Coproducts of Finite Groups

TL;DR

This paper proves that the coproduct of any two non-trivial finite groups does not exist within the category of finite groups, highlighting a fundamental limitation in the structure of finite groups.

Contribution

It establishes a non-existence result for coproducts in the category of finite groups, providing new insights into their categorical properties.

Findings

Coproducts of non-trivial finite groups are not representable.

The result clarifies limitations in categorical constructions within finite group theory.

Highlights the difference between finite and infinite group categories.

Abstract

We show that for any pair of non-trivial finite groups, their coproduct in the category of finite groups is not representable.

Equations

$$\exists X\in \mathcal{C},\exists ({\iota }_{X_1},{\iota }_{X_2})\in {\mathrm{Hom}}_{\mathcal{C}}(X_1,X)\times {\mathrm{Hom}}_{\mathcal{C}}(X_2,X)$$

$$\forall Y\in \mathcal{C},\forall (f_1,f_2)\in {\mathrm{Hom}}_{\mathcal{C}}(X_1,Y)\times {\mathrm{Hom}}_{\mathcal{C}}(X_2,Y),\exists !f\in {\mathrm{Hom}}_{\mathcal{C}}(X,Y)$$

$$X_1\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$${\iota }_G:G\to F\text{ and }{\iota }_H:H\to F$$

$$g\in G\setminus \{1\}\text{ and }h\in H\setminus \{1\},$$

$$q_m:G\ast H\to T_m$$

$$f_G:G\to T_m\text{ and }{f}_H:H\to T_m$$

$$G\to G\ast H\text{ and }H\to G\ast H$$

$$G\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\{f\in \mathrm{Hom}(F,T_m):f_G={\iota }_{G}f\text{ and }{f}_H={\iota }_{H}f\}$$

$$(f{\iota }_G(g))(f{\iota }_H(h))=f({\iota }_G(g){\iota }_H(h))$$

$$\rho :G\ast H\to {\mathrm{GL}}_n(K)$$

$${\mathrm{GL}}_n(S)\to {\mathrm{GL}}_n(K)$$

$$G\ast H\to {\mathrm{GL}}_n(S)\to {\mathrm{GL}}_n({\mathbb{F}}_s)$$

$${\rho }_s:G\ast H\to {\mathrm{GL}}_n(\mathbb{Q}).$$

$${\mathrm{GL}}_n(\mathbb{Z}[1/D_s])\to {\mathrm{GL}}_n(\mathbb{Q}),$$

$${\rho }_{s,p}:G\ast H\to {\mathrm{GL}}_n(\mathbb{Z}/p)$$

$$q_m:G\ast H\to T_m$$

$$q_m(w^{m!})=q_m(w_m)\ne 1,$$

Full text

We gratefully acknowledge Bob Guralnick for helpful discussions on this topic.

Given a category $\mathcal{C}$ and a pair of objects $X_{1},X_{2}\in\mathcal{C}$, the coproduct $X_{1}\coprod X_{2}$ is representable in $\mathcal{C}$ iff

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such that

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making the diagram

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commute.

Let $G,H$ be finite groups. If we regard them as members of the category of all groups, then $G\coprod H$ is representable: it is the free product $G*H$, and $G\to G*H$ and $H\to G*H$ are the canonical inclusions. The purpose of this note is to prove the following theorem:

The key will be the following proposition:

Before proving the proposition we show how it implies Theorem 1.

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be homomorphisms. Let

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let $m$ be the order of $\iota_{G}(g)\iota_{H}(h)$, and let

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be a homomorphism such that $|\langle\,q_{m}(gh)\,\rangle|>m$ as in Proposition 2. Let

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be the respective compositions of

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with $q_{m}$ so that

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commutes. Observe that

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is empty since $f_{G}(g)f_{H}(h)$ has order exceeding $m$ while

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has order $m$. Therefore one cannot find a morphism $F\to T_{m}$ to complete the above diagram, and hence $G\coprod H$ is not representable in the category of finite groups. ∎

The proof of the proposition will occupy the remainder of this note.

Recall that a group is residually finite iff every non-identity element is contained in the complement of a finite-index normal subgroup.

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be the corresponding representation given by Theorem 3. It is faithful and the image is finitely generated, so the corollary is a consequence of the fact that any finitely generated subgroup of $\mathrm{GL}_{n}(\mathbb{C})$ is residually finite. Rather than appeal to this general fact though, we prove directly that $G*H$ is residually finite.

Let $w\in G*H$ be a non-identity element. We must show that there is a finite-index normal subgroup of $G*H$ whose complement contains $w$.

The set $\rho(G\cup H)$ is finite and contained in $\mathrm{GL}(K)$, so the least common multiple of the denominators of all the entries of the matrices in $\rho(G\cup H)$ is a non-zero polynomial $D\in\mathbb{Q}[t]$. Moreover, the greatest common divisor of the numerators of all the entries of $\rho(w)-I_{n}$ is a non-zero polynomial $N\in\mathbb{Q}[t]$.

Let $Z\subset\mathbb{Q}$ be the finite set of zeros of $D\cdot N$ in $\mathbb{Q}$ and $U:=Q\smallsetminus Z$. Let $S$ be the polynomial ring $\mathbb{Q}[t][1/DN]$ and

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be the homomorphism induced by the inclusion $S\to K$.

Observe that the elements of $\rho(G\cup H)$ all have finite order and generate $\rho(G*H)$. Therefore the elements of $\det(\rho(G\cup H))$ are roots of unity and generate $\det(\rho(G*H))$, so the latter is contained in $S^{\times}$. In particular, $\rho$ factors through (1).

For each $s\in U$, let $\mathbb{F}_{s}$ be the quotient field $S/(t-s)S$ and $\rho_{s}$ be the composition

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where the last homomorphism is induced by the quotient $S\to\mathbb{F}_{s}$. Observe that $\mathbb{F}_{s}$ is (canonically) isomorphic to $\mathbb{Q}$, so we can regard $\rho_{s}$ as a homomorphism

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Observe also that the least common multiple of the denominators of the entries of all matrices in $\rho_{s}(G\cup H)$ is a positive integer $D_{s}\in\mathbb{N}$, and that $\det(\rho_{s}(G*H))\subseteq\mathbb{Z}[1/D_{s}]^{\times}$. Therefore the image of $\rho_{s}$ is contained in the image of the natural homomorphism

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and hence there is a homomorphism

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for each $p\nmid D_{s}$. By construction, $\rho_{s,p}(w)\neq I_{n}$ and the kernel of $\rho_{s,p}$ has finite index in $G*H$. In particular, the complement of the latter is a finite-index subgroup containing $w$, so $G*H$ is residually finite as claimed. ∎

Observe that $gh$ has infinite order in $G*H$. Therefore, for each $m\geq 1$, the element $w_{m}:=(gh)^{m!}$ is not the identity, hence Corollary 4 implies that there exists a finite-index normal subgroup $K_{m}\subseteq G*H$ whose complement contains $w_{m}$.

To complete the proof of the proposition, we let $T_{m}:=(G*H)/K_{m}$ and

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be the canonical quotient. By definition,

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hence $q_{m}(w)$ has order exceeding $m$ as claimed. ∎