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Let G* be the set of all one dimensional representations i.e all group homomorphism from G to C* and G' denotes the commutator subgroup of a finite group G. I need to prove that G* is isomorphic to the quotient group G/G'.
I can say that for $\phi\in$G* then G' lies inside Ker$(\phi$) just by property of commutator subgroup but I don't know how to proceed further.
Kindly help!!
Thanks & regards in advance
You can use the following two isomorphisms (which you can try to prove):
$Hom(G,\mathbb{C}^\times)\simeq Hom(G/G',\mathbb{C}^\times)$ for any group $G$.
$Hom(A,\mathbb{C}^\times)\approx A$ for any finite abelian group $A$.
(The difference in notations for the isomorphisms is because the first one is canonical, and the second one is not.)
This question as stated is a bit trickier than it ought to be, as it combines two facts without telling you:
The group of linear characters $\text{Pic}(G)$ under $\otimes$ is an abelian group, with a canonical isomorphism to the group $\text{Hom}_{Ab}(G^{\text{ab}},\mathbb C^*)$.
For any finite abelian group $A$, the groups $A$ and $\text{Hom}_{Ab}(A,\mathbb C^*)$ happen to be isomorphic.
These facts, taken together show that the groups $\text{Pic}(G)$ and $G^{\text{ab}}$ are isomorphic, but you won't in general find a "natural" isomorphism between them.
As for proving these facts, the first is closer to the spirit of your question, and you should try using the universal property of the abelianisation, along with the fact that one dimensional representations of a group $G$ are group homomorphisms into abelian groups.
If you also wish to prove the second fact, it is more involved, and is generally proven by reducing to the case of a cyclic $p$ group via the classification of finite abelian groups, then verifying that the dual group is also cyclic of the same order.
