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!

  • 3
    $\begingroup$ 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. $\endgroup$ – Gregory Grant Feb 17 '16 at 23:27
  • $\begingroup$ Could I define $\phi_2(gG') = \phi_1(g)$? And then show the composition is the identity ? $\endgroup$ – user40294 Feb 17 '16 at 23:40
  • $\begingroup$ 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. $\endgroup$ – 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 $G/G'$).

  • $\begingroup$ It's kind of hard to understand the set of group homorphism. Does that mean the set of group homomorphism between G to A? And G/G' to A? $\endgroup$ – user40294 Feb 17 '16 at 23:38
  • $\begingroup$ Yes that's exactly that. $\endgroup$ – Nitrogen Feb 17 '16 at 23:39
  • $\begingroup$ 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? $\endgroup$ – user40294 Feb 17 '16 at 23:42
  • $\begingroup$ 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. $\endgroup$ – Nitrogen Feb 17 '16 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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