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So we have $$\mathcal{Q}_8:=\{1,-1,i,-i,j,-j,k,-k\},$$ and I know all of its subgroups are $$\{\{1\},\{-1,1\},\{1,i,-i,-1\},\{1,j,-1,-j\},\{1,k,-1,-k\},\mathcal{Q}_8\}.$$ However, I am a bit puzzled on how to approach this question. All the homomorphisms will just be functions that map subgroups to subgroups right? Then, the isomorphisms are the functions that are homomorphisms from one subgroup to the other of the same order? So we just exclude those and get the final answer? Or am I barking up the wrong tree here?

OK, so let $\phi:\ \mathcal{Q}_8 \rightarrow \mathcal{Q}_8$ be a homomorphism. Then, by the first isomorphism theorem, we have that ker$(\phi)$ is a normal subgroup of $\mathcal{Q}_8$. We have that $\{-1,1\}$ is the centre of the group, so this is normal. The trivial group is also in the normal group and the indexes of the other subgroups are $2$, so they also are normal. Thus the kernel is just all of the subgroups of $\mathcal{Q}_8$.

The image is a subgroup of $\mathcal{Q}_8$ and so, the possible candidates for $im(\phi)$ is precisely the same as the list of candidates for the kernel.

But now I'm stumped and would appreciate a hint here.

thesmallprint
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    Homomorphisms do map subgroups to subgroups, but maps that map subgroups to subgroups need not be homomorphisms. I would check the definition of homomorphism if I were you. – Angina Seng Nov 21 '17 at 19:49
  • Just a remark: the isomorphisms from $Q_8$ to $Q_8$ are called automorphisms. We already know that ${\rm Aut}(Q_8)\cong S_4$, see here. – Dietrich Burde Nov 21 '17 at 19:50
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    Remember the isomorphism theorem: The image of a homomorphism is the quotient by the kernel. So you need to understand what possible subgroups and quotients there are of this group (not that many in common). – Tobias Kildetoft Nov 21 '17 at 19:52
  • Correction: the kernel must be one of the normal subgroups. Curiously, all those subgroups are normal... – Jyrki Lahtonen Nov 21 '17 at 21:05
  • Since $i$ and $j$ generate $\mathcal{Q}_8$, I would just examine the possible choices that $i$ and $j$ could be mapped to. While having to deal with 64 possibilities seems large, a lot of them can be dismissed fairly easily. In particular, if $f$ is a homomorphism and $\phi$ is an automorphism then $\phi \circ f$ is a homomorphism which will be isomorphic iff $f$ is. – Paul Sinclair Nov 22 '17 at 02:04

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