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Here is a simple question but I am trapped in solving the final part of it:

Show that $Z(A_4\times\mathbb Z_2)$ is characteristic subgroup of $A_4\times\mathbb Z_2$ but not a fully invariant subgroup.

I know that $Z(A_4\times\mathbb Z_2)=Z(A_4)\times Z(\mathbb Z_2)=1\times\mathbb Z_2\cong\mathbb Z_2$ and so for all $\phi\in Aut(A_4\times\mathbb Z_2); \phi(1\times\mathbb Z_2)=1\times\mathbb Z_2$. May I ask to notify me that magic endomorphism in second part of the question? Thanks.

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We'll need to map $\mathbb{Z}_2$ injectively to a 2-element subgroup of $A_4 \times 1$, since otherwise we'd be mapping to the identity, which is contained in $1\times \mathbb{Z}_2$.

We can avoid getting mixed up in any concerns about non-abelian groups by first applying $\pi_A$ and following in with any isomorphism of $\mathbb{Z}_2$ with a subgroup of $A_4$: a particular example $\varphi$ would send $\langle \sigma, n\rangle$ to $(1,2)^n$. You can verify it's a homomorphism either by composition or directly, $(1,2)^{m+n}= \varphi(\langle \sigma\tau, m+n\rangle)= \varphi(\langle \sigma, m\rangle)\varphi(\langle \tau, n \rangle)= (1,2)^m(1,2)^n$.

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    The easiest way to think of it is that your endomorphism should have the subgroup $A_4$ in its kernel. You can them map the central element of order 2 to any element of order 2 in the group.2012-09-22