# Equivalent matrix representations

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 $1$ .
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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

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 $\rho=\rho'\circ p$.

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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