Does $G\times H \cong G\times K$ imply $H\cong K$ in finite groups?

Question

The question is as in the title, i.e.

Does $G\times H \cong G\times K$ imply $H\cong K$, where $G,H,K$ are finite groups?

We can show that

(1) the abelianizations of $H$ and $K$ are isomorphic

(2) $H/M_S(H)\cong K/M_S(K)$, where $S$ is a non-abelian finite simple group and for a group $G$, $M_S(G)$ is the intersection of all normal subgroups $U$ with $G/U\cong S$; so $G/M_S(G)$ is the maximal quotient of the form $S\times \cdots \times S$.

I believe that your question was answered in the affirmative in this math.stackexchange post: http://math.stackexchange.com/questions/2193/cancellation-of-direct-products — , Feb 09 '17 at 15:50
I'm voting to close this question as off-topic because it's already been answered on MSE. — Jeremy Rouse, Feb 09 '17 at 15:57
Of related interest is http://mathoverflow.net/a/48781, although this is for relational structures. I suspect finite groups also cancel. Gerhard "Still Amazed After Many Years" Paseman, 2017.02.09. — Gerhard Paseman, Feb 09 '17 at 15:58
@Everyone: As a loyal fan of MO, I feel obliged to point out that this question was answered on MO two or three months before it was even asked on MSE: http://mathoverflow.net/a/26410/1593 — José Hdz. Stgo., Feb 09 '17 at 18:52

score 2 · Answer 1 · edited Apr 13 '17 at 12:19

2

Answer is yes.

Hope you can find interesting this too.

edited Apr 13 '17 at 12:19

Community

1

answered Feb 09 '17 at 16:09

Ivan Di Liberti

8,176

Does $G\times H \cong G\times K$ imply $H\cong K$ in finite groups?

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