Is this true for all groups

**dimuk** · September 30th 2008, 01:07 PM

Let H be a group and $\gamma \in Aut(H)$ . Then $\gamma (Ker \theta) =ker \theta$ for a group G and $\theta \in Hom(H, G)$ .

This is true if H is a cyclic group. I want to know whether is this true for any other finite groups.

**NonCommAlg** · September 30th 2008, 02:39 PM

Originally Posted by dimuk

Let H be a group and $\gamma \in Aut(H)$ . Then $\gamma (ker \theta) =ker \theta$ for a group G and $\theta \in Hom(H, G)$ .

the answer is no. let $N$ be any non-characteristic normal subgroup of a group $H.$ let $G=H/N$ and $\theta: H \longrightarrow H/N$ be the natural homomorphism. obviously $\ker \theta = N.$

since $N$ is not charactersitic, there exists $\gamma \in \text{Aut}(H)$ such that $\gamma(N) \neq N.$ thus $\gamma(\ker \theta) \neq \ker \theta. \ \ \Box$

**dimuk** · October 1st 2008, 12:08 PM

Then how about the finite abelian groups?

**ThePerfectHacker** · October 1st 2008, 04:52 PM

Originally Posted by dimuk

Then how about the finite abelian groups?

If $G$ is a finite abelian group then $G\simeq C_1\times ... \times C_k$ where $C_1,...,C_k$ are cyclic. You proved this result for cyclic groups. Now it remains to prove (or disprove) that if it is true for $H_1,H_2$ then it is true for $H_1\times H_2$ .

**NonCommAlg** · October 1st 2008, 07:33 PM

Originally Posted by dimuk

Then how about the finite abelian groups?

i don't know if you understood the idea in my previous post or not. as long as $H$ has a normal subgroup which is not characteristic, the claim in your problem would be false.

here's a simple example of finite abelian groups: let $H=\mathbb{Z}_2 \oplus \mathbb{Z}_2.$ define $\gamma: H \longrightarrow H$ by: $\gamma(a,b)=(b,a).$ then $\gamma \in \text{Aut}(H).$ let $N=\{(0,0),(1,0)\}$ and $G=\mathbb{Z}_2$ and define

$\theta: H \longrightarrow G$ by: $\theta(a,b)=b.$ clearly $\theta \in \text{Hom}(H,G),$ and $\ker \theta=\{(0,0),(1,0) \}=N \neq \{(0,0),(0,1)\}=\gamma(N)=\gamma(\ker \theta).$

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