Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.