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Suppose I have two groups $G$ and $H$ with no non-abelian quotients. Then does $G \times H$ have no non-abelian quotients?

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Did you mean Abelian quotients? If $G$ and $H$ each have no non-trivial Abelian quotient group, then each is a perfect group, so $G \times H$ is a perfect group, and has no non-trivial Abelian quotient group –  Geoff Robinson Aug 27 '12 at 13:45
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Every group is a quotient of itself, so if $G$ and $H$ have only abelian quotients then in particular $G$ and $H$ are abelian, and so is $G \times H$. Since every quotient of an abelian group is again abelian, $G \times H$ has only abelian quotients.

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really sorry, for some reason I'd put the non in there by accident. I've edited the question now –  Peter Aug 27 '12 at 13:22
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@Peter, you meant no proper (excluding dividing out by the trivial or whole group) non-abelian quotients? –  Nicky Hekster Aug 27 '12 at 13:44
    
Nicky, that's a more interesting question but one I suppose is difficult to answer. Do all groups have nontrivial normal subgroups? –  akkkk Aug 27 '12 at 13:52
    
@Auke see simple groups. –  JSchlather Aug 27 '12 at 14:11
    
@Auke, see Jacob's remark, if $G$ and $H$ would be non-abelian simple groups, then $G \times H$ would have non-abelian quotients ... –  Nicky Hekster Aug 27 '12 at 14:38
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