# If every irreducible representation of a finite group has dimension $1$, why must the group be abelian?

Suppose $G$ is a group and that every irreducible representation of $G$ has dimension $1$. Why does this mean that $G$ is abelian?

The number of $1$-dimensional representations of $G$ is given by $|G/G'|$, where $G'$ is the derived subgroup of $G$. So if every irreducible representation of $G$ has degree $1$, then the number of conjugacy classes of $G$ is equal to $|G/G'|$. I can't see how to conclude that $G$ is abelian (or if this is the right approach).

Any help appreciated. Thanks

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The regular representation is a direct sum of irreducible representations. If all of those representations are $1$-dimensional, then...? Alternately, recall that the sum of the squares of the dimensions of the irreducible representations is equal to $|G|$. – Qiaochu Yuan May 30 '12 at 22:55
@QiaochuYuan Got it. Then the regular representation has dimension equal to the number of conjugacy classes of $G$. But this must also be equal to the size of $G$. So each element is it's own conjugacy class, so $G$ is abelian. Thanks! – Rep May 30 '12 at 22:57
You don't even need to know that the number of conjugacy classes is the same as the number of irreducible representations. If the regular representation is a direct sum of $1$-dimensional representations, then every element of $G$ can be simultaneously diagonalized, and diagonal matrices commute with each other. – Qiaochu Yuan May 30 '12 at 22:58

The following steps lead to a solution. The argument I am giving here is essentially equivalent to the argument given by Qiaochu above (in the comments) but is more general in that it is also applicable for compact Lie groups. The main point in this connection is in the application of the Peter-Weyl theorem in the solution to Exercise 4 below for compact Lie groups.

Let $G$ be a group and let $G'$ denote the commutator subgroup of $G$. We wish to understand how $G'$ acts on one-dimensional representations of $G$:

Exercise 1: Let $(\pi,V)$ be a one-dimensional representation of $G$. Prove that the induced representation of $G'$ is trivial.

The following result is fundamental in representation theory and is applicable to finite groups or, more generally, compact Lie groups:

Exercise 2: Prove that every finite-dimensional representation of $G$ is a direct sum of irreducible representations.

We know that every irreducible representation of $G$ is one-dimensional by assumption. In particular, every finite-dimensional representation of $G$ is a direct sum of one-dimensional representations.

Exercise 3 Prove that $G'$ acts trivially on every finite-dimensional representation of $G$.

Let us recall that a representation $(\pi,V)$ of $G$ is said to be faithful if the kernel $\text{ker }\pi$ is the trivial subgroup of $G$. The following exercise might be a little difficult; you can assume it on faith if desired:

Exercise 4 Prove that if $G$ is a finite group or a compact Lie group, then $G$ possesses a finite-dimensional faithful representation. (Hint: If $G$ is finite, then you can prove this using a familiar representation of $G$. If $G$ is an arbitrary compact Lie group, then appeal to the Peter-Weyl theorem if you can.)

The result should now be clear:

Exercise 5: Prove that if every irreducible representation of $G$ is one-dimensional, then $G'=\{e\}$, the trivial subgroup of $G$, i.e., $G$ is abelian.

I hope this helps!

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