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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...

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The homomorphism is determined by where $(0,1)$ and $(1,0)$ are sent, which can be chosen independently. Done. – blue May 5 '14 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 '14 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 '14 at 3:07
right, great tnx – tmpys May 5 '14 at 3:14

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.

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