$H_1 \cong H_2$ doesn't mean $G / H_1 \cong G / H_2$!
11 Oct 2020  Tags: algebraicmisconceptions
I spend a lot of time on math stackexchange (mse), and I periodically see “simple” questions that totally shatter a misconception that I didn’t know I even held.
My first time experiencing this was when I learned that normal subgroups need not be transitive. That is,
If $N \vartriangleleft H$ and $H \vartriangleleft G$, it is not true (in general) that $N \vartriangleleft G$.
For cultural growth, we do have transitivity whenever $N$ is characteristically normal in $H$. Symbolically, $N \vartriangleleft^c H$.
In fact, not only is $\vartriangleleft^c$ transitive, we recover some amount of transitivity with the regular “normal subgroup” relation too. That is, in addition to transitivity
\[N \vartriangleleft^c H \vartriangleleft^c G \implies N \vartriangleleft^c G\]we also have a “weak transitivity”
\[N \vartriangleleft^c H \vartriangleleft G \implies N \vartriangleleft G\]I forget exactly when I first saw that normal subgroupness isn’t transitive, but I remember it was shocking. Even though in hindsight it is obvious how normalcy fails to be transitive, it was never something I explicitly saw. Moreover, the notation is so suggestive of a transitive relation, it felt more “obvious” than it probably should have. I had this realiziation long enough ago that I no longer have the mse post where I first saw it. However, I have had two similar realizations in recent memory, and I figured I would include them here as quick posts analogous to my “quick analysis trick” series.
The first fact is that simple groups need not be hereditary. That is, if $G$ is simple and $H \leq G$, it is not necessarily true that $H$ is simple.
This misconception is interesting to me because I think if you asked me “are simple groups hereditary?” I would have probably answered “I don’t think so?” as a gut reaction. Giving myself the benefit of the doubt, I’ll say I might even have come up with a counterexample after a minute or two.
Even though I think I would have said “no” when asked, when I first saw this written out it struck me as somewhat surprising. At the very least, it struck me as a mistake a younger me might have made. Even knowing it is false, though, I don’t think I would not have guessed how badly it fails! Again, I’m afraid I’ve lost the mse link to the question where I learned about this, but I remember the idea:
Every finite group embeds into some alternating group $A_n$.
We know by Cayley’s Theorem that every $G$ embeds into the symmetric group \(\mathfrak{S}_{G}\). Then it remains to show \(\mathfrak{S}_n\) embeds into the alternating group $A_{n+2}$. But we can do this by the map (which one easily checks to be an embedding)
\[f(\sigma) = \begin{cases} \sigma & \text{when } \sigma \text{ is even}\\ \sigma (n+1 \quad n+2) & \text{when } \sigma \text{ is odd} \end{cases}\]Then every $G$ embeds into \(A_{G+2}\). Pairing this with the fact^{1} that we can find a finite group $G_n$ so that every group of order $\leq n$ embeds in $G_n$, we see there is an alternating group $A_N$ (with $N \gg n$) which contains every group of order at most $n$ as a subgroup!
Once we know that every finite group embeds into some alternating group, it is natural to ask the same of other families of finite simple groups. This question has been asked before, and there are references on that post which show that some natural families of simple groups don’t have this property (such as the Suzuki Groups), but I don’t know enough about the classification to properly think about this myself. Someday, though!
The second fact is one that I saw today, in this question.
There are abelian groups $G$ with isomorphic subgroups $H_1 \cong H_2 \leq G$ such that $G/H_1 \not \cong G/H_2$.
This caught me super off guard. Unlike the last one, my gut reaction would be that whenever $H_1 \cong H_2$ we must have $G/H_1 \cong G/H_2$. It still feels weird to me, but at least I’m aware of it, so now I can start properly internalizing it. An example of this phenomenon was given by Andreas Blass in the comments of the post above.
\[G = (\mathbb{Z}/2) \times (\mathbb{Z}/4)\] \[H_1 = (\mathbb{Z}/2 \times \{ 0 \})\] \[H_2 = (\{ 0 \} \times 2 \mathbb{Z}/4)\]Here we have $H_1 \cong \mathbb{Z}/2 \cong H_2$, while $G/H_1 \cong \mathbb{Z}/4$ and $G/H_2 \cong \mathbb{Z}/2 \times \mathbb{Z}/2$.
One natural question is “when does $G/H_1 \cong G/H_2$”? This is explored in a different mse post. The asker notices that it is enough to have some automorphism $\varphi : G \to G$ so that $\varphi(H_1) = H_2$. Unfortunately for those looking for a clean characterization, the answer on that question shows this is not enough. However, this can also be taken as a comfort. There are lots of examples of nonautomorphic subgroups which still have $G/H_1 \cong G/H_2$, so perhaps it’s not so surprising that I found this surprising.

Enumerate the (finitely many) groups of order $\leq n$ (up to isomorphism) as $H_i$. Then we can simply product them all together to get a (rather large) finite group $G_n = \prod H_i$. This is obviously inefficient, but it seems like not much is known about how small we can make $G_n$. ↩