

(1.2) $\psi(w)=w^{3t+1}$ and $\psi(a)=b^{-1}$. It follows that $\psi(b)=\psi(waw^{-1})=\psi(w)\psi(a)\psi(w^{-1})=ba$. Hence, $\psi(wbw^{-1})=w^{3t+1}baw^{-(3t+1)}=abb=b^{2}a$, but $\psi(ab)=b^{-1}ba=a$. Thus, this is an impossible case.

(1.3) $\psi(w)=w^{3t+2}$ and $\psi(a)=a$. It follows that $\psi(b)=\psi(waw^{-1})=\psi(w)\psi(a)\psi(w^{-1})=b^{-1}a$. Hence, $\psi(wbw^{-1})=w^{3t+2}b^{-1}aw^{-(3t+2)}=a^{-1}ab=b$, but $\psi(ab)=ab^{-1}a=ba^{2}=b^{3}$. Thus, this is an impossible case.

(1.4) $\psi(w)=w^{3t+2}$ and $\psi(a)=b^{-1}$. It follows that $\psi(b)=\psi(waw^{-1})=\psi(w)\psi(a)\psi(w^{-1})=b^{2}a$. Here, $\psi=\psi_{1}^{k}$, where $2^{k}\equiv 3t+2\;(3^{q})$.

(2) $\ker\psi=D^{*}_{4\cdot 2}\times\mathbb{Z}_{3^{q-r}}$ with $0<r<q$. Then $\psi(a)=\psi(b)=1$. Since $\psi(w^{3^{r}})=1$, the order of $\psi(w)$ is a divisor of $3^{r}$. By Lemma 2.8, up to a conjugation, we may also assume that $\psi(w)=w^{3t}$. Note that $2$ is a generator of $\mathbb{Z}^{*}_{3^{q}}$. There is some $s$ such that $3t\cdot 2^{s}\equiv 3^{r}\;(3^{q})$. Thus, $\psi_{1}^{s}\psi(w)=w^{3^{r}}$. It follows that $\psi=\psi_{1}^{-s}\psi_{2}^{r}$ if $r>0$; $\psi=\psi_{2}^{-s}\psi_{3}$ ir $r=0$. The degree of $\psi_{2},\psi_{3}$ were given in [5, p.1250, Case VI].

(3) $\ker\psi=T^{\prime}_{8\cdot 3^{q}}$. Clearly, $\psi=\psi_{4}$.

Consider the degree of $\psi_{1}$. Since $\psi_{1}|_{\langle b,a\rangle}$ is an automorphism on $Q_{8}$, we have that $\overline{\deg}(\psi_{1}|_{\langle b,a\rangle})\equiv 1\;(8)$ by [14]. Thus, $\overline{\deg}(\psi_{1})\equiv 1\;(8)$. On the other hand, By Lemma 3.3, we have that $\overline{\deg}(\psi_{1}|_{\langle w\rangle})\equiv 2^{2}\;(3^{q})$. We obtain that $\overline{\deg}(\psi_{1})\equiv 4-3^{2q+1}\;(8\cdot 3^{q})$. ∎

Corollary 4.8**.**

The group $Out(T^{\prime}_{8\cdot 3^{q}})\cong\mathbb{Z}_{3^{q}-3^{q-1}}$ is generated by $\eta$, which is defined by $\eta(a)=b^{-1}$ and $\eta(w)=w^{2}$.

Theorem 4.9**.**

Let $\varphi$ be a surjective non-trivial homomorphism from $T^{\prime}_{8\cdot 3^{q}}$ to $G$ induced by a map from $S^{3}/T^{\prime}_{8\cdot 3^{q}}$ to some $3$-dimensional spherical manifold $S^{3}/G$. Then $\varphi=\eta^{\prime}\psi\eta^{\prime\prime}$, where $\eta^{\prime\prime}\in Aut(T^{\prime}_{8\cdot 3^{q}})$ and $\eta^{\prime}\in Aut(G)$. Moreover, all possible $G$ and $\psi$ are listed as follows:

[TABLE]

Proof.

(1) $\ker\psi=\langle w^{3^{r}}\rangle$, the homomorphism $\psi:T^{\prime}_{8\cdot 3^{q}}\to T^{\prime}_{8\cdot 3^{r}}$ can be given by $\psi(b)=\bar{b}$, $\psi(a)=\bar{a}$ $\psi(w)=\bar{w}$. By Lemma 3.2, $S^{3}/\langle w\rangle$ and $S^{3}/\langle\bar{w}\rangle$ are respectively Lens spaces $L(3^{q};2-3^{q},2+3^{q})$ and $L(3^{r};2-3^{r},2+3^{r})$. By Lemma 3.3,

[TABLE]

Since $\psi|_{\langle b,a\rangle}$ is an isomorphism, $\overline{\deg}\psi|_{\langle b,a\rangle}\equiv 1\;(8)$. It follows that

[TABLE]

(2) $\ker\psi=\mathbb{Z}_{2\cdot 3^{q-r}}$. Since its quotient $(\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{3^{r}}$ is not a subgroup in our list, this is an impossible case.

(3) $\ker\psi=D^{*}_{4\cdot 2}\times\mathbb{Z}_{3^{q-r}}$ ($0<r\leq q$). Then $G=Im\psi=\mathbb{Z}_{3^{r}}=\langle c\mid c^{3^{r}}\rangle$. The homomorphism $\psi:T^{\prime}_{8\cdot 3^{r}}\to\mathbb{Z}_{3^{r}}$ can be given by $\psi(a)=1$ and $\psi(w)=c$. By identification $\iota:G\to\langle w^{3^{q-r}}\rangle$, we have that $\iota\psi=\psi_{2}^{q-r}$, where $\psi_{2}^{r}$ is listed in Lemma 4.7. Since $\overline{\deg}\psi_{2}^{q-r}=(9-3^{2q+2})^{q-r}$, we obtain that $\overline{\deg}\psi=(9-3^{2q+2})^{q-r}/(8\cdot 3^{q-r})=\frac{1}{8}\times(3-3^{2q+1})^{q-r}\;(3^{r})$. ∎

Lemma 4.10**.**

Each endomorphism of $D^{\prime}_{n\cdot 2^{q}}$ is conjugate to a composition of some endomorphisms in the following table

[TABLE]

Proof.

By Lemma 2.11, $\ker\psi$ must be $1$, $\langle u^{k}\rangle$ ($k|n,1\leq k<n$), $\langle uw^{2^{r}}\rangle$ ($0<r<q$), $\langle u^{k}w^{2^{r}}\rangle$ ($k|n,1<k<n$, $1<r<q$), or $D^{\prime}_{n\cdot 2^{q}}$.

(1) $\ker\psi=1$. Since $\psi(u)$ has order $n$ and $\langle u\rangle$ is the commutator and hence is a fully invariant subgroup, $\psi(u)=u^{s}$ for some $s$ with $(s,n)=1$. Since $\psi(w)$ has order $2^{q}$, we have $\psi(w)=u^{m}w^{t}$ for some odd $t$. Note that $u(u^{m}w^{t})u^{-1}=u^{m+2}w^{t}$. Up to a conjugation, we may assume that $\psi(w)=w^{t}$. Thus, $\psi=\psi_{1,s}\psi_{2,t}$. Moreover, since $w(u^{s}w^{t})w^{-1}=u^{-s}$, we may assume that $1\leq s<\frac{n}{2}$.

(2) $\ker\psi=\langle u^{k}\rangle$ with $k|n$ and $1\leq k<n$. Then $\psi(u)=u^{\frac{n}{k}}$. Since $\psi(w)$ has order $2^{q}$, by Lemma 2.10, we have $\psi(w)=w^{t}$ for some odd $t$. Thus, $\psi$ is $\psi_{2,t}\psi_{3,k}$.

(3) $\ker\psi=\langle uw^{2^{r}}\rangle$ with $0<r<q$. Then $\psi(u)=1$. Since $\psi(w)$ has order $2^{r}$, by Lemma 2.10, we have $\psi(w)=w^{t\cdot 2^{r}}$ for some odd $t$. Thus, $\psi$ is $\psi_{2,t}\psi_{4}^{r}$ if $0<r<q$.

(4) $\ker\psi=D^{\prime}_{n\cdot 2^{q}}$. Clearly, $\psi=\psi_{4}^{q}$.

Now we consider the degree $\overline{\deg}(\psi)$. Note that both $\langle u\rangle$ and $\langle w\rangle$ are invariant subgroups of $\psi$. Assume that $\psi(u)=u^{s}$ and $\psi(w)=w^{t}$. By Lemma 3.3, we have that $\overline{\deg}(\psi|_{\langle u\rangle})=s^{2}$ and $\overline{\deg}(\psi|_{\langle w\rangle})=t^{2}$. Thus, $\overline{\deg}(\psi)$ is the unique solution module $n\cdot 2^{q}$ of the system of congruences:

[TABLE]

By Euler-Fermat theorem, we have that $n^{\varphi(2^{q})}=n^{2^{q-1}}\equiv 1\;(2^{q})$. Thus, we have $n\cdot n^{2^{q-1}-1}+2^{-q}(1-n^{2^{q-1}})2^{q}=1$. By the Chinese remainder theorem, we have $\overline{\deg}(\psi)=(1-n^{2^{q-1}})s^{2}+n^{2^{q-1}}t^{2}$. We obtain the all $\overline{\deg}(\psi)$’s. ∎

Corollary 4.11**.**

The group $Out(D^{\prime}_{n\cdot 2^{q}})$ is isomorphic to $(\mathbb{Z}_{n}^{*}/\mathbb{Z}_{2})\times\mathbb{Z}_{2^{q}}^{*}$.

Proof.

Here we give an alternate proof of this reuslt as [8, Theorem 2.4(2)] contains an error which will be corrected. From Proposition 1.3 of [8] we have a short exact sequence,

[TABLE]

where $Der_{\alpha}(\mathbb{Z}_{2^{q}},\mathbb{Z}_{n})\cong\mathbb{Z}_{n}$, and $\eta(\mathbb{Z}_{n}\rtimes_{\alpha}\mathbb{Z}_{2^{q}})\cong\mathbb{Z}_{n}^{*}\times\mathbb{Z}_{2^{q}}^{*}$. Since the group $\mathbb{Z}_{n}^{*}\times\mathbb{Z}_{2^{q}}^{*}$ embeds naturally in $Aut(\mathbb{Z}_{n}\rtimes_{\alpha}\mathbb{Z}_{2^{q}})$ we have a section. Thus, we obtain

[TABLE]

Denote by $(a,(b,c))$ a typical element of $Aut(\mathbb{Z}_{n}\rtimes_{\alpha}\mathbb{Z}_{2^{q}})$ where $a\in\mathbb{Z}_{n},(b,c)\in\mathbb{Z}_{n}^{*}\times\mathbb{Z}_{2^{q}}^{*}$. Here, we write $\mathbb{Z}_{k}$ as an additive group and $\mathbb{Z}_{k}^{*}$ as a multiplicative group. Now let us factor this group by the inner automorphisms. Write $\langle x\rangle=\mathbb{Z}_{n}$ and $\langle y\rangle=\mathbb{Z}_{2^{q}}$.

The conjugation of $(x,0)$ by $(0,y)$ yields $(-x,0)$ thus the corresponding inner automorphism is the element $(0,(-1,1))$. The conjugation of $(0,y)$ by $(x,0)$ yields $(2x,y)$ thus the corresponding inner automorphism is the element $(2,(1,1))$. It follows that the inner automorphisms form the subgroup $\mathbb{Z}_{n}\rtimes(\mathbb{Z}_{2}\times\{1\}).$ Therefore the quotient is isomorphic to $(\mathbb{Z}_{n}^{*}/\mathbb{Z}_{2})\times\mathbb{Z}_{2^{q}}^{*}$ and the result follows. ∎

Theorem 4.12**.**

Let $\varphi$ be a surjective non-trivial homomorphism from $D^{\prime}_{n\cdot 2^{q}}$ to $G$ induced by a map from $S^{3}/D^{\prime}_{n\cdot 2^{q}}$ to some $3$-dimensional spherical manifold $S^{3}/G$. Then $\varphi=\eta^{\prime}\psi\eta^{\prime\prime}$, where $\eta^{\prime\prime}\in Aut(D^{\prime}_{n\cdot 2^{q}})$ and $\eta^{\prime}\in Aut(G)$. Moreover, all possible $G$ and $\psi$ are listed as follows:

[TABLE]

Proof.

If $\ker\psi=\langle u^{k}w^{2^{r}}\rangle\cong\mathbb{Z}_{\frac{n}{k}\cdot 2^{q-r}}$ ($k|n$, $k>1$ and $1<r<q$). The homomorphism $\psi:D^{\prime}_{n\cdot 2^{q}}\to D^{\prime}_{k\cdot 2^{r}}$ can be given by $\psi(u)=\bar{u}$, $\psi(w)=\bar{w}$. By Lemma 3.2, $S^{3}/\langle w\rangle$ and $S^{3}/\langle\bar{w}\rangle$ are respectively lens spaces $L(2^{q};1+2^{q-1},1)$ and $L(2^{r};1+2^{r-1},1)$. By Lemma 3.3, $\overline{\deg}\psi|_{\langle w\rangle}\equiv 2^{q-r}(1+2^{r-1})[(1+2^{q-1})]_{(2^{q})}^{-1}=2^{q-r}(1+2^{r-1})(1-2^{q-1})\equiv 2^{q-r}\;(2^{r})$. On the other hand, by Lemma 3.2 again, $S^{3}/\langle u\rangle$ and $S^{3}/\langle\bar{u}\rangle$ are respectively lens spaces $L(n;1,-1)$ and $L(k;1,-1)$. By Lemma 3.3, $\overline{\deg}\psi|_{\langle u\rangle}\equiv 1\;(k)$. It follows that

[TABLE]

The other cases come from Lemma 4.10, by using Lemma 4.1. ∎

Next Lemma deals with the case where the domain is in the form of a direct product.

Lemma 4.13**.**

Let $\varphi:\mathbb{Z}_{m}\times G_{1}\to G^{\prime}$ be a surjective homomorphism. Then $G^{\prime}=\mathbb{Z}_{k}\times G_{2}$ and $\varphi=\varphi_{1}\times\varphi_{2}$. Moreover, both of $\varphi_{1}:\mathbb{Z}_{m}\to\mathbb{Z}_{k}$ and $\varphi_{2}:G_{1}\to G_{2}$ are surjective, and $\overline{\deg}\varphi=(|G_{1}|\overline{\deg}\varphi_{2}+m\overline{\deg}\varphi_{1})[|G_{1}|+m]^{-1}_{(m|G_{1}|)}$.

5. Computations

We conclude this paper with some computations of the mapping degree of certain spherical $3$-manifolds.

In order to compute $D(S^{3}/G_{1},S^{3}/G_{2})$, our strategy is as follows: (1) look for all normal subgroups $H$ of $G_{1}$ such that $G_{1}/H$ is isomorphic to a subgroup of $G_{2}$; (2) find the degree of $\varphi_{H}:G_{1}\to G_{1}/H$; (3) look for all embedding $\xi:G_{1}/H\to G_{2}$. The degrees of epimorphisms $\varphi_{H}:G_{1}\to G_{1}/H$ were computed in Lemma 3.3 if $G_{1}$ is cyclic, and in Theorems 4.4, 4.5, 4.6, 4.9, 4.12 if $G_{1}$ is one of the groups listed in Table 2.1.

Thus Lemma 4.1 will compute the set $\{\overline{\deg}(\xi)\overline{\deg}(\varphi_{H})\}$. Here, $\overline{\deg}(\xi)=\overline{\deg}(\eta^{\prime})|G_{2}||H|/|G_{1}|$ for some $\eta^{\prime}\in Out(G_{1}/H)$. The set $D(M,N)$ can be determined once $\overline{\deg}(\psi)$ is for all $\psi$.

Example 5.1**.**

Let $G_{1}=D^{\prime}_{n\cdot 2^{q}}$ and $G_{2}=D^{\prime}_{m\cdot 2^{r}}$ with $r<q$.

Suppose $f:M_{1}\to M_{2}$ is a map between two spherical $3$-manifolds with $\pi_{1}(M_{1})=G_{i}$ for $i=1,2$ and $\varphi=f_{\sharp}:G_{1}\to G_{2}$. By Lemma 2.11, if $\ker\varphi$ is non-trivial, then $\ker\varphi$ is either $\mathbb{Z}_{\frac{n}{k}\cdot 2^{q-r}}$ with $k|(n,m)$ or $\mathbb{Z}_{n\cdot 2^{l}}$ with $1<l\leq r$.

(1) $\ker\varphi=\langle u^{k}w^{2^{r}}\rangle\cong\mathbb{Z}_{\frac{n}{k}\cdot 2^{q-r}}$ with $k|(n,m)$. By Theorem 4.12, we have

[TABLE]

where $\eta^{\prime\prime}\in Aut(D^{\prime}_{n\cdot 2^{q}})$ and $\eta^{\prime}\in Aut(D^{\prime}_{q\cdot 2^{r}})$.

(2) $\ker\varphi=\langle uw^{2^{q-l}}\rangle\cong\mathbb{Z}_{n\cdot 2^{l}}$ with $1<l\leq r$. By Theorem 4.12, we have

[TABLE]

where $\eta^{\prime\prime}\in Aut(D^{\prime}_{n\cdot 2^{q}})$ and $\eta^{\prime}\in Aut(D^{\prime}_{q\cdot 2^{r}})$.

Clearly, if $\ker\varphi=D^{\prime}_{n\cdot 2^{q}}$, then $\overline{\deg}(\varphi)\equiv 0\;(m\cdot 2^{r})$. Thus, we can write down the set $D(S^{3}/D^{\prime}_{n\cdot 2^{q}},S^{3}/D^{\prime}_{m\cdot 2^{r}})$.

Example 5.2**.**

Let $G_{1}=\mathbb{Z}_{120}$ and $G_{2}=O^{*}_{48}\times\mathbb{Z}_{5}$.

By Lemma 3.2, the manifold $S^{3}/(O^{*}_{48}\times\mathbb{Z}_{5})$ is covered by three lens spaces $L(30;1,11)$, $L(40;3,13)$ and $L(40;-7,23)$. Thus, any map from $L(120;r_{1},r_{2})$ can be lifted to a map into one of these three lens spaces.

Note that the degrees from these three lens spaces to $S^{3}/(O^{*}_{48}\times\mathbb{Z}_{5})$ are respectively $8$, $6$ and $6$. By Theorem 3.4, $D(L(120;r_{1},r_{2}),S^{3}/(Q_{48}\times\mathbb{Z}_{5}))$ consists of the following three sets:

[TABLE]

Note that any lens space $L(120;r_{1},r_{2})$ must be homeomorphic to one of the form $L(120;1,r)$, where

[TABLE]

We list all the elements of $D(L(120;r_{1},r_{2}),\ S^{3}/(O^{*}_{48}\times\mathbb{Z}_{5}))$ as follows.

[TABLE]