Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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, $(1,2)^{m+n}= \varphi(\langle \sigma\tau, m+n\rangle)= \varphi(\langle \sigma, m\rangle)\varphi(\langle \tau, n \rangle)= (1,2)^m(1,2)^n$.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.