Let $G,K$ be groups with $H\le G$. It is not always possible to extend a homomorphism $\sigma:H\to K$ to $G$

Question

This is Exercise 4.2 of Roman's, "Fundamentals of Group Theory: An Advanced Approach". According to this MSE search and this Approach0 search, it is new to MSE.

The Question:

Let $G$ and $K$ be groups with $H\le G$. Show that it is not always possible to extend a homomorphism $\sigma:H\to K$ to $G$.

Thoughts:

This looks, prima facie, like a job for the extreme cases, like $K$ (and hence $H$) being trivial, although I haven't found anything; the solution seems more elusive than that.

My second idea came in the form of supposing $G$ and $K$ are finite, then employing some trick - I don't know what yet - with the orders of the groups involved, like Lagrange's Theorem for the subgroup $H$ in $K$, aiming to get $\lvert \sigma'(G)\rvert\nmid \lvert K\rvert$ somehow, for some potential extension $\sigma':G\to K$. This seems feasible at first but, after a while, I have doubts.

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Besides, I'm not entirely sure whether the extension is from $H$ to $G$ or from $K$ to $G$ (if you see what I mean). The word "extension" is defined only once prior to the exercise ibid. but it is in an entirely different context. For completeness, here is that definition:

Let $G$ be a group. We refer to subgroups $H$ and $K$ of $G$ for which $H\le K$ as an extension.

Please help :)

Answer 1

The division-by-$2$ map $2\mathbb{Z}\rightarrow\mathbb{Z},\,k\mapsto k/2$ is a bona fide group homomorphism, but it cannot be extended to a group homomorphism $\mathbb{Z}\rightarrow\mathbb{Z}$, for such an extension $\tilde{f}$ would have to satisfy $$1=f(2)=\tilde{f}(2)=2\tilde{f}(1),$$ yet no such integer $\tilde{f}(1)$ exists.

Answer 2

The other commenter's example was better but here's one using simple groups. Assume $G$ is a non-cyclic simple group and let $n$ be the order of $\sigma \in G$ and take the natural isomorphism $f: \langle \sigma \rangle \to \mathbb{Z}/n\mathbb{Z}$. Then if $f$ extended to a homomorphism $F: G \to \mathbb{Z}/n\mathbb{Z}$, it would be an isomorphism since its kernel would have to be trivial by the simplicity of $G$, contradicting the fact that $G$ was not cyclic.