1

Let $G$, $H$, and $K$ be groups. Under what conditions does $G\times K\cong H\times K$ imply $G\cong H$?

This can easily be shown to be true when they are finitely generated abelian groups, and we can find counterexamples when $K$ is not finitely generated ($\{0\}\times\mathbb{R}\cong\mathbb{R}\times\mathbb{R}$ but $\{0\}\ncong\mathbb{R}$), but under what other circumstances is it true or false?

Anonymous
  • 2,630
  • 8
  • 27
  • 1
    See previous related questions, e.g., https://math.stackexchange.com/questions/349826/counterexample-g-times-k-cong-h-times-k-implies-g-cong-h and https://math.stackexchange.com/questions/835731/g-times-h-cong-g-times-k-then-k-cong-h and https://math.stackexchange.com/questions/1042904/cancellation-of-direct-product-in-grp and https://math.stackexchange.com/questions/2389376/is-there-a-non-trivial-group-c-such-that-if-ac-cong-bc-then-a-cong-b – Gerry Myerson May 06 '18 at 06:09
  • 2
    Related questions, perhaps, but not the same, since this question asks "under what conditions..." which is neither asked nor answered in those threads. – C Monsour May 06 '18 at 06:12
  • 2
    And if you keep editing your comment as you find more old threads you aren't being helpful, Gerry, especially if you don't delete the irrelevant links you are posting. – C Monsour May 06 '18 at 06:13
  • Thanks, Gerry. I've looked at some of those questions already. My question is more about the specific conditions under which the implication is true or false. – Anonymous May 07 '18 at 11:14
  • You could have saved me some work, Julian, if you had mentioned what you had already done when you posted your question. – Gerry Myerson May 07 '18 at 12:58
  • I never said they were the same, @CMonsour, and I'll edit and post as I see fit. Even if you and Julian aren't interested in the older related questions, someone who stumbles across this question in the future may well be, and I've done her the service of gathering a fair number of them in one place. – Gerry Myerson May 07 '18 at 13:04

1 Answers1

2

It is true whenever $G\times K$ has a composition series. (It does not need to be abelian.) It is also true whenever it has an $X$-composition series for operators $X$. (In particular, it is also true under the weaker condition that $G\times K$ has a chief series.)

This is a consequence of the Krull-Schmidt theorem for operator groups.

C Monsour
  • 8,228
  • Interesting. I'll have to go back and reread about the Krull-Schmidt Theorem. – Anonymous May 06 '18 at 06:08
  • You'll want to be careful which discussion of the Krull-Schmidt theorem you use, since it is often stated only for modules (where the proof is easier). What you want is a full statement and proof for groups with operators, e.g., the one in Chapter 10 of Isaacs' Algebra: A Graduate Course. – C Monsour May 06 '18 at 06:22
  • I'll keep that in mind. The book I have right now is Hungerford and the section on the Krull-Schmidt Theorem comes a couple chapters before modules, so it should be okay. – Anonymous May 07 '18 at 11:15