The question is:
Knowing that $\displaystyle f_{12}( g) \ =\ g^{12}$,
- Find a group $\displaystyle G$ non-abelian so that $\displaystyle f_{12}$ is a homomorphism and prove.
- Find $\displaystyle |ker\ f_{12} |$ and $\displaystyle |img\ f_{12} |$.
Tip: Langrange can help.
My intution:
I feel like the permutation group $\displaystyle S_{3}$ could help. I know $\displaystyle S_{3}$ is non-abelian and I can easily prove it.
To prove that $\displaystyle f_{12}$ is a homomorphism I would have to prove that for all $\displaystyle \sigma ,\ \tau \in S_{3}$:
$\displaystyle ( \sigma \tau )^{12} \ =\ \sigma ^{12} \tau ^{12}$.
I know lagrage stastes that for all $\displaystyle g\in G\ :\ o( g) \ |\ \ |G|$.
And since I know that $\displaystyle |G|\ =\ 3!\ =\ 6$. I can conlcude that the order of each $\displaystyle \sigma \in S_{3}$ is either $\displaystyle 1,2,3,6$. Since $\displaystyle 12$ is a multiple of each, by taking the power of $\displaystyle 12$, no matter which element, I will always get $\displaystyle e$. From here that:
$\displaystyle ( \sigma \tau )^{12} \ =e\ =\ ee\ =\ \sigma ^{12} \tau ^{12} \ $
And I prove homomorphism.
From the same logic, I would get $\displaystyle |ker\ f_{12} |\ =\ 6$. Since it's the whole group. Since, as I said, each element to the power of $\displaystyle 12\ $should be $\displaystyle e$.
From here also that $\displaystyle img\ f_{12} \ =\ e$ and $\displaystyle |img\ f_{12} |\ =\ 1$ . Since the only element you get by multipying to the power of $\displaystyle 12$, as I said, it's $\displaystyle e$.
Is my answer correct?