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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?2012-04-15
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    You can use [either *commutator subgroup* or *derived subgroup*](http://en.wikipedia.org/wiki/Commutator_subgroup). It looks like there is a mistake in part b), for the result is false unless you restrict yourself to degree 1 representations.2012-04-15
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    Yes you are right , the second part has to be restricted to degree 1 .2012-04-15

1 Answers 1

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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 ?2012-04-15
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    $K^*$ is the multiplicative group of the field $K$, i.e. all the non-zero elements of $K$. Adding another hint.2012-04-15
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    I can't very clearly see , but i suppose that it should be commutative. help me if i am being stupid!2012-04-15
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    If $\rho(a)$ and $\rho(b)$ commute, and $\rho$ is a homomorphism, what can you say about $\rho(aba^{-1}b^{-1})$?2012-04-15
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    it maps to the unity of K*, isn't it ?2012-04-15
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    Correct. So what can you say about all of $\rho(G')$?2012-04-15
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3105/discussion-between-vedananda-and-jyrki-lahtonen)2012-04-15
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    since G' is a commutator subgroup , it will also map to the unity of K* !2012-04-15
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    Correct again. As the next step you should think, whether $\rho$ factors via the quotient group $G/G'$. Sorry, can't chat now. I gotta go do some car maintenance and take missus shopping.2012-04-15
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    sure , i will go through it and see . catch you later . and thanks a lot .2012-04-15
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    Jyrki , can you help me to wrap up with the problem ? i can't still see much !2012-04-17
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    @Vedananda: A couple more hints :-)2012-04-17