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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 $|G|$.2012-05-30
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    @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!2012-05-30
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    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.2012-05-30

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