Suppose I have two groups $G$ and $H$ with no non-abelian quotients. Then does $G \times H$ have no non-abelian quotients?
Products of groups with no non-abelian quotients
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group-theory
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0Did 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 – 2012-08-27
1 Answers
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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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0really sorry, for some reason I'd put the non in there by accident. I've edited the question now – 2012-08-27
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1@Peter, you meant no *proper* (excluding dividing out by the trivial or whole group) non-abelian quotients? – 2012-08-27
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0Nicky, 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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0@Auke see [simple groups](http://en.wikipedia.org/wiki/Simple_group). – 2012-08-27
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0@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