# How do you show that there are only finitely many isomorphism classes of representations of $G$ on $\mathbb Z^n$?

These are exercises in A Course in Finite Group Representation Theory, Chapter 1. Could you help me solve the 11st exercise below? Any help is appreciated and the result of exercise 10 is very likely needed, I think.

1. One form of the Jordan–Zassenhaus theorem asserts that for each $$n$$, $$GL(n,\mathbb Z)$$ (that is, $$Aut(\mathbb Z^n )$$) has only finitely many conjugacy classes of subgroups of finite order. Assuming this, show that for each finite group $$G$$ and each integer $$n$$ there are only finitely many isomorphism classes of representations of $$G$$ on $$\mathbb Z^n$$.

This is my thought: In the case of exercise 11, if $$α$$ in exercise 10 is proved to be inner, then it's easy to see that (2) ⇒ (3) ⇒ (1). Thus infinitely many isomorphism classes of representations of $$G$$ on $$\mathbb Z^n$$ will result in infinitely many conjugacy classes of subgroups of finite order, a controdiction.

Edit (2018/7/2): I attempt to prove that $$\rho_1(G) \sim \rho_2(G) \implies \rho_1 \cong \rho_2$$ (where $$\sim$$ means conjugate and $$\cong$$ means isomorphic) but find a counter example. Consider the two representations $$\rho_1 : V_4 \to \operatorname{GL}_2(\mathbb{Z})$$ $$\rho_2 : V_4 \to \operatorname{GL}_2(\mathbb{Z})$$ where $$V_4 := \{1,a,b,ab\}$$ is the Klein-4 group and $$\rho_1(a)=\begin{bmatrix} 1&0\\0&-1\\ \end{bmatrix},\rho_2(a)=\begin{bmatrix} -1&0\\0&-1\\ \end{bmatrix}, \rho_1(b)=\rho_2(b)=\begin{bmatrix} -1&0\\0&1\\ \end{bmatrix}.$$ Clearly, $$\rho_1(V_4) = \rho_2(V_4)$$ so they are conjugate but one can check that $$\rho_1 \not \cong \rho_2.$$

Edit (2019/6/25): I reconsidered this exercise and found out that the pigeonhole principle should be applied twice. I write my proof as an answer below. Thanks to all comments.

This is the exercise 10

1. Let $$ρ_1 : G → GL(V )$$ $$ρ_2 : G → GL(V )$$ be two representations of $$G$$ on the same $$R$$-module $$V$$ that are injective as homomorphisms. (We say that such a representation is faithful.) Consider the three properties

(1) the $$RG$$-modules given by $$ρ_1$$ and $$ρ_2$$ are isomorphic,
(2) the subgroups $$ρ_1(G)$$ and $$ρ_2(G)$$ are conjugate in $$GL(V)$$,
(3) for some automorphism $$α \in Aut(G)$$ the representations $$ρ_1$$ and $$ρ_2α$$ are isomorphic.

Show that (1) ⇒ (2) and that (2) ⇒ (3). Show also that if $$α \in Aut(G)$$ is an inner automorphism (i.e. one of the form ‘conjugation by $$g$$’ for some $$g \in G$$) then $$ρ_1$$ and $$ρ_1α$$ are isomorphic.
• Is it true/helpful to note that if there were infinitely many isomorphism classes of representations, then by the pigeonhole principle there would have to be at least one conjugacy class of subgroups such that $\rho_n(G)$ and $\rho_2(G)$ are both in that conjugacy class, with $\rho_1$ and $\rho_1$ not being isomorphic? Jun 5, 2018 at 19:49
• @Acccumulation Thanks for your comment. Did you mean "...such that $ρ_1(G)$ and $ρ_2(G)$ are both in that conjugacy class, with $ρ_1$ and $ρ_2$ not being isomorphic"? Your deduction is right. However, the hard and important part is to show it is impossible (i.e. to show that "$ρ_1(G)$ and $ρ_2(G)$ conjugate" implies "$ρ_1$ and $ρ_2$ are isomorphic" in the specific case of Exercise 11. Such implication is not true in general.) Jun 6, 2018 at 5:11
• When the $R$-module $V$ is free of rank $n$ (which is the case here), $GL(V)\cong GL(n,R)$ so, the following modification of the Exercise 12.1 in Groups and Representations by Alperin & Bell holds: Let $R$ be a ring, $G$ be a finite group, $\rho_{1,2}:G\rightarrow GL(n,R)$ be two representations. Then, $\rho_1\cong\rho_2$ iff $\exists M\in GL(n,R)\ \forall g\in G\ \rho_1(g)=M^{-1}\rho_2(g)M$ But the last condition is stronger then saying $\rho_1\sim \rho_2$, so I guess this observation does not help for this exercise. Mar 28, 2019 at 21:14

We only consider faithful representations of $$G$$, a finite group, because any representation $$\rho$$ of $$G$$ can be regarded as a faithful representation of $$G/\operatorname{Ker}(\rho)$$. Denote by $$P$$ the complete set of representatives of isomorphism classes of faithful representations of $$G$$ on $$\mathbb{Z}^n$$. In other words, any faithful representation is isomorphic to a unique representation in $$P$$.
Now we prove by contradiction. Assume that $$P$$ is infinite. We define an equivalence relation $$\sim_1$$ on $$P$$ as $$\varphi \sim_1 \rho \iff \varphi(G) \sim \rho(G).$$ Since $$GL(n,\mathbb Z)$$ has only finitely many conjugacy classes of subgroups of order $$|G|$$, by pigeonhole principle there is an infinite equivelance class $$S$$ of $$P$$. Now let $$\rho \in S$$ be a fixed representation. By the implication $$(2) \Rightarrow (3)$$ in Exercise 10, any $$\varphi \in S$$ is isomorphic to $$\rho\alpha$$ for some $$\alpha \in \operatorname{Aut}(G)$$. We define an equivalence relation $$\sim_2$$ on $$S$$ as $$\rho_1 \sim_2 \rho_2 \iff \exists \beta \in \operatorname{Aut}(G), \quad \rho_1 \cong \rho\beta \cong \rho_2.$$ Obviously, there are no more than $$|\operatorname{Aut}(G)|$$ equivalence classes. Again, since $$\operatorname{Aut}(G)$$ is finite and $$S$$ is infinite, by pigeonhole principle there is an infinite equivalence class of $$S$$. It follows that there are at least two distinct but isomorphic representations in $$S \subset P$$, which is contradict to the definition of $$P$$. Therefore the assumption is false and $$P$$ is finite.