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 $S\times \cdots \times S$.

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    $\begingroup$ I believe that your question was answered in the affirmative in this math.stackexchange post: math.stackexchange.com/questions/2193/… $\endgroup$
    – user1073
    Feb 9, 2017 at 15:50
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    $\begingroup$ I'm voting to close this question as off-topic because it's already been answered on MSE. $\endgroup$ Feb 9, 2017 at 15:57
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    $\begingroup$ I agree to close. $\endgroup$ Feb 9, 2017 at 15:58
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    $\begingroup$ Look at this easy answer. $\endgroup$
    – Watson
    Feb 9, 2017 at 16:35
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    $\begingroup$ @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: mathoverflow.net/a/26410/1593 $\endgroup$ Feb 9, 2017 at 18:52

1 Answer 1


Answer is yes.

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