This paper fully characterizes local conjugacy among subgroups of GL_2(Z/p^2Z) for odd primes p, building on prior classifications over Z/pZ and analyzing subgroup intersections and images.
Contribution
It provides a complete classification of local conjugacy in GL_2(Z/p^2Z), extending previous work on GL_2(Z/pZ) and detailing subgroup interactions.
Findings
01
Characterized conditions for local conjugacy in GL_2(Z/p^2Z)
02
Identified subgroup intersection and image structures for conjugacy classification
03
Developed casework approach for full categorization
Table 1. Table 1. Representatives of the Similarity Classes of Mat 2 β‘ ( β€ / p β β€ ) subscript Mat 2 β€ π β€ \operatorname{Mat}_{2}(\mathbb{Z}/p\mathbb{Z})
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TopicsFinite Group Theory Research Β· Algebraic Geometry and Number Theory Β· Advanced Algebra and Geometry
Full text
Local Conjugacy in GL2β(Z/p2Z)
SPUR Final Paper, Summer 2016, Revised August 2017
Given an elliptic curve E over a number field K and an integer n, the action of the absolute Galois group of K on the n-torsion subgroup E[n]β(Z/nZ)2 determines a subgroup of Aut(E[n])βGL2β(Z/nZ). Sutherland [6] gives an efficient algorithm for computing the images of the Galois representation associated to an elliptic curve in GL2β(Z/pZ) that determines the image up to local conjugacy. One needs to understand locally conjugate subgroups to determine the image up to conjugation. [6] categorizes local conjugacy of subgroups in GL2β(Z/pZ), see Theorem 2. A categorization of local conjugacy of subgroups in GL2β(Z/p2Z) is a first step in extending the categorization in GL2β(Z/pZ) to the full p-adic image in GL2β(Zpβ).
Locally conjugate subgroups H1β and H2β of a group G form what is known as a Gassman triple (G,H1β,H2β), which arise in the study of arithmetically equivalent number fields. Such number fields have the
same Dedekind zeta function but need not be isomorphic. Gassmann triples of the form (GL2β(Z/nZ),H1β,H2β) can be used to explicitly construct arithmetically equivalent number fields K1β and K2β as subfields of the n-torsion field Q(E[n]) of an elliptic curve E/Q, see [2]. Non-conjugate subgroups H1β and H2β give rise to non-isomorphic number fields K1β and K2β.
Propositions 10 and 13 yield all of the non-conjugate locally conjugate subgroups, which we refer to as nontrivially locally conjugate subgroups, of GL2β(Z/p2Z) up to conjugation, see Theorem 4. One can choose generators of such subgroups to resemble each other. In fact, they are expressible in forms resembling the nontrivially locally conjugate subgroups of GL2β(Z/pZ).
2. Subgroups of GL2β(Z/pkZ)
Throughout this paper, p is an odd prime and Ο΅ is taken to be some nonsquare in Z/pZ. Let GL2β(R), SL2β(R) and PGL2β(R) denote the general, special and projective linear groups of 2Γ2 matrices over a ring R.
Define the following subgroups of GL2β(Z/pkZ) for kβ₯1:
[TABLE]
They are respectively called the center, Cartan-split subgroup, Cartan-nonsplit subgroup, and Borel subgroup of GL2β(Z/pkZ). For Hβ€GL2β(Z/pkZ), let N(H) denote the normalizer of H in GL2β(Z/pkZ). In particular,
[TABLE]
3. Properties of Locally Conjugate Subgroups
This section defines locally conjugate subgroups and discusses some properties of local conjugacy. For a group G and an element gβG, let gG denote the conjugacy class of g in G.
Definition 1**.**
Let G be a group and H1β,H2ββ€G. If there is a bijection f:H1ββH2β such that h and f(h) are conjugate in G for all hβH1β, then H1β and H2β are locally conjugate in G. If H1β and H2β are locally conjugate in G but not conjugate in G, then H1β and H2β are nontrivially locally conjugate in G.
Conjugate subgroups of G are always locally conjugate in G. Hence, conjugate subgroups are considered to be βtriviallyβ locally conjugate. Moreover, it is possible for two subgroups H1β and H2β of a group Gβ², which is in turn a subgroup of G, to be locally conjugate in G but not in Gβ². It is therefore important to emphasize the parent group in which two subgroups are locally conjugate in. Nevertheless, the parent group will be clear in context even if it is not explicitly stated.
Local conjugacy of subgroups in a fixed group G is an equivalence relation. Furthermore, given that subgroups H2β and H3β of G are conjugate, a subgroup H1β of G which is locally conjugate to H2β is conjugate to H2β if and only if H1β is conjugate to H3β. With this in mind, we will categorize local conjugacy between subgroups of GL2β(Z/p2Z) up to conjugation of the subgroups.
The following gives an alternate definition to local conjugacy:
The following two propositions give necessary conditions for local conjugacy on a finite group G in terms of local conjugacy in a normal subgroup of G and quotient groups of G.
Let G,Gβ² be finite groups, H1β,H2ββ€G subgroups of G and Ο:GβGβ² a surjective homomorphism. If H1β and H2β are locally conjugate in G, then Ο(H1β) and Ο(H2β) are locally conjugate in Gβ².
Proof.
Let Cβ² be any conjugacy class of Gβ² and let U=βxβCβ²β(Οβ1(x))G. We claim that Οβ1(Cβ²)=U. If dβΟβ1(Cβ²), then Ο(d)βCβ², in which case Ο(d)β(Οβ1(Ο(d))GβU. Therefore, Οβ1(Cβ²)βU. Conversely, if dβ(Οβ1(x))G for some xβCβ², then d=gygβ1 for some gβG and yβΟβ1(x). It follows that Ο(d)=Ο(g)Ο(y)Ο(g)β1=Ο(g)xΟ(g)β1, and so dβΟβ1(Cβ²). Hence, UβΟβ1(Cβ²) as desired. In particular, Οβ1(Cβ²) is the union of conjugacy classes of G.
4. The Kernel of the Natural Homomorphism Ο:GL2β(Z/p2Z)βGL2β(Z/pZ)
For the rest of this paper, let Ο denote the natural homomorphism GL2β(Z/p2Z)βGL2β(Z/pZ). An element ΞΊ of kerΟ is of the form ΞΊ=I+Ap, where A is identifiable as an element of Mat2β(Z/pZ) and A uniquely determines ΞΊ. We will refer to A as the p-part of ΞΊ and define p(ΞΊ) to be A.
Note that kerΟ is isomorphic to the 4 dimensional Z/pZ vector space because (I+A1βp)(I+A2βp)=I+(A1β+A2β)p for all A1β,A2ββMat2β(Z/pZ).
Lemma 1 below determines when two elements of kerΟ are conjugate in GL2β(Z/p2Z).
Lemma 1**.**
Let ΞΊ1β,ΞΊ2ββkerΟ, A1β=p(ΞΊ1β) and A2β=p(ΞΊ2β). Then, ΞΊ1β and ΞΊ2β are conjugate in GL2β(Z/p2Z) if and only if A1β and A2β are conjugate by an element of GL2β(Z/pZ).
Proof.
If ΞΊ1β=I+A1βp is conjugate to ΞΊ2β=I+A2βp via gβGL2β(Z/p2Z), i.e. g(I+A1βp)gβ1=I+A2βp, then I+gA1βgβ1p=I+A2βp and so A1β is conjugate to A2β via Ο(g). Conversely, if A1β is conjugate to A2β are conjugate via some gβ²βGL2β(Z/pZ), then I+A1βp is conjugate to I+A2βp via any gβΟβ1(gβ²).
β
Furthermore, Lemma 2 below yields a well defined group action of GL2β(Z/pZ) on kerΟ in which gβGL2β(Z/pZ) sends ΞΊβkerΟ to g^βΞΊg^ββ1, where g^β is any element of Οβ1(g).
Lemma 2**.**
Let g,gβ²βGL2β(Z/p2Z) with Ο(g)=Ο(gβ²), i.e. g and gβ² are congruent modulo p. For any ΞΊβkerΟ, gΞΊgβ1=gβ²ΞΊgβ²β1.
Proof.
Let A=p(ΞΊ) and let gβ²=g+Bp for some BβMat2β(Z/pZ). It is not difficult to see that gβ²β1=gβ1βgβ1Bgβ1p. Therefore,
For kβ₯1, the similarity classes of Mat(Z/pkZ) are defined as the orbits of Mat(Z/pkZ) under conjugation by elements of GL2β(Z/pkZ). The similarity classes of Mat(Z/pkZ) extend the conjugacy classes of GL2β(Z/pkZ) in that the conjugacy classes are themselves similarity classes.
[1, Theorem 2.2] yields a way to categorize the similarity classes of Mat2β(Z/pkZ) for kβ₯1. Just as in [1, Section 1.2], fix a section Z/pZβͺZ/pkZ with image K1ββZ/pkZ. Further fix compatible sections Z/plZβZ/pkZ for 1β€l<k. [1, Lemma 2.1] asserts that Ξ±βMat2β(Z/pkZ) can be written in the form
[TABLE]
with lβ{0,β¦,k} maximal such that Ξ± is congruent to a scalar matrix modulo pl, with unique dβKlβ and unique nonscalar Ξ²βMat2β(Z/pkβlZ). [1, Theorem 2.2] concludes the following:
Theorem 1**.**
With Ξ±βMat2β(Z/pkZ) expressed in the form Ξ±=Id+Ξ²pl as above, lβ{0,β¦,k}, dβKlβ, and trace(Ξ²),det(Ξ²)βZ/pkβlZ completely determine the conjugacy class of g.
4.2. Orbits of Mat2β(Z/pZ) under conjugation by elements of GL2β(Z/pZ)
[6, Table 3.1] lists representatives for all the distinct conjugacy classes of GL2β(Z/pZ). Table 1 below uses Theorem 1 to extend [6, Table 3.1] to include the representatives of the similarity classes of Mat2β(Z/pZ). By Lemma 1, representatives of the conjugacy classes of elements of kerΟ can be given as I+Ap, where A is one of the matrices in Table 1.
4.3. An Equivalent Condition for Local Conjugacy Between Subgroups of kerΟ
The similarity class of a nonscalar element M of Mat2β(Z/pZ) is uniquely determined by trace(M),det(M)βZ/pZ.
We define Ο below to use Lemma 4 as a way to find an equivalent condition for local conjugacy between subgroups of kerΟ.
Definition 2**.**
For a subgroup H of kerΟ and for t,dβZ/pZ, let Ο(H,t,d)=β£{kβHβ£k=I+ApΒ whereΒ trace(A)=t,det(A)=d}β£.
Let H1β and H2β be subgroups of kerΟ. Say that H1β and H2β have equal trace-determinant distribution if Ο(H1β,t,d)=Ο(H2β,t,d) for all t,dβZ/pZ.
By Lemmas 1 and 4, H1β and H2β are locally conjugate.
β
4.4. Preliminary Results for Local Conjugacy in GL2β(Z/pZ) among Subgroups of kerΟ
The definition of locally conjugate subgroups yields the following result:
Lemma 5**.**
If H1β,H2ββ€kerΟ are locally conjugate in GL2β(Z/pZ), then dimH1β=dimH2β.
Proof.
Locally conjugate subgroups are in bijection and Ο is a finite dimensional vector space over the finite field Z/pZ.
β
Lemma 6 categorizes local conjugacy of finite cyclic subgroups.
Lemma 6**.**
Let G be a group and let H1β,H2β be finite locally conjugate subgroups of G. If H1β is cyclic, then H2β is cyclic and H1β and H2β are conjugate.
Proof.
Say that h1β generates H1β. There is some h2ββH2β which is conjugate to h1β in G. The orders of h2β and h1β are equal, H1β and H2β are finite and β£H1ββ£=β£H2ββ£, and so h2β generates H2β. Therefore, H2β is conjugate to H1β.
β
Lemma 7**.**
The subgroups of kerΟ of dimension [math] or 1, i.e. the cyclic subgroups, are conjugate in GL2β(Z/pZ) to one of the following subgroups of kerΟ:
No two distinct subgroups among these are locally conjugate.
Proof.
Let H be a subgroup of kerΟ generated by h=I+Ap for some AβMat2β(Z/pZ). H can be replaced with a conjugate such that A is one of the matrices in Table 1. The categorization of cyclic subgroups of kerΟ is finished by determining an alternative generator of H and conjugating H if necessary.
It is not difficult to see that for all h1ββH1β and h2ββH2β where H1β and H2β are distinct two groups among the ones listed in the statement of the lemma, h1β and h2β are not conjugate in GL2β(Z/p2Z) unless h1β=h2β=I. Thus, no two of the subgroups listed are locally conjugate in GL2β(Z/p2Z).
β
All dimension [math] and 1 subgroups of kerΟ are categorized up to conjugation in Lemma 7. Moreover, the only dimension 3 subgroup of T is T itself. Lemma 9 categorizes the 2 dimensional subgroups of T up to conjugacy as well as local conjugacy among them. It will be useful to consult Lemmas 15 and 16 in Section 7 for several of the upcoming lemmas.
Lemma 9**.**
The subgroups of T of dimension 2 are conjugate in GL2β(Z/p2Z) to one of the following:
No two distinct subgroups among these are locally conjugate.
Proof.
Let Hβ€T have dimension 2. Suppose, for contradiction, that det(p(h))ξ =βa2 for every hβH and any nonzero aβZ/pZ. If there is some nonidentity hβH such that det(p(h))=0, then h is conjugate to u1β=I+(00β10β)p by Lemmas 1 and 4. Replace H with a conjugate so that u1ββH. Since H is 2 dimensional, there is some u2ββH of the form u2β=I+(acβ0βaβ)p for some a,cβZ/pZ where a and c are not both [math]. a must be [math] because det(p(u2β))=βa2. c is therefore nonzero and so I+(01β00β)pβH. By extension, I+(01β10β)pβH, but det(p(I+(01β10β)p))=β1, which is a contradiction.
Otherwise, det(p(h))ξ =βa2 for every nonidentity hβH and any aβZ/pZ. For every nonidentity hβH, βΟ΅det(p(h))β is a nonzero square in Z/pZ. Thus, h is conjugate to v1β=I+(0yβΟ΅y0β)p for some nonzero yβZ/pZ. Replace H with a conjugate so that v1ββH. Since H is 2 dimensional, there is some nonidentity v2ββH of the form v2β=I+(acβ0βaβ)p for some a,cβZ/pZ. However, det(p(v2β))=βa2, which is a contradiction.
Hence, there is some nonidentity hβH such that det(p(h))=βa2 for some nonzero aβZ/pZ. The p-part of ha1β has determinant β1 and trace [math], and so ha1β is an element of H which is conjugate to I+(10β0β1β)p.
Replace H with a conjugate so that I+(10β0β1β)pβH and let w1β=I+(10β0β1β)p. Since H is 2 dimensional, there is some nonidentity w2ββH of the form w2β=I+(0cβb0β)p. If c=0, then bξ =0 and H=H1β. If b=0, then cξ =0 and H is conjugate to H1β via (01β10β). Now assume that b,cξ =0. If bc is a square, then H is conjugate to H2β via (10β0cbβββ). Otherwise, bc is not a square, in which case H is conjugate to H3β via (10β0cΟ΅bβββ).
It remains to show that H1β,H2β and H3β are not locally conjugate to one another. The p-parts of the elements of H1β,H2β and H3β are respectively of the form
[TABLE]
where xiβ,yiββZ/pZ. These p-parts have trace [math] and have determinants βx12β,βx22ββy22β and βx32ββΟ΅y32β respectively. Since Z/pZ has nonsquares, there is some x2β for which x22β+1 is a nonsquare. Setting x2β to be such a value and y2β=1 makes βx22ββy22β=β(x22β+1), which shows that H1β and H2β are not locally conjugate by Proposition 4. Letting x3β=0 and y3β=1 shows that H1β and H3β are not locally conjugate as well. Moreover, βx22ββy22β=0 has solutions such that (x2β,y2β)ξ =(0,0) exactly when β1 is a square in Z/pZ, which is exactly when βx32ββΟ΅y32β does not have solutions such that (x3β,y3β)ξ =(0,0). H2β and H3β are therefore not locally conjuguate.
β
4.6. Final Results for Local Conjugacy in GL2β(Z/pZ) among Subgroups of kerΟ
In particular, H2β and H3,0β are nontrivially locally conjugate. For d,dβ²βZ/pZ where d,dβ²ξ =β1, H3,dβ and H3,dβ²β are nontrivially locally conjugate if dξ =dβ² and ddβ²=1. For a1β,a2β,b1β,b2ββZ/pZ, H6,a1β,b1ββ and H6,a2β,b2ββ are conjugate if a12ββΟ΅b12β=a22ββΟ΅b22β. All other pairs of distinct subgroups listed above are not locally conjugate.
Suppose that u=I+(10β0β1β)p. Choose a nonidentity element hβH to be of the form I+(0cβbdβ)p. Since trace(p(h))=dξ =0, replacing h with hd1β makes d=1. If b=c=0, then H=H5β. If bξ =0, then H is conjugate to H4,bcββ via (10β0bβ). Otherwise, b=0 and cξ =0, but conjugating H via (01β10β) reduces H to the case where bξ =0.
Suppose that u=I+(01βΟ΅0β)p. Choose a nonidentity element hβH so that trace(p(h))=2, i.e. h is of the form h=I+(1+acβ²βbβ²1βaβ)p. Letting b=2Ο΅βbβ²+Ο΅cβ²β, compute
[TABLE]
Replacing h with I+(1+abββbΟ΅1βaβ)p shows that H=H6,a,bβ.
An element of H1β with trace 1 can have any determinant, whereas a trace 1 element of H3,dβ can only have determinant (d+1)2dβ. H1β and H3,dβ are therefore not locally conjugate by Proposition 4. This fully categorizes local conjugacy of H1β with the other subgroups.
i.e. d1β=d2β or d1βd2β=1. One can check that the latter condition implies the former condition. Hence, H3,d1ββ and H3,d2ββ are locally conjugate exactly when d1β=d2β or when d1βd2β=1.
The trace 1 elements of H4,cβ are of the form I+(xcβ11βxβ)p for xβZ/pZ, and such an element has determinant βx2+xβc. Compute βx2+xβc=β(xβ21β)2βc+41β, and so βx2+xβc takes the value βc+41β exactly once and all other values in Z/pZ exactly 2 or [math] times. Therefore, if c,cβ²βZ/pZ are distinct, then H4,cβ and H4,cβ²β are not locally conjugate.
Suppose a1β,b1β,a2β,b2ββZ/pZ satisfy a12ββΟ΅b12β=a22ββΟ΅b22β. Conjugating H6,a,bβ via (βΟ΅ββΟ΅βββϡϡβ)βGL2β(Fp2β) results in the group
[TABLE]
Further conjugating this group by (Ξ±0β0Ξ΄β)βGL2β(Fp2β) results in
[TABLE]
Therefore, H6,a1β,b1ββ is conjugate to H6,a2β,b2ββ via
[TABLE]
which is a scalar multiple of
[TABLE]
H6,a1β,b1ββ and H6,a2β,b2ββ are thus conjugate in GL2β(Z/p2Z).
The trace 1 elements of H6,a,bβ are of the form I+(21+aβ2bβ+xββ2Ο΅bβ+Ο΅x21βaββ)p, whose determinant is 41βa2+Ο΅b2ββΟ΅x2. This expression takes the value 41βa2+Ο΅b2β exactly once and all values of Z/pZ exactly two or zero times. Therefore, if a1β,b1β,a2β,b2ββZ/pZ satisfy a12ββΟ΅b12βξ =a22ββΟ΅b22β, then H6,a1β,b1ββ and H6,a2β,b2ββ are not locally conjugate. This fully categorizes local conjugacy of H6,a,bβ with the other subgroups.
β
Lemma 11**.**
The subgroups of kerΟ of dimension 3 that are not subgroups of T are conjugate in GL2β(Z/p2Z) to one of the following:
For nonzero KβZ/pZ, xβ²2 and Kβyβ²2 each take 2p+1β distinct values over xβ²βZ/pZ and yβ²βZ/pZ respectively. By the pigeonhole principle, xβ²2+yβ²2=K has at least one solution (xβ²,yβ²). In particular, (Β±xβ²,Β±yβ²) yields at least two distinct solutions to xβ²2+yβ²2=K even if one of xβ² and yβ² is [math]. If pβ‘1(mod4), then the equation xβ²2+yβ²2=K is equivalent to (xβ²+jy)(xβ²βjy)=K where jβZ/pZ satisfies j2=β1. The number of solutions to xβ²2+yβ²2=K in this case is therefore pβ1.
The number of solutions to det(p(hiβ))=41+ci2ββ is the number of solutions to xβ²2+yβ²2=0, which is 1 if pβ‘3(mod4) and 2pβ1 if pβ‘1(mod4). Since c12βξ =c22β, the number of solutions to det(p(h2β))=41+c12ββ is at least 2 and exactly pβ1 if pβ‘1(mod4). H3,c1ββ and H3,c2ββ are thus not locally conjugate. Similarly, H4,c1ββ and H4,c2ββ are not locally conjugate.
β
Local conjugacy in kerΟ can be summarized as follows:
Proposition 5**.**
Let H1β,H2ββ€GL2β(Z/p2Z) be nontrivially locally conjugate in GL2β(Z/p2Z). H1β and H2β are conjugate, in some order, to the following subgroups for some dβZ/pZ such that dξ =Β±1:
Dickson [3] classifies the subgroups of GL2β(Z/pZ) based on their images in PGL2β(Z/pZ):
Proposition 6**.**
Let p be an odd prime and let G be a subgroup of GL2β(Z/pZ) with image H in PGL2β(Z/pZ). If G contains an element of order p then GβB(p) or SL2β(Z/pZ)βG. Otherwise, one of the following holds:
(1)
H* is cyclic and a conjugate of G lies in Csβ(p) or Cnsβ(p).*
2. (2)
H* is dihedral and a conjugate of G lies in N(Csβ(p)) or N(Cnsβ(p)), but no conjugate of G lies in Csβ(p) or Cnsβ(p).*
3. (3)
H* is isomorphic to A4β,S4β or A5β and no conjugate of G lies in N(Csβ(p)) or N(Cnsβ(p)).*
Sutherland [6] uses this classification to identify local conjugacy among the subgroups of GL2β(Z/pZ).
Theorem 2**.**
Let H1β,H2ββ€GL2β(Z/pZ) be nontrivially locally conjugate in GL2β(Z/pZ). H1β and H2β are, in some order, conjugate to the following groups:
[TABLE]
where Dβ€Csβ(p), Dβ²=(01β10β)D(01β10β)β1222Since Dβ€Csβ(p2), (z0β0wβ)βDβ² if and only if (w0β0zβ)βD by Lemma 16. and Dξ =Dβ², i.e. there is some (w0β0zβ)βD such that (z0β0wβ)ξ βD.
A special case of the Schur-Zassenhaus Theorem, which is stated below in Theorem 3, gives a sufficient condition for certain pairs of subgroups of GL2β(Z/p2Z) to be conjugate. Recall that a Hall subgroup of a finite group is a subgroup whose order is relatively prime to its index.
Theorem 3** (Schur-Zassenhaus).**
If K is an abelian normal Hall subgroup of a finite group G, then there is a splitting Ο:G/KβG which is unique up to conjugation.
Since kerΟ is a 4 dimensional Z/pZ vector space, kerΟ is abelian and β£kerΟβ£=p4. Moreover, p does not divide β£Ο(Hiβ)β£ by assumption, and so the Schur-Zassenhaus Theorem yields a splitting Ο:Ο(Hiβ)βΟβ1(Ο(Hiβ)), which is unique up to conjugation. Similarly, there is a splitting Οiβ:Ο(Hiβ)βHiβ that arise from the short exact sequence
This section lists algebraic computations and facts resulting from such computations that are used in previous sections.
From this point on and unless stated otherwise, R will denote the ring (Zpβ[Ο΅β])/(p2Zpβ[Ο΅β]), which is isomorphic to (Z/p2Z)[Ο΅β]. The elements of R are identifiable as the sums a+bΟ΅β where a,bβZ/p2Z. Let Ο~β denote the natural homomorphism
[TABLE]
In particular, Ο~β is an extension of Ο.
There is a splitting (Z/pZ)Γβ(Z/p2Z)Γ given by xβ¦xp, where xβ(Z/pZ)Γ and x is any lift of x in Z/p2Z. This map is well defined because (x+ap)pβ‘xp(modp2) for all x,aβZ/p2Z by the bionamial theorem. It is a splitting as xpβ‘x(modp). Similarly, there is a splitting (Z/pZ[Ο΅β])ΓβFp2ΓββRΓ given by xβ¦xp2 where xβFp2Γβ and x is any lift of x in R.
Let S denote the image of the map (Z/pZ[Ο΅β])ΓβRΓ. For xβFp2β, we will often abuse notation and let x also denote the lift of xβFp2β in S. In particular, given w,x,y,z,a,b,c,dβZ/pZ, write
Since (w0β0wβ) multiplicatively commutes with (acβbdβ), the Binomial Theorem applies.
β
Corollary 1**.**
Let H be a subgroup of GL2β(Z/p2Z). Then, H contains h=(w0β0wβ)+(acβbdβ)p, where wβ(Z/pZ)Γ and a,b,c,dβZ/pZ, if and only if H contains h1β=(w0β0wβ) and h2β=I+(acβbdβ)p.
Proof.
Suppose that hβH. h1ββH by Lemma 12 and hh1β1β=I+w1β(acβbdβ)p. Therefore, (hh1β1β)w=I+(acβbdβ)p=h2β, and so h2ββH.
Conversely, if h1β,h2ββH, then H contains h1βh2w1ββ=h1β(I+w1β(acβbdβ)p)=h.
β
Lemma 13**.**
For w,zβ(Z/pZ)Γ and a,b,c,dβZ/pZ such that wξ =z,
[TABLE]
and
[TABLE]
For w,zβRΓ and a,b,c,dβZ/pZ such that wξ =z,
[TABLE]
and
[TABLE]
Proof.
Suppose that w,zβ(Z/pZ)Γ and a,b,c,dβZ/pZ. By expanding,
[TABLE]
Since wξ β‘z(modp), βk=0pβ2βwkzpβ2βk and βk=0pβ2βwpβ2βkzk are both geometric series evaluating to [math]. Therefore,
[TABLE]
From here,
[TABLE]
can be immediately calculated. The claims made for w,zβRΓ and a,b,c,d,βR such that wξ =z can be proved similarly.
β
Corollary 2**.**
Let H be a subgroup of GL2β(R). H contains h=(w0β0zβ)+(acβbdβ)p, where w,zβ(Z/pZ)Γ and a,b,c,dβZ/pZ with wξ =z, if and only if H contains h1β=I+(βwaβ0β0βzdββ)p and (w0β0zβ)+(0cβb0β)p
Let R be a ring. If (w0β0zβ)βGL2β(R) and (acβbdβ)βMat2β(R), then
[TABLE]
Lemma 16**.**
Let R be a ring. If (0yβx0β)βGL2β(R) and (acβbdβ)βMat2β(R), then
[TABLE]
In particular,
[TABLE]
Lemma 17**.**
For a,b,c,dβZ/pZ and nβZ,
[TABLE]
Proof.
Expand
[TABLE]
Since p is an odd prime, βk=0nβ1βk=2(nβ1)nβ, and so
[TABLE]
β
Lemma 18**.**
Let R be a ring. For a,b,c,dβR,
[TABLE]
Lemma 19**.**
Let Hβ€GL2β(Z/p2Z) with t=(10β11β)βΟ(H). Suppose that k=I+(acβbdβ)pβH.
(1)
If aξ =d, then I+(00β10β)pβH.
2. (2)
If cξ =0, then I+(00β10β)p,I+(10β0β1β)pβH.
Proof.
(1)
By Lemma 3, tktβ1βH. Let kβ²=tktβ1kβ1, which must be in H as well. Using Lemma 18, compute
[TABLE]
If c=0, then a power of kβ² is I+(00β10β)p because aξ =d by assumption. If cξ =0, then I+(00β10β)pβH because
[TABLE]
is an element of H.
2. (2)
By 1, I+(00β10β)pβH. Furthermore, since tktβ1kβ1=I+(c0ββaβc+dβcβ)pβH, I+(c0β0βcβ)pβH as well. Thus, I+(10β0β1β)pβH.
β
Lemma 20**.**
Let Hβ€GL2β(Z/p2Z) with t=(10β11β)βΟ(H). If p>3, then I+(00β10β)pβH.
Proof.
There is some element hβH of the form (10β11β)+(acβbdβ)p. By Lemma 17, the pth power of h is
[TABLE]
Since p>3, βk=0pβ1βk2=6(pβ1)p(2pβ1)β, which is [math] modulo p. Therefore, I+(00β10β)pβH,
β
Lemma 21**.**
For a,b,c,d,Ξ²βZ/pZ,
[TABLE]
Lemma 22**.**
For a,b,c,d,Ξ±,Ξ΄βZ/pZ,
[TABLE]
Lemma 23**.**
For a,b,c,d,Ξ±,Ξ³,Ξ΄βZ/pZ,
[TABLE]
Lemma 24**.**
Let Hβ€GL2β(Z/p2Z). If Ο=(10β11β)+(acβbdβ)βH and h=(w0β0zβ)βH where wξ β‘z(modp), then a=d or there is some element of H of the form
If cξ =0 and zwβξ =Β±1, then c(wzββzwβ)ξ =0 and so I+(10β0β1β)pβH by Lemma 19.
If c=0, then compute
[TABLE]
If aξ =d, then I+(00β10β)pβH by Lemma 19 and since zwβξ =1 by assumption, I+(a(1βzwβ)0β0d(1βzwβ)β)pβH with a(1βzwβ)ξ =d(1βzwβ).
If zwβ=β1, then compute
[TABLE]
If aξ =d, then I+(2aβc0β02dβcβ)pβH with 2aβcξ =2dβc.
β
This section categorizes the subgroups, up to conjugation, of GL2β(Z/p2Z) whose images under Ο are subgroups of Csβ(p) or Cnsβ(p) but not subgroups of Z(p).
The notation below will make the Cartan split and Cartan nonsplit cases similar to each other.
Definition 3**.**
Let Hβ€GL2β(R). Say that H is of type Csβ if Hβ€GL2β(Z/p2Z) and Ο(H)β€Csβ(p). Say that H is of type N(Csβ) if Hβ€GL2β(Z/p2Z) and Ο(H)β€N(Csβ(p)).
Let Hβ² be the conjugate of H via (βΟ΅ββΟ΅βββϡϡβ)β1. Say that H is of type Cnsβ if Hβ²β€GL2β(Z/p2Z) and Ο~β(Hβ²)β€Cnsβ(p). Say that H is of type N(Cnsβ) if Hβ²β€GL2β(Z/p2Z) and Ο~β(Hβ²)β€N(Cnsβ(p)).
Say that H is in diagonalized form if H is of type N(Csβ) or of type N(Cnsβ).
Suppose Hβ€GL2β(R) is of type N(Cnsβ). The elements of Ο~β(H) must be of the form (w+Ο΅βy0β0wβΟ΅βyβ) or of the form (0w+Ο΅βyβwβΟ΅βy0β) by Lemma 14. From now on, let K denote the conjugate of kerΟ via (βΟ΅ββΟ΅βββϡϡβ). Note that K is a 4 dimensional Z/pZ vector space just as kerΟ is. By Lemma 14,
Lemma 26 and Corollaries 3 and 4 below show that local conjugacy in GL2β(Z/p2Z) between two subgroups H1β,H2β of GL2β(Z/p2Z) such that Ο(Hiβ)β€N(Cnsβ(p)) is equivalently determined by local conjugacy between H1β²β and H2β²β in GL2β(R), where Hiβ²β is the conjugate of Hiβ via (βΟ΅ββΟ΅βββϡϡβ).
Lemma 26**.**
If g1β,g2ββGL2β(Z/p2Z) are conjugate via gβGL2β(R), then g1β and g2β are conjugate via some gβ²βGL2β(Z/p2Z), whose value is only dependent on g.
Proof.
Express g in the form
[TABLE]
where Ξ±iβ,Ξ²iβ,Ξ³iβ,Ξ΄iββZ/p2Z for i=1,2. Since the entries of g1β and g2β are in Z/p2Z,
[TABLE]
Therefore, if (Ξ±iβΞ³iββΞ²iβΞ΄iββ) is invertible for i=1 or 2, then h2β=gβ²h1βgβ²β1, where gβ²=(Ξ±iβΞ³iββΞ²iβΞ΄iββ). Otherwise, Ξ±iβΞ΄iββΞ²iβΞ³iβ=0 for i=1,2, in which case
[TABLE]
Since g is invertible, Ξ±1βΞ΄2β+Ξ±2βΞ΄1ββΞ²1βΞ³2ββΞ²2βΞ³1βξ =0. Let gβ²=(Ξ±1β+Ξ±2βΞ³1β+Ξ³2ββΞ²1β+Ξ²2βΞ΄1β+Ξ΄2ββ). Note that gβ² is invertible because
[TABLE]
Furthermore, gβ²g1β=g2βgβ², and so g2β=gβ²g1βgβ²β1 as desired.
β
The idea behind Lemma 26 can be immediately extended to the following corollaries:
Corollary 3**.**
Two subgroups of GL2β(Z/p2Z) that are conjugate in GL2β(R) are conjugate in GL2β(Z/p2Z).
Corollary 4**.**
Two subgroups of GL2β(Z/p2Z) that are locally conjugate in GL2β(R) are locally conjugate in GL2β(Z/p2Z).
Lemma 27**.**
Let Hβ€GL2β(R) be in diagonialized form. Suppose that there is some hβH such that Ο~β(h)=(w0β0zβ) where wξ =z. Then, H has some element k=I+(acβbdβ)p if and only if H has both l=I+(a0β0dβ)p and m=I+(0cβb0β)p.
Proof.
Since k=lm, all three of l,m,k are in H if two of them are. In particular, if l,m,βH, then kβH.
Conversely, assume that kβH. Use Lemma 15 to compute
[TABLE]
If w=βz, then hkhβ1kβ1=I+(0β2cββ2b0β)p. Moreover, m is a power of hkhβ1kβ1, and so mβH. Thus, lβH as well.
Now assume that wξ =Β±z. If b=0, then hkhβ1kβ1=I+(0c(wzββ1)β00β)p. If H is of type N(Cnsβ) as well, then c=0 by Lemma 14, and there is nothing to prove. Otherwise, H is of type N(Csβ), in which case wzββ1 is in Z/p2Z and nonzero modulo p, and so a power of hkhβ1kβ1 is m and we are done. The case where c=0 is similar.
to see that hkhβ1kβ1 and h(hkhβ1kβ1)hβ1 are linearly independent as Z/pZ-vectors. Moreover, both have p-parts whose diagonal entries are [math]. Note that the subspaces of K and kerΟ consisting of the matrices whose p-parts have [math] as their diagonal entries are both 2 dimensional. Therefore, mβH as desired.
β
This section categorizes the subgroups H of GL2β(R) of type Csβ or of type Cnsβ such that Ο~β(H)ξ β€Z(p) up to conjugation to understand local conjugacy among such subgroups. For Hβ€GL2β(R) let DHβ denote the group consisting of all elements of H that are diagonal matrices.
Proposition 8**.**
Let Hβ€GL2β(R) be of type Csβ or of type Cnsβ such that Ο~β(H)ξ β€Z(p). There is a conjugate Hβ² of H satisfying the following:
(1)
Hβ²* is of type Csβ if H is of type Csβ and Hβ² is of type Cnsβ if H is of type Cnsβ*
2. (2)
Ο~β(Hβ²)=Ο~β(H)**
3. (3)
Hβ²* is the internal semidirect product ΞHβ²β₯ββDHβ²β*
The case where H is of type Cnsβ is similar.
β
Before proceeding, we introduce a definition which will be useful for understanding the structure of nontrivially locally conjugate subgroups of GL2β(Z/p2Z).
Definition 4**.**
Let R be a ring and let D1β and D2β be two subgroups of GL2β(R) consisting only of diagonal matrices. Say that D1β and D2β are diagonal swaps if (w0β0zβ)βD1β exactly when (z0β0wβ)βD2β. Equivalently, D1β and D2β are conjugate to each other via (01β10β).
It will turn out that up to conjugation, pairs of nontrivially locally conjugate subgroups of GL2β(Z/p2Z) are generated by some equal generators along with two subgroups of Csβ(p2) which are unequal diagonal swaps.
Proposition 9**.**
Let H1β,H2ββ€GL2β(R) both be of type Csβ or of type Cnsβ. Suppose that H1β and H2β are locally conjugate in GL2β(R), Ο~β(Hiβ)ξ β€Z(p), Ο~β(H1β)=Ο~β(H2β) and Hiβ=ΞHiββ₯ββDHiββ.
(1)
The groups DH1ββ and DH2ββ are equal or are diagonal swaps.
2. (2)
Suppose that H1β and H2β are both of type Csβ. Elements of Hiβ are expressible as the product kh for unique kβΞHiββ₯β and hβDHiββ. Fix hβDHiββ so that Ο(h)=(w0β0zβ), where w,zβ(Z/pZ)Γ are unequal. Express h in the form h=(w0β0zβ)+(a0β0dβ)p. For kβΞHiββ₯β, express k in the form k=I+(0Ξ³βΞ²0β)p. Compute
[TABLE]
and
[TABLE]
By Theorem 1, kh and h are in the same conjugacy class, i.e. the conjugacy class of kh does not depend on k. The above calculation also shows that the only matrices that could be elements of DHiββ and conjugate to h are h itself and (z0β0wβ)+(d0β0aβ)p.
Suppose that there is some (w0β0zβ)βΟ(H1β) such that (z0β0wβ)ξ βΟ(H1β). Since H1β=ΞH1ββ₯ββDH1ββ, (w0β0zβ)βH1β by Corollary 2. If I+(a0β0dβ)βH1β, then
[TABLE]
is an element of H1β. Since (z0β0wβ)ξ βΟ(H2β), (w0β0zβ)+(aw0β0dzβ)pβH2β, and so I+(a0β0dβ)pβH2β as well. Therefore, DH1βββDH2ββ and DH2βββDH1ββ by symmetry.
Now suppose that for all (w0β0zβ)βΟ(H1β), (z0β0wβ)βΟ(H1β) as well. For any I+(a0β0dβ)pβH1β, (w0β0zβ)+(aw0β0dzβ)pβH1β, and so (w0β0zβ)+(aw0β0dzβ)p or (z0β0wβ)+(dz0β0awβ)p is in H2β. In the former case, I+(a0β0dβ)pβH2β. In the latter, I+(d0β0aβ)pβH2β. If ΞH1ββ is 2 dimensional, i.e. it is generated by I+(10β01β)p and I+(10β0β1β)p, then ΞH2ββ contains both I+(10β01β)p and I+(10β0β1β)p, and so ΞH1ββ=ΞH2ββ. If ΞH1ββ is 1 dimensional, then say that it is generated by I+(a0β0dβ)p. In particular, ΞH2ββ is not 2 dimensional. If I+(a0β0dβ)pξ βΞH2ββ, then I+(d0β0aβ)pβΞH2ββ. In this case, let H2β²β be the conjugate of H2β via (01β10β). By Lemma 16, I+(a0β0dβ)pβΞH2β²ββ, and so ΞH1ββ=ΞH2β²ββ. If ΞH1ββ is [math] dimensional, then so is ΞH2ββ, concluding the case where H1β and H2β are both of type Csβ.
The case where H1β and H2β are both of type Cnsβ is similar.
2. (2)
Suppose that H1β and H2β are nontrivially locally conjugate and DH1ββ=DH2ββ. Further suppose that H1β and H2β are both of type Csβ. By Proposition 2 and Lemma 5, dim(ΞH1ββ₯β)=dim(ΞH2ββ₯β). Note that dim(ΞHiββ₯β) is not 2 or [math] because H1β=H2β otherwise. Hence, dim(ΞHiββ₯β)=1. Say that I+(0ciββbiβ0β)p generates ΞHiββ₯β. In particular, biβ or ciβ is nonzero. By Lemma 15, there is some diagonal matrix (Ξ±iβ0β0Ξ΄iββ) such that
Now suppose that H1β and H2β are both of type Cnsβ. Note that ΞHiββ₯β consists only of elements of the form I+(0a+Ο΅βcβaβΟ΅βc0β)p. Similarly as in the type Csβ case, dim(ΞH1ββ₯β)=dim(ΞH2ββ₯β)=1. Say that I+(0aiββΟ΅βciββaiβ+Ο΅βciβ0β)p generates ΞHiββ₯β. At least one of aiβ and ciβ is nonzero modulo p, and so (a2β+Ο΅βc2β0β0a1β+Ο΅βc1ββ) is invertible. Conjugating H1β by (a2β+Ο΅βc2β0β0a1β+Ο΅βc1ββ) yields H2β by Lemma 15, and so H1β and H2β are conjugate to begin with, a contradiction. Hence, H1β and H2β must both be of type Csβ.
β
Corollary 6 below categorizes local conjugacy for the subgroups of GL2β(Z/p2Z) whose images under Ο are contained in Cnsβ(p) but not Z(p).
Corollary 6**.**
Let H1β,H2ββ€GL2β(Z/p2Z) with Ο(Hiβ)β€Cnsβ(p) but Ο(Hiβ)ξ β€Z(p). If H1β and H2β are locally conjugate in GL2β(Z/p2Z), then they are conjugate in GL2β(Z/p2Z).
Proof.
It suffices to show that H1β and H2β are conjugate in GL2β(R) by Corollary 3. Replace Hiβ with its conjugate via (βΟ΅ββΟ΅βββϡϡβ) so that Hiβ is of type Cnsβ. Using Proposition 8, further replace Hiβ with a conjugate such that Hiβ is still of type Cnsβ, Ο~β(Hiβ) is preserved and Hiβ=ΞHiββ₯ββDHiββ. Proposition 9 shows that H1β and H2β are conjugate in GL2β(R) as desired.
β
Proposition 10 below categorizes local conjugacy for the subgroups of GL2β(Z/p2Z) whose images under Ο are contained in Csβ(p).
Proposition 10**.**
Let H1β,H2ββ€GL2β(Z/p2Z) with Ο(Hiβ)β€Csβ(p). Then, H1β and H2β are nontrivially locally conjugate if and only if they are conjugate to the groups
[TABLE]
in some order, where D and Dβ² are subgroups of Csβ(p2), D and Dβ² are diagonal swaps, and Dξ =Dβ².
Proof.
Suppose that H1β and H2β are nontrivially locally conjugate. By Proposition 3, Ο(H1β) and Ο(H2β) are locally conjugate. By Theorem 2, Ο(H1β) and Ο(H2β) are conjugate, and so H2β can be replaced with a conjugate so that Ο(H1β)=Ο(H2β).
in some order. Conjugating the second of these groups by (01β10β) yields
[TABLE]
Note that H1β and H2β are conjugate if DHiββ=(01β10β)DHiββ(01β10β)β1. It is also not difficult to see that H1β and H2β are nontrivially locally conjugate otherwise.
β
9.2. The Normalizer of Cartan Cases
This section categorizes the subgroups, up to conjugation, of GL2β(Z/p2Z) whose images via Ο are subgroups of N(Csβ(p)) or N(Cnsβ(p)) that are not contained in Csβ(p) or Cnsβ(p).
Lemma 28**.**
Let Hβ€GL2β(R) be of type N(Csβ(p)) or N(Cnsβ(p)). Suppose that Ο~β(H) contains some element of the form (w0β0zβ) where wξ =z and another element of the form (0yβx0β).
(1)
Suppose that H is of type N(Csβ(p)). Then, ΞHβ is generated by both, one or neither of I+(10β01β)p and I+(10β0β1β)p.
2. (2)
Suppose that H is of type N(Cnsβ(p)). Then, ΞHβ is generated by both, one or neither of I+(10β01β)p,I+(Ο΅β0β0βΟ΅ββ)p.
3. (3)
Suppose that H is of type N(Csβ(p)). Then, ΞHβ₯β is generated by both, one or neither of I+(0yβx0β)p and I+(0βyβx0β)p.
4. (4)
Suppose that H is of type N(Cnsβ(p)). Then, ΞHβ₯β is generated by both, one or neither of I+(0yβx0β)p and I+(0βyΟ΅ββxΟ΅β0β)p.
5. (5)
If wξ =Β±z, then dim(ΞHβ₯β)=2 or [math].
Proof.
(1)
If I+(a0β0dβ)pβH, then I+(d0β0aβ)pβH by Lemma 16. Suppose that dim(ΞHβ)=1. Note that ΞHβ must only contain elements of the form I+(a0β0aβ)p or only contain elements of the form I+(a0β0βaβ)p.
2. (2)
This follows from the same argument as the last part.
3. (3)
If I+(0cβb0β)pβH, then I+(0bxyββcyxβ0β)pβH be Lemma 16. Suppose that dim(ΞHβ₯β)=1 and assume that b or c is nonzero. In this case, c2yxβ=b2xyβ. Since x and y are nonzero and at least one of b and c is nonzero, both b and c are nonzero. Thus, yxβ=Β±cbβ. Since b,c,x,yβZ/pZ, we are done.
4. (4)
Likewise, suppose that dim(ΞHβ₯β)=1 and take I+(0cβb0β)pβH such that b or c is nonzero. Note that x,y,b,c are of the forms
[TABLE]
for some Ξ±,Ξ²,Ξ³,Ξ΄βZ/pZ by Lemma 14. If yxβ=cbβ, then cx=by, in which case
[TABLE]
Thus, Ξ±Ξ΄=Ξ²Ξ³, i.e. bxβ is in Z/pZ. In this case, ΞHβ₯β is generated by I+(0yβx0β)p. If yxβ=βcbβ, then similarly compute Ξ±Ξ³=ϡβδ. Note that bxΟ΅βββZ/pZ, and so ΞHβ₯β is generated by I+(0βyΟ΅ββxΟ΅β0β)p.
5. (5)
Let H1β,H2ββ€GL2β(R) be locally conjugate and both of type N(Csβ) or both of type N(Cnsβ) but not of types Csβ or Cnsβ. If one of H1β or H2β has an element of the form (w0β0zβ) where wξ =Β±z, then H1β and H2β are conjugate.
Let H1β,H2ββ€GL2β(R) be locally conjugate and both of type N(Csβ) or both of type N(Cnsβ) but not of types Csβ or Cnsβ. H1β and H2β are conjugate.
Proof.
If all of the diagonal elements of Ο~β(Hiβ) are scalar, then there is some rβRΓ such that yxβ=r for all (0yβx0β)βΟ~β(Hiβ). Furthermore, whether or not Hiβ is of type N(Csβ), it is not difficult to see that r is a square in R. Replace Hiβ by its conjugate via (rβ1βrββ1β)β1. All elements of Ο~β(Hiβ) are then diagonal and it is not difficult to see that Hiβ is in fact of type Csβ or of type Cnsβ. This possibility was already considered in Section 9. Assume that some diagonal elements of Ο~β(Hiβ) are nonscalar.
By Lemma 29, it suffices to show that H1β and H2β are conjugate given that all nonscalar diagonal elements of Ο~β(Hiβ) are of the form (w0β0βwβ). In this case, there is some rβRΓ such that yxβ=Β±r for all (0yβx0β)βΟ(Hiβ)~β. Moreover,
Suppose that there is some (w0β0βwβ)βΟ~β(Hiβ) which is not conjugate to any element of Hiβ of the form (0yβx0β). By Corollary 2, Hiβ has an element hiβ of the form hiβ=(w0β0βwβ)+(0ciββbiβ0β)p. For k=I+(Ξ±Ξ³βΞ²Ξ΄β)pβH1β, compute
[TABLE]
An element of H2β which is conjugate to h1βk must be of the form (w0β0βwβ)+(Ξ±wββββΞ΄wβ)p or (βw0β0wβ)+(βΞ΄wβββΞ±wβ)p. Thus, if I+(Ξ±0β0Ξ΄β)pβH1β then I+(Ξ±0β0Ξ΄β)p or I+(Ξ΄0β0Ξ±β)p is an element of H2β. By Lemma 28, ΞHiββ is generated by both, one or neither of I+(10β01β)p and I+(a0β0βaβ)p where a=1 or Ο΅β. Thus, ΞH1ββ=ΞH2ββ and so dim(ΞH1ββ₯β)=dim(ΞH2ββ₯β). If dim(ΞHiββ₯β)=2 or [math], then ΞH1ββ₯β=ΞH2ββ₯β, in which case H1β=H2β.
Fix an element (0yβx0β) of Ο~β(Hiβ) and suppose that dim(ΞHiββ₯β)=1. Fix giββHiβ to be of the form giβ=(0yβx0β)+(aiβciββbiβdiββ)p. An element of H2β which is conjugate to g1β must be an element of Ο~ββ1((0yβ²βxβ²0β)) for some (0yβ²βxβ²0β)βΟ~β(Hiβ). In particular, βxy=det(0yβx0β)=det(0yβ²βxβ²0β)=βxβ²yβ² and xyβ=Β±xβ²yβ²β, and so (0yβ²βxβ²0β) is one of (0Β±yβΒ±x0β) or (0βjyβΒ±jx0β) where j is a square root of β1. If Hiβ is of type N(Csβ), then jβZ/pZ. Moreover, using Lemma 14 shows that jξ βZ/pZ if Hiβ is of type N(Cnsβ).
Suppose that one of (0βjyβΒ±jx0β) is an element of Ο~β(Hiβ). Since βIβΟ~β(Hiβ), both of them are elements of Ο~β(Hiβ). Moreover,
[TABLE]
and so for all (0yβ²βxβ²0β)βΟ~β(Hiβ), (0βjyβ²βΒ±jxβ²0β) are elements of Ο~β(Hiβ). Note that Lemma 15 shows that Ο~β(Hiβ) is preserved under conjugation via (j0β01β). Further suppose that ΞH1ββ₯βξ =ΞH2ββ₯β. If Hiβ is of type N(Csβ), then say that, without loss of generality, ΞH1ββ₯β is generated by I+(0yβx0β)p and that ΞH2ββ₯β is generated by I+(0βyβx0β)p using Lemma 28. Replace H2β with its conjugate via (j0β01β). Since jβZ/pZ, ΞH1ββ₯β and ΞH2ββ₯β are now equal. By Proposition 7, H1β and H2β are conjugate. Similarly, if Hiβ is of type N(Cnsβ), then one can replace H2β with its conjugate via (j0β01β). Since jΟ΅β is an element of Z/pZ, ΞH1ββ₯β and ΞH2ββ₯β are now equal and so H1β and H2β are conjugate by Proposition 7.
Suppose that neither of (0βjyβΒ±jx0β) is in Ο~β(Hiβ). For k=I+(Ξ±Ξ³βΞ²Ξ΄β)pβH1β, compute
[TABLE]
If ΞHiββ₯β is generated by I+(0yβx0β)p, then trace(g1βk) is of the form (aiβ+diβ+2nxyu)p where nβZ/pZ and u=1 or Ο΅β, in particular, when Ξ²=nxu and Ξ³=nyu. Otherwise, trace(g1βk) can only have trace (a1β+d1β)p. Since the only elements of Ο~β(Hiβ) that are conjugate to (0yβx0β) are (0Β±yβΒ±x0β), one can see that ΞH1ββ₯β=ΞH2ββ₯β. Thus, H1β and H2β are conjugate by Proposition 7. This concludes the case where there is some (w0β0βwβ)βΟ~β(Hiβ) which is not conjugate to any element of Ο~β(Hiβ) of the form (0yβx0β).
Now assume that every (w0β0βwβ)βΟ~β(Hiβ) is conjugate to an element of Ο~β(Hiβ) of the form (0yβx0β). In particular, xy=w2. Replace Hiβ with its conjugate via (xwβ0β0wxββ) so that (0wβw0β)βΟ~β(Hiβ) instead of (0yβx0β).
Thus, replacing H2β with its conjugate via (11βjβjβ) makes all diagonal elements of Ο(H2β) scalar, which is a case that was already discussed. If (0βjyβΒ±jx0β)βΟ(Hiβ), then further compute
This section categorizes the subgroups, up to conjugation, of GL2β(Z/p2Z) whose images under Ο are subgroups of B(p) containing (10β11β).
As with Lemma 20, the types of such subgroups that can arise differ between the case when p>3 and the case when p=3. We consider the case where p>3 first.
Now assume that Ο is of the form Ο=(10β11β)+(acβbdβ)p where aξ =d. By Lemma 24, there is some I+(Ξ±0β0Ξ΄β)pβH where Ξ±ξ =Ξ΄. Moreover,
[TABLE]
by Lemma 22. One can further show that aβΞ±Ξ±βΞ΄aβdβ=dβδαβΞ΄aβdβ. By Lemma 21, there is some Οβ²βH of the form Οβ²=(10β11β)+(aβ²cβ²β0aβ²β)p, which yields the desired result.
β
and so (10β11β)+(aβ²1β0aβ²β)pβHβ². If c is a nonsquare, then one can similarly replace Hβ² with a conjugate so that a matrix of the form (10β11β)+(aβ²Ο΅β0aβ²β)pβHβ² and so Hβ² is therefore of the desired form.
β
Similarly, the only elements of B(p) that are conjugate to (w0βxwβ)p where xξ =0 are of the form (w0βxβ²wβ) where xβ²ξ =0. Furthermore, the only elements of B(p) that are conjugate to (w0βxzβ)p where wξ =z are of the form (w0βxβ²zβ) or (z0βxβ²wβ). Subgroups of Csβ(p) are generated by at most two elements, one of which can be chosen to be in Z(p). It is then not difficult to see that CH1ββ and CH2ββ are equal or diagonal swaps.
β
Lemma 33**.**
The conjugacy class of an element (10βn1β)+(acβbdβ)p in GL2β(Z/p2Z), where 0<n<p, is determined completely by a+d and cn.
Proof.
The determinant of the matrix is 1+(a+dβcn)p whereas the trace is 2+(a+d)p. Apply Theorem 1.
β
Proposition 13**.**
Suppose that p>3. Let H1β,H2ββ€GL2β(Z/pZ/p2Z) be nontrivially locally conjugate with (10β11β)βΟ(Hiβ). H1β and H2β are conjugate to
[TABLE]
in some order, where Ο is one of
(1)
(10β11β)**
2. (2)
(10β11β)+(10β01β)p**
3. (3)
(10β11β)+(a1β0aβ)p* for some aβZ/pZ*
4. (4)
(10β11β)+(aΟ΅β0aβ)p* for some aβZ/pZ,*
k* is one of*
(1)
I**
2. (2)
I+(01β00β)p**
D* is a subgroup of Csβ(p), D and Dβ² are diagonal swaps but Dξ =Dβ².*
In the case that niβξ =0(modp), Lemma 33 asserts that its conjugacy class is determined by the values of 2miβ+2niβ+miβniβΞ³iβ and miβniβΞ³iβ. Alternatively, the conjugacy class is determined by the values of miβ+niβ and miβniβΞ³iβ. Parametrize m1β as m1β=rn1β. Note that m1β+n1β=(r+1)n1β and m1βn1βΞ³1β=rn12βΞ³1β, and so there must be some m2β and n2β such that m2β+n2β=(r+1)n1β and m2βn2β=rn12βΞ³2βΞ³1ββ, i.e. there is a solution to the quadratic equation x2β(r+1)n1βx+rn12βΞ³2βΞ³1ββ=0. This happens only when the discriminant, which is (r+1)2n12ββ4rn12βΞ³2βΞ³1ββ, is a square in Z/pZ. When n1βξ =0, (r+1)2β4rΞ³2βΞ³1ββ must be a square. Completing the square shows that
[TABLE]
This value needs to be a square for all rβZ/pZ. Therefore, for all squares sβZ/pZ, s+1β(1β2Ξ³2βΞ³1ββ)2 is also a square. Let c=1β(1β2Ξ³2βΞ³1ββ)2 and suppose that cξ =0. The case where s=0 shows that c must be a square, say c=t2. In this case, s/t2+1 is a square for all squares sβZ/pZ. However, s/t2 spans all squares in Z/pZ, which is a contradiction. Hence, 1β(1β2Ξ³2βΞ³1ββ)2=0 and so Ξ³1β=Ξ³2β. H1β and H2β are therefore equal.
For p=3, there are 40 pairs of nontrivially locally conjugate subgroups of GL2β(Z/p2Z) up to conjugation. They can be expressed similarly as the pairs of nontrivially locally conjugate subgroups of GL2β(Z/p2Z) for p>3 as described in Proposition 13. In particular, the pairs are expressible in the form
[TABLE]
where Ο satisfies Ο(Ο)=(10β11β), kβ€kerΟ and D and Dβ² are unequal diagonal swaps. The main difference between the cases where p=3 and p>3 is in Lemma 20: whereas I+(00β10β)pβH if p>3 and H is a subgroup of GL2β(Z/p2Z) with (10β11β)βΟ(H), then I+(00β10β)pβH, but if p=3, then this is not necessarily true.
11. The SL2β(Z/pZ) case
This section categorizes the subgroups, up to conjugation, of GL2β(Z/p2Z) whose images via Ο contain SL2β(Z/pZ).
Lemma 34**.**
Suppose that p>3. Let Hβ€GL2β(Z/p2Z). If SL2β(Z/pZ)β€Ο(H), then Tβ€H.
Proof.
A computation similar to that of Lemma 20 shows that I+(01β00β)pβH. By Lemma 19, Tβ€H.
β
Lemma 35**.**
Suppose that p>3. If Hβ€GL2β(Z/p2Z) with Ο(H)=SL2β(Z/pZ), then H=SL2β(Z/p2Z) or Οβ1(SL2β(Z/pZ)).
Proof.
By Lemma 34, Tβ€H. Therefore, H has an element Ο of the form Ο=(10β11β)+(a0β0aβ)p. Since p>3, there is some wβZ/pZ such that w2ξ =1. In particular, H has an element of the form (w0β0w1ββ)+(0cβb0β)p by Corollary 2. Since I+(00β10β)p,I+(01β00β)pβH, H has (w0β0w1ββ) as well. If aξ =0, then a computation in Lemma 24 shows that I+(10β01β)p, and so kerΟβ€H. In this case, H=Οβ1(SL2β(Z/pZ)). Similarly, H has an element of the form (10β11β)+(aβ²0β0aβ²β)p. If aβ²ξ =0, then kerΟβ€H. If a=aβ²=0, then (10β11β),(11β01β)βH and so SL2β(Z/p2Z)β€H.
It now suffices to show that if SL2β(Z/p2Z)ξ =H, then H=Οβ1(SL2β(Z/pZ)). Choose some hβH that is not in SL2β(Z/p2Z). Since Ο(H)=SL2β(Z/pZ), det(h)β‘1(modp). On the other hand, det(h)ξ =1 because hξ βSL2β(Z/p2Z). Choose some sβSL2β(Z/p2Z) such that Ο(s)=Ο(h). Note that shβ1βkerΟ, and det(shβ1)ξ =1, and so shβ1 is of the form I+(acβbdβ)p where a+dξ =0. Hence, kerΟβ€H, in which case H=Οβ1(SL2β(Z/pZ)).
β
Proposition 14**.**
Suppose that p>3. Let H1β,H2ββ€GL2β(Z/p2Z). If H1β and H2β are locally conjugate with SL2β(Z/pZ)β€Ο(Hiβ), then H1β=H2β.
Let H denote the image of H in PGL2β(Z/pZ). Note that H contains a subgroup which is isomorphic to A4β. Therefore, H has an element h1ββ of order 3 and an element h2ββ of order 2 which do not commute. In fact, h1ββh2ββ has order 3. Furthermore, H is generated by h1ββ and h2ββ. Choose h1β,h2ββΟ(H) such that the images of h1β and h2β in H are h1ββ and h2ββ respectively. By Proposition 6, Ο(H) has no element of order p. In particular, Ο(H) has no element which is conjugate to a matrix of the form (w0β1wβ) where wβ(Z/pZ)Γ. Thus, h1β is conjugate to an element in Csβ(p) or an element in Cnsβ(p). Assume for the rest of the proof that h1β is conjugate to an element in Csβ(p). The case where h1β is conjugate to an element of Cnsβ(p) works similarly. Replace H with a conjugate so that h1β is replace with a matrix of the form (w0β0zβ)βCsβ(p). Note that wξ =Β±z. Express h2β as h2β=(Ξ±Ξ³βΞ²Ξ΄β). Compute
[TABLE]
Since h2ββ is an element of PGL2β(Z/pZ) of order 2, Ξ±2=Ξ΄2 and (Ξ±+Ξ΄)Ξ²=(Ξ±+Ξ΄)Ξ³=0. If Ξ±=Ξ΄, then h2β=(Ξ±0β0Ξ±β) or (0Ξ³βΞ²0β). However, H would be cyclic or dihedral, which is a contradiction to Proposition 6. Hence, Ξ±=βΞ΄ξ =0 and we express h2β as h2β=(Ξ±Ξ³βΞ²βΞ±β), where not both Ξ² and Ξ³ are [math]. Suppose that Ξ³=0. Compute
[TABLE]
Since h1ββh2ββ has order 3, Ξ±3w3=βΞ±3z3. However, w3=z3 because h1ββ has order 3, which is a contradiction. Hence, Ξ³ξ =0 and similarly, Ξ²ξ =0.
An argument similar to that of Lemma 27 shows that I+(acβbdβ)βH if and only if I+(a0β0dβ)p,I+(0cβb0β)pβH. Moreover, if I+(0cβb0β)pβH with b and c nonzero, then I+(0cβb0β)p and h1β(I+(0cβb0β)p)h1β1β=I+(0cwzββbzwβ0β)p are linearly independent because wξ =Β±z, in which case I+(00β10β)p,I+(01β00β)pβH.
Suppose that H has an element of the form I+(0cβb0β)p where b or c is nonzero. Compute
[TABLE]
cΞ²2βbΞ±2 and βcΞ±2+bΞ³2 are in the same ratio as b and c, i.e. c(cΞ²2βbΞ±2)=b(βcΞ±2+bΞ³2), when b2Ξ³2=c2Ξ²2. If this holds, then b and c are both nonzero. Otherwise, I+(0cβb0β)p and I+βΞ±2βΞ²Ξ³(0βcΞ±2+bΞ³2βcΞ²2βbΞ±20β)pβ are linearly independent. In either case, I+(00β10β)p,I+(01β00β)pβH. Moreover,
[TABLE]
and since Ξ± and Ξ³ are nonzero, I+(10β0β1β)pβH.
Suppose that there is some I+(a0β0dβ)pβH such that aξ =d. Compute
[TABLE]
Since Ξ±ξ =0 and Ξ² and Ξ³ are nonzero, H has an element of the form I+(0cβb0β)p where b and c are nonzero.
Let H1β,H2ββ€GL2β(Z/p2Z) be locally conjugate and suppose that the image of Hiβ in PGL2β(Z/pZ) is isomorphic to A4β, S4β or A5β. H1β and H2β are conjugate.
Note that β£Ο(Hiβ)β£ divides 12(pβ1), 24(pβ1) or 60(pβ1) because the image of Hiβ in PGL2β(Z/pZ) is isomorphic to A4β, S4β or A5β and the kernel of the natural homomorphism GL2β(Z/pZ)βPGL2β(Z/pZ) is Z(p), which has order pβ1. Therefore, if p>5, then p does not divide β£Ο(Hiβ)β£ [6, Table 2] shows that the image of Hiβ in PGL2β(Z/pZ) cannot be isomorphic to A4β, S4β or A5β if p=3 and cannot be isomorphic to A5β if p=5. In any case, p does not divide β£Ο(Hiβ)β£ and so H1β and H2β are conjugate by Proposition 7.
β
13. Conclusion
Theorem 4 summarizes the categorization of pairs of nontrivially locally conjugate subgroups of GL2β(Z/p2Z).
Theorem 4**.**
The pairs of nontrivially locally conjugate subgroups of GL2β(Z/p2Z) are, up to conjugation, those listed in Propositions 10 and 13 and for p=3, the ones described at the end of Section 10.
Acknowledgements
I would like to thank my mentor Atticus Christensen for his invaluable insight on approaching the main problem, his daily help on the complicated details towards solving it, and his guidance in constructing this paper. I would also like to thank Professor David Jerison, Professor Ankur Moitra, and Dr. Andrew Sutherland for their encouragement and experienced advice on working with mathematics. In particular, Dr. Sutherland motivated this project and guided me on it as my supervisor for the MIT Undergraduate Research Opportunities Program. I give my final thanks to Dr. Slava Gerovitch and MIT Mathematics for making the Summer Program in Undergraduate Research possible.
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