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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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    You mean that we search for an element in $A_4$ of order $\neq 2$?2012-09-22
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    Nope, define a map $\varphi$ from $A_4\times \mathbb{Z}_2$ to $A_4$ by $\varphi(\langle \sigma,n\rangle)=\sigma,$ where $\sigma \in A_4, n \in \mathbb{Z}_2$. It's unfortunate from a certain perspective if you haven't run into these maps yet. One can axiomatize products of two objects as objects with projections onto the objects and product maps going in-where a product map $\phi\times\psi:C\rightarrow A\times B$ takes $c$ to $\langle \phi(c),\psi(c)\rangle$, and can show that these requirements force us to pick groups isomorphic to the product as constructed "by hand."2012-09-22
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    You're looking for an endomorphism of $A_4\times \mathbb{Z}_2$ that does not fix $Z(A_4\times \mathbb{Z}_2)$, right? What's the problem?2012-09-22
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    He is looking for an endomorphism $\phi(1\times\mathbb Z_2)\not \subset 1\times\mathbb Z_2$2012-09-22
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    As you noted: Taking $$\phi:=A_4\times\mathbb Z_2\to A_4,\phi(\langle x,y\rangle)=x$$ we have $$\phi(\langle 1,y\rangle)=1$$ which $|y|=2$ and so $|\langle 1,y\rangle|=2$. A contradiction! Right?2012-09-22
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    It's true that the projection I initially suggested sends $1\times \mathbb{Z}_2$ into itself, and so I've proposed a different solution above. The problem isn't that $\phi$ sends an order-2 subgroup to identity, though, or I don't understand your point.2012-09-22
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    Thanks Kevin. My first comment came into my mind exactly for that you noted in your first paragraph in the answer. Sorry for misunderstanding. Thanks again. You got the my point correctly. :-)2012-09-22
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    Ah, yep, that's the problem I was pointing out in my first paragraph, exactly.2012-09-22
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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