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Let $G'$ be the derived subgroup of a finite group $G$.

We have a correspondence $\{\mathrm{reps \ of \ G/G'}\} \longleftrightarrow \{\mathrm{reps \ of \ G \ with \ kernel \ containing \ G' }\} $

If we restrict to 1-dimensional reps, we get:

$\{\mathrm{1\ dimensional \ reps \ of \ G/G'}\} \longleftrightarrow \{\mathrm{1 \ dimensional \ reps \ of \ G \ with \ kernel \ containing \ G' }\} $

Now my notes say that there are $|G/G'|$ 1-dimensional reps of $G$. Since there are $|G/G'|$ 1-dimensional reps of $G/G'$, this must mean that all 1-dimensional reps of $G$ have kernel containing $G'$. Why is this so?

Thanks

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This is being discussed in a thread near you:-). This is almost a duplicate of that question, but your title is asking more, so not voting to close at this time. –  Jyrki Lahtonen Apr 17 '12 at 20:44

1 Answer 1

up vote 4 down vote accepted

The derived group $G^\prime$ is generated by the commutators, i.e. the elements of the form $ghg^{-1}h^{-1}$.

A 1-dimensional representation is a character, i.e. an homomorphism $$ \rho:G\longrightarrow\Bbb C^\times. $$ Since $\Bbb C^\times$ is abelian, $G^\prime<\ker(\rho)$.

Also, $G/G^\prime$ is abelian, so the number of its characters coincides with the number of its elements.

Putting all things together, $G$ has $|G/G^\prime|$ one dimensional representations.

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