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Question:

Let $V$ be a finite dimensional vector space over a field $k$. Let $G$ be a finite group. Let $\varphi:G\to GL(V)$ be an irreducible representation of $G$. Suppose that $H$ is a finite abelian subgroup of $GL(V)$ such that $H$ is contained in the centralizer of $\varphi(G)$. Show that $H$ is cyclic.

Answer

First of all since $V$ is finite dimensional, we can consider $GL_n(k)$ instead of $GL(V)$. Now, I know that there is a corollary of Schur's lemma saying that if a representation $\varphi$ is irreducible if and only if every matrix $A$ satisfying $\varphi(g)A=A\varphi(g)$ for all $g\in G$ is of the form $\lambda I_n$. Thus, if $k$ were to be $\mathbb{C}$, then $H$ should consist of matrices of the form $\lambda I_n$. However, even in this case, I couldn't see why $H$ must be cyclic. For this case or the general, any help would be appreciated. Thanks in advance....

confused
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1 Answers1

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Let $D$ be the centralizer of $\phi(G)$ in $\mathrm{End}_k(V)$. As $V$ is irreducible, Schur's lemma implies that $D$ is a division algebra. Inside $D$ we can form $M$, the subalgebra generated by $k$ and $H$. As $H$ is abelian, and $D$ is finite dimensional over $k$, it follows that $M$ is a commutative finite dimensional $k$-algebra with no zero divisors. Such beasts are easily seen to be fields.

Therefore $H$ is a finite subgroup of the multiplicative group of a field, and hence must be cyclic.

Jyrki Lahtonen
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