$\newcommand{iotaLift}{\iota\!\uparrow\!(S, H)}$ Say we have finite groups $S\leq G$ and $H$. If we have a homomorphism $\phi\colon S\to H$, it does not always have to be the case that we find a homomorphism from $G\to H$ that agrees with $\phi$ on $S$. For instance, the identity homomorphism $G \geq S\to S$ cannot be extended if (surjective) homomorphisms $G\to S$ don't even exist at all. This happens for instance if there is no $N\lhd G$ such that $S \simeq G/N$.
Definition Let $S, G, H$ be finite groups, $\iota\colon S\to G$ a monomorphism (that is, an injective group homomorphism). Define $$ \iotaLift := \{\phi\colon S\to H \mid \exists \overline\phi\colon G\to H: \overline\phi\circ \iota = \phi\} $$ to be the set of all such extendable homomorphisms.
- For which $H$ is $\iotaLift$ trivial (contains only the zero homomorphism)?
- For which $H$ is $\iotaLift = \operatorname{Hom}_{\underline{\operatorname{Grp}}}(S, H)$?
- Does this set have any obvious algebraic structure, e.g. a partial order, or an algebraic operation that constructs new extendable functions from old ones?
- Does the complement of this set?
(I know that the last two points are rather open-ended, and hope this is still acceptable for M.SE.)
Examples
- Assume the injection $\iota$ “splits”, i.e., admits an inverse $\sigma: \sigma\circ\iota = \operatorname{id}_S$. Then every $\phi$ can be extended to $\phi\circ\sigma$, which indeed satisfies $\phi\circ\sigma\circ\iota=\phi$. So $\iotaLift = \operatorname{Hom}_{\underline{\operatorname{Grp}}}(S, H)$.
- Similar to the case mentioned above, Let $G$ be simple, $S$ any subgroup, and $H$ a group such that $G$ does not embed into $H$. Therefore, the only homomorphisms $G\to H$ must be trivial, and so is every restriction onto $S$. Ergo, $\iotaLift=\{(s\mapsto 1)\}$.
Consider the (only nontrivial) embedding $C_2\to C_4$ with arbitrary $H$. Since morphisms $C_n\to H$ correspond with elements $h\in H$ of order dividing $n$, extendable morphisms $C_2\to H$ correspond to elements who have a square root (including the identity).
If we have an extendable function, we can compose it with any element of $\operatorname{Aut}(H)$ that is not trivial on its image to obtain a new function, whose extandability is easily checked.
