# The center of $A_4\times\mathbb Z_2$

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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Edited

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 $\neq 2$? – Babak S. Sep 22 '12 at 8:03
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." – Kevin Carlson Sep 22 '12 at 8:05
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? – Kevin Carlson Sep 22 '12 at 8:12
He is looking for an endomorphism $\phi(1\times\mathbb Z_2)\not \subset 1\times\mathbb Z_2$ – clark Sep 22 '12 at 8:21
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? – Babak S. Sep 22 '12 at 8:29