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Finding all homomorphisms from $\mathbb{Z}_6$ to $\mathbb{Z}_2 \times \mathbb{Z}_3$.

Consider a homomorphism $\phi: \mathbb{Z}_6 \rightarrow \mathbb{Z}_2 \times \mathbb{Z}_3$. Since all homomorphisms maps the identity of the first group to that of the second, $\phi([0])=([0],[0]), \forall\phi$. This implies that $\phi$ is of the form $\phi ([x])=([mx],[nx])$. Therefore, $m\equiv 0$ or $1 \mod{2}$ and $n\equiv 0,1$ or $2 \mod{3}$. Therefore, there are $6$ homomorphisms between the two groups.

From here, my thought is that $\phi $ is an isomorphism $\iff$ $m\equiv 1 \mod{2}$ and $n\equiv 1$ or $2\mod{3}$. This implies that there are $2$ isomorphisms between the groups and $4$ homomorphisms that are not isomorphisms.

This is my first time doing a problem of this sort, so I would love some feedback.

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you are doing fine! note that $G=\mathbb{Z}_6$ is cyclic, so choose a generator, say $g$. a homomorphism from $G$ is entirely determined by the image of $g$.

in a homomorphism $g$ can be mapped to any element $h$ whose order divides the order of $g$. it is important to see why this restriction must be imposed.

in the case at hand the co-domain is $\mathbb{Z}_2 \times \mathbb{Z}_3$ so you can see that its elements must have order $1,2,3 $ or $6$. these orders all divide $6$ which is the order of $g$, the chosen generator of $\mathbb{Z}_6$. in consequence you may choose any element of $\mathbb{Z}_2 \times \mathbb{Z}_3$ to be the image of $g$, each choice generating a distinct homomorphism.

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    Thank you for the in depth answer! Building off of what you say, am I correct to put it that mapping $g$, the generator, to any element of order $6$, will produce an isomorphism? – 2017-02-19
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    yes indeed, i was going to leave that as a question for you, but you seem to be thinking for yourself! (+1) – 2017-02-19