# Number of homomorphisms from $\mathbb{Z}/4\mathbb{Z} \to \{\pm{1},\pm{i}\}$

I want to count the number of non-trivial group homomorphisms from $\mathbb{Z}/4\mathbb{Z} \to \{\pm{1},\pm{i}\}$.

• Since we want a homomorphism first thing is we want to map $\bar{0} \to 1$.

• Once the above is mapped we are left with elements $\bar{1},\bar{2},\bar{3}$. Now I can map $1 \mapsto i$, $2 \mapsto -1$ and $3 \mapsto -i$. This is one non trivial homomorphism.

• And by the same reasoning I can map $1 \to -i$, $2\mapsto -1$ and $3\mapsto i$.

So I have got 2 homomorphisms (infact they are isomorphisms as well). Are these the only ones. Is this the way one generally counts homomorphisms?

• Domain is generated by $1$ so once you know where $1$ goes you know the homomorphism. – Gerry Myerson Apr 27 '13 at 5:06
• You will need to examine mappings that are not necessarily one-to-one. – André Nicolas Apr 27 '13 at 5:09

You’re doing fine so far, but you’re not done: you can also map $\overline1\mapsto -1$, producing a homomorphism with kernel $\{\overline 0,\overline 2\}$.
Added: Since $\Bbb Z/4\Bbb Z$ is cyclic with generator $\overline 1$, every homormorphism $h:\Bbb Z/4\Bbb Z\to G$ is completely determined by $h(1)$. Thus, you need only ask yourself what the possible values of $h(\overline 1)$ are. Since the order of $\overline 1$ is $4$, $h(\overline 1)$ must have the property that $\left(h(\overline 1)\right)^4=1$, i.e., that the order of $h(\overline 1)$ divides $4$. In this case that allows $h(\overline 1)$ to be any element of the target group, so there are four homomorphisms. (The remaining one is of course the trivial homomorphism collapsing all of $\Bbb Z/4\Bbb Z$ to the identity.)