$G$ be a finite group and $G'$ be its commutator subgroup and $\widehat G$ be the character group of $G$ ; then $G/G' \cong \widehat G$?

Question

Let $G$ be a finite group and $G'$ be its commutator subgroup and $\widehat G$ be the character group of $G$ ; then is it true that $G/G' \cong \widehat G$ ? I am completely stuck , please help . thanks in advance

Answer 1

$Step$ $0$: $\widehat{\mathbb{Z_n}}$ $\cong \mathbb{Z_n}$ for all $n$.

$Step$ $1$: If $G\cong H\times K$ then $\widehat{G}\cong \widehat{H\times K}$.

This was discussed here The character group of $G$ for an abelian group $G$.

$Step$ $2$: If $G$ is finite abelian then $G\cong \widehat{G}$.

This follows by application of structure theorem on $G$ and $Step$ $0$ and $1$.

$Step$ $3$: $\widehat{G/G'} \cong \widehat{G}$.

For any $\chi$ in $\widehat{G}$ there is unique $\chi'$ in $\widehat{G/G'}$ (by universal property of quotient) such that $\chi'(\bar{g})$ $=$ $\chi(g)$. Then $\chi$ goes to $\chi'$ gives an isomorphism from $\hat{G}$ onto $\widehat{G/G'}$.

Now we know that $G/G'$ is abelian as $G'$ is the commutator subgroup and so $Step$ $2$ gives the result.