# Commutator subgroup and bijective representation

Let $G$ be a finite group and $G' = [G,G]$ be its commutator subgroup, which is defined to be the subgroup generated by elements $[g,h] = g^{-1}h^{-1}gh$ for all $g,h \in G$, where $G'$ is a normal subgroup of $G$. And $G/G'$ is abelian.

Prove that for any field $\mathbb{K}$, the degree $1$ representations of $G$ over $\mathbb{K}$ are in bijection with the degree $1$ representations of $G/G'$ over $\mathbb{K}$.

attempt: Suppose $\phi_1: G → GL(\mathbb{K})$ be a representation of degree 1. is similar to $\phi_1: G → \mathbb{K}^{\times}$.

And define $\phi_2 : G/G'→ \mathbb{K}^{\times}$

Do I have to show $\phi_1 → \phi_2$ and show it's a bijective function? I am not sure what I have to show. Can someone please help? Thank you!

• Isn't it just because $\Bbb K^{\times}$ is abelian, so the kernel of any homomorphism from $G$ to $\Bbb K^{\times}$ must contain the commutator subgroup? In other words you can always factor $G\to\Bbb K^{\times}$ through $G/G'$. That gives you the bijective map in one direction. And you can lift any homomorphism from $G/G'$ to $\Bbb K^{\times}$ to a homomorphism from $G$ to $\Bbb K^{\times}$. That gives you the other direction. Just show the composition is the identity, in both directions, to conclude it's a bijection. – Gregory Grant Feb 17 '16 at 23:27
• Could I define $\phi_2(gG') = \phi_1(g)$? And then show the composition is the identity ? – user40294 Feb 17 '16 at 23:40
• Yes, that gives you the way to take a map from $G/G'$ to $\Bbb K$ to get a map from $G$ to $\Bbb K$. You have to show it's well defined. – Gregory Grant Feb 18 '16 at 0:18

Show that for every abelian group $A$, $Hom(G,A)\cong Hom(G/G',A)$ (where $Hom(-,-)$ is the set of group homomorphisms). Since $\Bbb K^\times$ is abelian, you're done.
(I think it's important to see that for general $A$ since this is a fundamental property of $G/G'$).
• So I have to define $f : G → A$ , and $f_1 : G/G' → A$? And show it's a hormomorphism? Then show this is isomorphic? – user40294 Feb 17 '16 at 23:42
• Not quite. From a map $f:G\to A$, you have to construct a map $\bar{f}:G/G' \to A$ and vice-versa. Furthermore, you have to show that these constructions are inverse to each other. – Nitrogen Feb 17 '16 at 23:44