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 $K$ be a field, $G$ a group and $G'=[G,G]$ the commutator subgroup of $G$. Show

  1. Two matrix representations of $G$ over $K$ of degree $1$ are equivalent only if they are identical.
  2. The group $G$ and the factor group $G/G'$ have the same number of matrix representations over $K$ of degree $1$ .
share|cite|improve this question
This is a little hard to follow but what I'm guessing you are trying to get at is the fact that the one-dimensional irreps of $G$ are in bijective correspondence with irreps of $G^\text{ab}$. What have you tried? – Alex Youcis Apr 15 '12 at 8:41
You can use either commutator subgroup or derived subgroup. It looks like there is a mistake in part b), for the result is false unless you restrict yourself to degree 1 representations. – Jyrki Lahtonen Apr 15 '12 at 8:44
Yes you are right , the second part has to be restricted to degree 1 . – Theorem Apr 15 '12 at 9:09
up vote 2 down vote accepted

Hint #1: $GL_1(K)\cong K^*$ is commutative.

Hint #2: If $\rho: G\to K^*$ is a group homomorphism, what can you say about $\rho(G')$ in light of the first hint?

Hint #3: Let $p:G\to G/G'$ be the projection homomorphism. Show that if $\rho_1'$ and $\rho_2'$ are two distinct representations of $G/G'$, both of degree 1, then $\rho_1=\rho_1'\circ p$ and $\rho_2=\rho_2'\circ p$ are two distinct representations of $G$, both of degree 1.

Hint #4: Show that Hint #2 implies that if $\rho$ is any representation of $G$ of degree 1, then there exists a degree 1 representation $\rho'$ of $G/G'$ such that $\rho=\rho'\circ p$.

share|cite|improve this answer
Thanks Jyrki, but that will only help me to see the answer to the first question , isn't it ? and btw i didn't get what your meaning to $K*$ is ? – Theorem Apr 15 '12 at 9:28
$K^*$ is the multiplicative group of the field $K$, i.e. all the non-zero elements of $K$. Adding another hint. – Jyrki Lahtonen Apr 15 '12 at 9:33
I can't very clearly see , but i suppose that it should be commutative. help me if i am being stupid! – Theorem Apr 15 '12 at 9:47
If $\rho(a)$ and $\rho(b)$ commute, and $\rho$ is a homomorphism, what can you say about $\rho(aba^{-1}b^{-1})$? – Jyrki Lahtonen Apr 15 '12 at 10:34
Correct. So what can you say about all of $\rho(G')$? – Jyrki Lahtonen Apr 15 '12 at 11:03

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.