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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,

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  • $\begingroup$ 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
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    $\begingroup$ 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

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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.

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