If a group is formed by an extension of a central subgroup by a cyclic group generated by some generator (so the short exact sequence splits, with now being a subgroup of as well as a quotient), then and both commute with (as is central), and of course commutes with itself. Since and together generate the group , the entire group is then abelian.

]]>I am in my final year of my bachelors and I am doing my dissertation on subgroup lattices. While working through some proofs I have become stuck on results that use group extensions. In particular the book I am using often justifies a subgroup being abelian because it is “a cyclic extension of a central subgroup”.

I don’t really understand what this means (this post has been a big help!) or why it would lead to the subgroup being abelian.

Any help would be greatly appreciated, thank you for your time! ]]>

This result does not directly invoke Goursat’s lemma (which derives a stronger conclusion assuming stronger hypotheses) but the two results are in the same spirit and use similar tools in the proof.

]]>Thank you for your response!

I was desperate trying to someway connect the group extension problem with Goursat’s lemma without assuming my extension was a direct product, I did not came that far.

Thinking about the analogous problem with vector spaces helped!

For a K-by-H extension G I get a homomorphism f from G with Im(f)=H and Ker(f)=K. So for a subgroup G’ from G I call the intersection of K and G’ K’ and this is the kernel of f restricted to G’. For H’ I take the image of G’ under f.

I wrote it down and thought it through and it seems to work, assuming I did not make an embarrassing mistake.

But is it that simple? What does that have to do with Goursat’s lemma?

]]>What have you tried so far, and what did you get?

If you are more comfortable with linear algebra than with group theory, one can work with the analogous problem involving vector spaces over a fixed field (e.g. real vector spaces). If are vector spaces, a -by- vector space is a space which has a linear map onto with kernel . The assertion for groups has an exact analogue in this category (replacing “subgroup” by “subspace” of course), and you may find it easier to prove from your linear algebra intuition.

Another category where the analogous result is true is that of sets, although here the claim is somewhat degenerate and perhaps *too* easy to prove. If are disjoint sets, a -by- set would be a set that contains a subset such that the set-theoretic difference is equal to . (This notion is rather degenerate since there is precisely one such extension, namely .) The corresponding claim is then rather trivial, but it may still give you a bit of intuition, in particular with regards to how and are actually related to , , and .

at the moment I am working on my bachelor thesis with the topic group extensions. Your overview is great and an interesting result is that a subgroup of an K-by-H extension is an K’-by-H’ extension with K’ and H’ subgroups of K and H. You say the proof is essentially Goursat’s Lemma but I just can’t figure out how this should work. Perhaps it’s just missing experience and know-how in group theory…

Maybe you can start me off?

Thanks a lot!

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