Are Subgroups' Quotients also Quotients' Subgroups?

Question

Let $G$ be a group, $G'\le G$ a subgroup, and $\varphi:G'\to H$ a surjective group homomorphism. Must there exist a surjective group homomorphism $\psi:G\to H'$ such that $H'\ge H$?

I know the converse is true: Take $G'=\psi^{-1}(H)$ and let $\varphi=\psi$.

I know if $G$ is abelian then the question is true: Take $H'=G/\ker(\varphi)$.

Whether the full direction asked above is true or false is unclear. I have asked some friends for their help and we couldn't reach a conclusion.

If anyone can point to a reference with a full solution to the problem, that is also great. Thanks in advance!

Answer 1

Your question is basically: does every subquotient of $G$ embed into a quotient of $G$?

In other words, if $K \trianglelefteq H \leq G$, does there exist $N \trianglelefteq G$ such that $H/K$ is isomorphic to a subgroup $G/N$?

The answer is no in general, but I cannot think of any examples which would be easy to describe. A computer calculation shows that smallest finite examples occur for $|G| = 64$, for example SmallGroup(64,18) has SmallGroup(16,13) as a subquotient, but not as a subgroup of a quotient.

In the case where $G$ is simple the question is whether every subquotient of $G$ is isomorphic to a subgroup of $G$. Here too you can find many examples. For example the Mathieu group $G = M_{24}$ has $A_6$ has a subquotient, but not as a subgroup.

There are probably easier examples, which perhaps others could suggest.

Answer 2

In their (absolutely correct) answer, spin writes:

The answer is no in general, but I cannot think of any examples which would be easy to describe.

In fact, there is a very beautiful construction of such examples due to Bergman (HSP$\not=$SHPS for metabelian groups, and related results). Recall that for a class of algebras (e.g. groups) $\mathbb{K}$, we let $\mathsf{H}(\mathbb{K}),\mathsf{S}(\mathbb{K})$, and $\mathsf{P}(\mathbb{K})$ denote the classes of homomorphic images of, subalgebras of, and products of families of algebras from $\mathbb{K}$ respectively, and that Birkhoff's HSP theorem says that $\mathsf{HSP}(\mathbb{K})$ is always the class of all algebras satisfying all equations true in all elements of $\mathbb{K}$.

Bergman's example is geometric. Consider the group of orientation-preserving isometries of the complex plane $$G=\{\lambda z. \alpha z+\beta: \alpha,\beta\in\mathbb{C}, \vert\alpha\vert=1\}.$$ For each prime $p$ let $G_p$ be the subgroup of $G$ consisting of elements such that $\alpha$ is a $p^n$th root of unity for some natural number $n$. Bergman shows that for distinct primes $p,q$ we have $$G_q\in\mathsf{HSP}(\{G_p\})\setminus \mathsf{SHPS}(\{G_p\}).$$ A fortiori, for a sufficiently large cardinal $\kappa$ we get that $G_q$ is a quotient of a subgroup of $(G_p)^\kappa$ but is not a subgroup of a quotient of $(G_p)^\kappa$.

(That $G_q\in \mathsf{HSP}(\{G_p\})$ is true for topological reasons: $G_p$ and $G_q$ are each dense in $G$, and so by looking at how the topology plays with the algebra we see that the three groups have the same equational theories. The harder part is showing that $G_q\not\in\mathsf{SHPS}(\{G_p\})$, and this takes some group theory.)

Interestingly, the $G_p$s are each metabelian groups; as you observe in the OP, no abelian groups will do the job, so this is algebraically as close-to-abelian as possible.

Answer 3

For some reason this old question came up in my feed, and I felt compelled to find an example with finite groups which could be proved "by pure thought".

Let $p$ be an odd prime which is $2 \bmod 3$. Since $p \equiv 2 \bmod 3$, the group $G = \mathrm{SL}_3(\mathbf{F}_p)$ is a simple group. I claim that it always has a subgroup $H$ with a non-trivial quotient which is not a subgroup of $G$.

The quotient will be given by the group $\Gamma = \mathrm{PSL}_2(\mathbf{Z}_p) \times C$ where $C$ is a non-trivial cyclic group of order dividing $p-1$ (and so prime to $3$ and $p$). Suppose there was an inclusion

$$\Gamma = \mathrm{PSL}_2(\mathbf{F}_p) \times C \rightarrow \mathrm{SL}_3(\mathbf{F}_p).$$

The image of a generator of $C$ will be semi-simple because it has order prime to $p$, and its eigenvalues will lie in $\mathbf{F}^{\times}_p$ because it has order dividing $p-1$, so up to conjugation it will be diagonal. Since $C$ has order prime to $3$, this diagonal matrix will not be scalar. But the image of $\mathrm{PSL}_2(\mathbf{Z}_p)$ has to commute with the image of $C$, and the commutator of a non-scalar diagonal matrix either consists of scalar matrices (if all eigenvalues are different), or decomposes into block form as $2 \times 2$ and $1 \times 1$ matrices. Since $\mathrm{PSL}_2(\mathbf{F}_p)$ is perfect (it is simple!), that means the map to the $1 \times 1$ block is trivial, and so there would have to be an injection

$$\mathrm{PSL}_2(\mathbf{F}_p) \rightarrow \mathrm{SL}_2(\mathbf{F}_p).$$

But there is no such injection, since the group on the right is also perfect and so has no (normal) subgroups of index $2$.

Now it suffices to show that $\Gamma$ is a quotient of a subgroup of $G$. There is an inclusion $\mathrm{GL}_2(\mathbf{F}_p) \subset \mathrm{SL}_3(\mathbf{F}_p)$ by sending $g$ to $g$ in the first $2 \times 2$ block and $\det(g)^{-1}$ in the next $1 \times 1$ block. Now let $H$ be the subgroup of $\mathrm{GL}_2(\mathbf{F}_p)$ of elements whose determinants are squres. This is just the group generated by $\mathrm{SL}_2(\mathbf{F}_p)$ together with the element

$$ \gamma = \left(\begin{matrix} \varepsilon & 0 \\ 0 & \varepsilon \end{matrix} \right)$$

of order $p-1$ where $\varepsilon \in \mathbf{F}^{\times}_p$ is a primitive root. But now the quotient of this group by the element $\mathrm{diag}(-1,-1) = \gamma^{(p-1)/2} \in \mathrm{SL}_2(\mathbf{F}_p)$ is just

$$\mathrm{PSL}_2(\mathbf{F}_p) \times C$$

where $|C| = (p-1)/2$, and we are done.