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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 group2012-08-27

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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 now2012-08-27
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    @Peter, you meant no *proper* (excluding dividing out by the trivial or whole group) non-abelian quotients?2012-08-27
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    Nicky, that's a more interesting question but one I suppose is difficult to answer. Do all groups have nontrivial normal subgroups?2012-08-27
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    @Auke see [simple groups](http://en.wikipedia.org/wiki/Simple_group).2012-08-27
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    @Auke, see Jacob's remark, if $G$ and $H$ would be non-abelian simple groups, then $G \times H$ *would* have non-abelian quotients ...2012-08-27