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 $G_2$


  • $\begingroup$ 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. $\endgroup$ May 17, 2019 at 20:07
  • $\begingroup$ @ArturoMagidin Thanks. I have edited the post. Please have a look. $\endgroup$ May 17, 2019 at 20:10
  • 1
    $\begingroup$ Also note that normal subgroups are not very interesting here because the groups are abelian. (so any subgroup is normal) $\endgroup$
    – Mark
    May 17, 2019 at 20:15
  • $\begingroup$ Q1: Yes; Q2: Not really. $\endgroup$ May 17, 2019 at 20:16

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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