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.

I am trying to figure out all the homomorphisms from $\mathbb{Z}_2\times\mathbb{Z}_2$ to $\mathbb{Z}_2$.

Is there a good process for doing such a think? I am getting lost...

share|improve this question
    
The homomorphism is determined by where $(0,1)$ and $(1,0)$ are sent, which can be chosen independently. Done. –  blue May 5 at 2:40
    
So we have 4 of them? $\\(1,0)\rightarrow 1, (0,1)\rightarrow 1\\$ and $\\(1,0)\rightarrow 1, (0,1)\rightarrow 0\\$ and $\\(1,0)\rightarrow 0, (0,1)\rightarrow 1\\$ and $\\(1,0)\rightarrow 1, (0,1)\rightarrow 0\\$? and –  tmpys May 5 at 3:00
    
You write $(1,0)\mapsto1,(0,1)\mapsto0$ twice. Your fourth should be $(1,0),(0,1)\mapsto 0$. –  blue May 5 at 3:07
    
right, great tnx –  tmpys May 5 at 3:14

1 Answer 1

Look at the possible kernels. What are the normal subgroups of $G=\mathbb{Z}_2\times \mathbb{Z}_2$?

$1$ and $G$ are always normal subgroups of any group $G$. There are three proper subgroups, generated by $(1,0)$, $(0,1)$, and $(1,1)$, respectively, and all are normal because $\mathbb{Z}_2\times\mathbb{Z}_2$ is abelian. Let's call those $K_1,K_2,$ and $K_3$, respectively.

What quotient groups do they yield?

For each $i$, $\mathbb{Z}_2\times\mathbb{Z}_2/K_i \cong \mathbb{Z}_2$ by simple order arguments. Of course, for any group $G$, $G/G$ is the trivial group, and $G/1=G$.

Which of these groups could fit into $\mathbb{Z}_2$ as images?

The $G/K_i$ and the $G/G$ do, and $G/1$ doesn't.

Now use the first isomorphism theorem to finish.

share|improve this answer

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.