Are there general procedures to find group isomorphisms?

Question

This question is actually a series of related questions that have been bothering me while studying group theory. It seems to me that if we have two groups $G,H$ of same order and $G$ has a subgroup of order $n$ and $H$ has no subgroup of order $n$, then we can't have an isomorphism.

Now, suppose that we have two subgroups $G,H$ and they have the same quantity of subgroups $G_1,\dots,G_n$ and $H_1,\dots,H_n$ and $|G_i|=|H_i|$ does this means we have an isomorphism? Or are there groups where these conditions are true but there is no isomorphism?
I've been thinking about mappings from generating sets of groups, if we can map each generator of $G$ to another generator of $H$ such that each generator from $H$ and $G$ have the same order, do we have an isomorphism? Perhaps this is equivalent to the previous one but I am not really sure.

This is bothering me because in my lectures, we speak about what is an isomorphism but we never go in depth to actually talk about general procedures to find isomorphisms. Perhaps it's too early? I don't know.

For the first bullet item, the answer is no. A counterexample is mentioned here: https://math.stackexchange.com/a/1440984/169852 — , Feb 12 '21 at 23:05
For the second bullet point, the answer is no, too: take $G = C_2 \times C_2$ and $H = C_2$, then $G$ and $H$ are not isomorphic, but there is a group homomorphism of $G$ onto $H$ that maps generators of $G$ onto generators of $H$. — Rob Arthan, Feb 12 '21 at 23:11
To find isomorphisms, you construct them, you exhibit them explicitly, you write down a formula for them. — Lee Mosher, Feb 12 '21 at 23:18
Does this answer your question? Does the order, lattice of subgroups, and lattice of factor groups, uniquely determine a group up to isomorphism? — freakish, Feb 12 '21 at 23:28
@freakish Perhaps, but at the moment, I don't understand much of what they are saying there. — Red Banana, Feb 12 '21 at 23:44
I'm surprised to see the negative reception this question has garnered. The only issue I can think of is that it's asking multiple questions at once, but that seems minor to me. — Noah Schweber, Feb 13 '21 at 01:16
To the OP: note that while the question @Bungo cites doesn't require the subgroups to have matching sizes, the counterexample in the answer does, so it is indeed a counterexample to your first bulletpoint. — Noah Schweber, Feb 13 '21 at 01:18
@Billy Rubina the point is that two non isomorphic groups can have all proper subgroups pairwise isomorphic. In particular subgroups can be pairwise of the same order, while groups are non isomorphic. — freakish, Feb 13 '21 at 07:43

score 4 · Accepted Answer · answered Feb 13 '21 at 01:25

Determining whether two (finite) groups are isomorphic is quite difficult in general. There are lots of obstacles to isomorphism - basically, any non-silly property you can think of is preserved by isomorphisms, so any non-silly disagreement between two groups witnesses their non-isomorphicness - but it's difficult to guarantee an isomorphism without explicitly constructing one. (In particular, per the comments above neither of your conditions is sufficient to conclude that the two groups are isomorphic.)

That said, there are some tools we can use.

In the abelian case, group isomorphism is actually easy. The fundamental theorem of finite abelian groups gives a canonical way to describe any finite abelian group such that two finite abelian groups are isomorphic iff they have the same description. And the analysis required to get that description isn't very hard.
Even in the non-abelian case, we can show that a putative isomorphism between two groups $G$ and $H$ must satisfy certain constraints, and so narrow the range of possibilities we have to consider. For example, it has to send Sylow subgroups to Sylow subgroups appropriately, so breaking down the two groups via Sylow theory can make it easier to find an isomorphism if one exists (or prove that there is no isomorphism, for that matter).

Thanks for your answer. Do you know some groups which are tricky to prove they are isomorphic but understandable by an undergraduate? I'd like to see how complicated things could really get. — Red Banana, Feb 13 '21 at 01:37
@BillyRubina There are some surprising isomorphisms mentioned here. However, note that many of them concern infinite groups; that may or may not be desired. — Noah Schweber, Feb 13 '21 at 01:39
@BillyRubina The examples of groups you see in a textbook are somewhat misleading on how hard isomorphism is, and the ``bad'' examples are somewhat harder to describe. The paper https://arxiv.org/pdf/1010.5466.pdf constructs a large number of groups (as quotients of upper triangular matrices) that are virtually indistinguishable through invariants. — ahulpke, Feb 13 '21 at 17:36

Are there general procedures to find group isomorphisms?

1 Answers1