# Extending homomorphism on subgroup to the whole group

I was working on homomorphisms and related concepts and I was wondering whether there exist some good criteria involving $$H$$ and $$G_{1}$$ which guarantees that all homomorphism $$\phi$$ on the subgroup $$H$$ of $$G_{1}$$ into an arbitrary group $$G_{2}$$ can be extended to a homomorphism $$\bar{\phi}$$ from all of $$G_{1}$$ to $$G_{2}$$.

And just to be clear: When I speak of an extension, I mean that I require that $$\bar{\phi}$$ restricted to H reduces to $$\phi$$.

​ My current conjecture is that this is possible for all homomorphism on $$H$$ if and only if there exists a normal subgroup $$N$$ of $$G_{1}$$ such that $$H$$ is isomorphic to $$G_{1}/N$$

The criteria is certainly sufficient since then $$G_{1}$$ is the semidirect product of H and N and by mapping all elements in N to the identity in $$G_{2}$$ a homomorphism is constructed. I've not been able to produce a proof that it is necessary also. Any ideas or arguments in favour or against the conjecture?

Edit: As stated in the answer below my criteria for it being a semidirect product is actually wrong. You need the canoncial projection h $$\mapsto$$ [h] to be an isomorphism. It then guarantees trivial intersection and G = NH

• This is the first thing I thought of but sadly it doesn't really work that way. Some products from outside H may map into H and outside of its kernel. Subgroups are sort of "semipermeable" Commented Jun 7, 2021 at 18:45
• Nevermind. Sorry. Commented Jun 7, 2021 at 18:46
• It is not necessary. For example, $G_2=G_1$ and every subgroup $H$ with $\operatorname{id}\colon H\to H\subset G_2$ extends to $G_1$. Commented Jun 7, 2021 at 18:51
• Are you looking for a criterion on $H$ and $G_1$ such that all $\phi$ with domain $H$ can be extended to $G_1$? I.e., characterize all $(H,G_1)$ such that the restriction $\operatorname{Hom}(G_1,\cdot)\to \operatorname{Hom}(H,\cdot)$ is always onto? Commented Jun 7, 2021 at 18:54
• In commutative algebra we have injective modules, which are the modules such that $X\rightarrow 0$ has the right lifting property against monomorphisms. This notion should adapt to the setting of groups giving other sufficient conditions (on $G_2$!) for your extension problem. This is of cause a tautology then, but maybe injective objects in groups are well studied and have different interesting characterizations? Commented Jun 7, 2021 at 18:57

Your conjecture is almost true. But you need more than just $$G/N$$ isomorphic to $$H$$. See below the line.

Explicitly:

The group $$G_1$$ and subgroup $$H$$ satisfy the stated property if and only if there exists a normal subgroup $$N$$ of $$G_1$$ such that $$G_1=NH$$ and $$N\cap H=\{e\}$$. That is, $$H$$ must be a retract of $$G_1$$ (h/t to Moishe Kohan).

Sufficiency follows as you indicate.

For necessity, as usual with this kind of statements, the key is to pick a particular (clever?) choice of $$G_2$$ and $$\phi$$ to force the desired conclusion.

Suppose $$H$$ and $$G_1$$, with $$H\leq G_1$$, has the property that for every group $$G_2$$ and every morphism $$\phi\colon H\to G_2$$ there exists a morphism $$\psi\colon G_1\to G_2$$ such that $$\psi|_{H} = \phi$$.

Take $$G_2=H$$, and $$\phi=\mathrm{id}_H$$. Then there exists $$\psi\colon G_1\to H$$ such that $$\psi|_{H}=\mathrm{id}_H$$. In particular, if we compose $$\iota\colon H\hookrightarrow G_1$$ with $$\psi$$, we get $$\psi\circ\iota\colon H\to H$$ and $$\psi\circ\iota(h) = \psi(h) = h$$ for all $$h\in H$$. That is, $$\psi$$ splits the embedding $$\iota\colon H\hookrightarrow G_1$$.

Let $$N=\mathrm{ker}(\psi)$$. Then $$N\cap H=\{e\}$$. Given $$g\in G_1$$, we have $$g(\iota(\psi(g))^{-1}\in N$$, since $$\psi(g)\psi(\iota(\psi(g)))^{-1} = \psi(g)\psi(g)^{-1}=e;$$ thus, $$g\in NH$$. Hence, $$G_1=NH$$, $$N\triangleleft G_1$$, and $$N\cap H=\{e\}$$. Therefore, $$G_1/N\cong H$$, as desired.

The error in your argument for sufficiency is that you are asserting that $$G/N\cong H$$ implies that $$G$$ is an internal semidirect product of $$N$$ by $$H$$. This is not true in general.

For example, take $$G=Q_8$$, the quaternion group of order $$8$$, and let $$H=\{1,-1\}$$. Then $$G$$ contains four different normal subgroups $$N$$ with $$G/N\cong H$$, but we know that $$G$$ is not a semidirect product. You cannot extend the identity map of $$H$$ to a homomorphism $$Q_8\to C_2$$; you can map $$Q_8$$ to $$C_2$$, but it won't extend the identity map of $$H$$ because $$H$$ is contained in all the nontrivial normal subgroups of $$Q_8$$. You cannot decompose $$Q_8$$ as an internal semidirect product.

So you need more than just $$G/N$$ isomorphic to $$H$$; you need that projection to split. That is, you need $$G$$ to be the internal semidirect product of $$N$$ by $$H$$.

• Incidentally, such a subgroup is called a retract. Commented Jun 7, 2021 at 19:41