Find untrivial homomorphism : $\mathbb{Z}_2 \oplus \mathbb{Z}_2\rightarrow \mathbb{Z}_4 $ I have the following question :
Find untrivial homomorphism : $\mathbb{Z}_2 \oplus \mathbb{Z}_2\rightarrow \mathbb{Z}_4 $
I know that $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is not cyclic group since $gcd(2,2)=2\neq 1$, and I also know that $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is abelian group for all $a,b\in \mathbb{Z}_2$ we get $a+b=b+a$ (since operator add +).
Now, I need to find function $$f:\mathbb{Z}_2 \oplus \mathbb{Z}_2\rightarrow \mathbb{Z}_4 $$
Such that : $f(ab)=f(a)f(b)$ now I'm a bit confused since $f$ works on two variables so should I need to find a function $f$ so that $f(a,b)=f(ab)??$ Any ideas how to find such function, tips how to approach such questions?
Thank you!
 A: Hint: you won't be able to find a homomorphism of $\Bbb{Z}_2 \oplus \Bbb{Z}_2$ onto $\Bbb{Z}_4$ (as the two groups have the same number of elements and aren't isomorphic). So the homomorphism you want will factor through the inclusion of a  non-trivial subgroup of $\Bbb{Z}_4$ in $\Bbb{Z}_4$. $\Bbb{Z}_4$ only has one non-trivial subgroup, the subgroup $H = \langle[2]\rangle$ generated by the equivalence class of $2$, which is isomorphic to $\Bbb{Z}_2$. So you need to find a non-trivial homomorphism from $\Bbb{Z}_2 \oplus \Bbb{Z}_2$ to $\Bbb{Z}_2$ and compose it with the isomorphism defined by $[1] \mapsto [2]$ from $\Bbb{Z}_2$ to $H$.
A: You can also get all non trivial such homomorphism. Let $\varphi:\mathbb Z_2\oplus\mathbb Z_2\longrightarrow \mathbb Z_4$ a  non-trivial homomorphism $\varphi:\mathbb Z_2\oplus \mathbb Z_2\longrightarrow \mathbb Z_4$, i.e. $\text{Im}(\varphi)\neq\{0\}$. Therefore, $\text{Im}(\varphi)=\mathbb Z_2$, $\text{Im}(\varphi)=2\mathbb Z_2$ or $\text{Im}(\varphi)=\mathbb Z_4$. If $\text{Im}(\varphi)=\mathbb Z_4$, then $\ker(\varphi)=\{0\}$ and thus $\varphi$ is bijective which is impossible since those group are not isomorphic. If $\text{Im}(\varphi)=\mathbb Z_2$, in particular $$\mathbb Z_2\oplus\mathbb Z_2/\ker \varphi\cong \mathbb Z_2,$$
and thus $\ker(\varphi)\in \{\left<(1,1)\right>,\left<(1,0)\right>,\left<(0,1)\right>\}.$ But if $\ker (\varphi)=\left<(1,1)\right>$, then $$0=\varphi((1,1))=\varphi(0,1)+\varphi(1,0)=2$$
which is impossible in $\mathbb Z_4$. Therefore, you only have two such homomorphisms that are $$\varphi_1:(1,0)\longmapsto 1$$
and $$\varphi_2:(0,1)\longmapsto 1.$$
Since $\mathbb Z_2\cong 2\mathbb Z_2$, you get two other homomorphism that are $$\varphi_3: (1,0)\longmapsto 2$$
and$$\varphi_4:(0,1)\longmapsto 2,$$
and one more, the one where $\ker(\varphi)=\left<(1,1)\right>$ (here it's possible). Then, you also have $$\varphi_5:(1,0)\longmapsto 2,(0,1)\longmapsto 2.$$
