# The number of group homomorphisms from $\mathbb{Z_2}\times \mathbb{Z_2}\rightarrow \mathbb{Z}_4$

Find the number of group homomorphisms from $$G_1:=\mathbb{Z_2}\times \mathbb{Z_2}\rightarrow G_2:=\mathbb{Z}_4$$

Let $$a,b$$ be the generators of $$G_1$$.

Then $$0=f(0)=2f(0)$$ so we have $$f(a)\in \{0,2\}$$ and similarly $$f(b)\in \{0,2\}$$ so there are four possible homomorphisms.

Question 1: Am I right?

Question 2: Are there other ways to solve the problem. I mean for example using the knowledge of Normal Subgroups and the fact that Kernel is the normal subgroup of $$G_2$$

Thanks,

• Note: $\mathbb{Z}_4$ is an additive group; the element $1$ has order $4$, not two; the element $-1\equiv 3\pmod{4}$ has order $4$, not two. Modulo that, and specifying what "the generators of $G_1$" are (there are many choices, so you should specify yours), the idea is correct. May 17, 2019 at 20:07
• @ArturoMagidin Thanks. I have edited the post. Please have a look. May 17, 2019 at 20:10
• Also note that normal subgroups are not very interesting here because the groups are abelian. (so any subgroup is normal)
– Mark
May 17, 2019 at 20:15
• Q1: Yes; Q2: Not really. May 17, 2019 at 20:16

The other way to do it is for every subgroup $$G$$ of $$\mathbb Z_4$$ to count number of normal subgroups $$H$$ s.t. $$(\mathbb Z_2 \times \mathbb Z_2) / H \cong G$$, multiply it by number of automorphisms of $$G$$. In this case there are just $$3$$ variants for $$G$$ - $$\mathbb Z_4$$, $$\{0, 2\}$$ and $$\{0\}$$. The first isn't image of any homomorphism, the second requires $$H \cong \mathbb Z_2$$ (three variants) and has only one automorphism, the last requires $$H = \mathbb Z_2 \times \mathbb Z_2$$ (one variant) and has only one automorphism.