# Thread: kernel of group

1. ## kernel of group

Let G=Sym(n). Suppose that $\theta \in Hom(C, G)$, where C is a cyclic group. Then how to show that $\gamma (Ker \theta)=Ker \theta$ for $\gamma \in Aut(C).$

2. Originally Posted by dimuk

Let G=Sym(n). Suppose that $\theta \in Hom(C, G)$, where C is a cyclic group. Then how to show that $\gamma (Ker \theta)=Ker \theta$ for $\gamma \in Aut(C).$
the claim is true for any group G. let $C=$ and $\ker \theta = .$ let $\gamma(x)=x^r$ and $x = \gamma(x^s).$ then for any integer $k$:

$\subseteq: \ \ \theta(\gamma(x^{km}))=(\theta(x^{km}))^r = 1_G.$

$\supseteq: \ \ x^{km}=\gamma(x^{skm}) \in \gamma(\ker \theta).$

3. ## kernel of group

Thanks.

But I can't understand what you mean.

4. Originally Posted by dimuk
Thanks.

But I can't understand what you mean.
forget about that proof! it can be done much easier and in general case:

let $N=$ be any subgroup of $C=$ and $\gamma \in \text{Aut}(C).$ let $\gamma(x)=x^r.$ then $\gamma(x^{km})=x^{rkm} \in N.$ thus $\gamma(N) \subseteq N.$ replacing $\gamma$ with $\gamma^{-1}$ gives us $N \subseteq \gamma(N).$ thus $\gamma(N)=N.$