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In set theory we have $(A\times B)\cap {(C\times D)}=(A\cap C)\times(B\cap D)$, is this true in group theory if we replace the $\times$ by products of subgroups i.e.
$$(AB)\cap (CD)=(A\cap C)(B\cap D)$$ I know that $(A\cap C)(B\cap D)$ is contained in $AB\cap CD$ but I could not prove the other inclusion.

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Just some clarification: are all of these groups subgroups of some larger group? Any assumption about normality? –  you Apr 1 '12 at 16:00
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up vote 5 down vote accepted

No. Take some non trivial group $G$ and consider the product group $G\times G$. Set $A=D=\mathbf{1}\times G$ and $B=C=G\times \mathbf{1}$ where $\mathbf{1}\in G$ is the identity element. Then $AB=G\times G=CD$ and $A\cap C=B\cap D= \mathbf{1}$ so $(A\cap C)(B\cap D)= \mathbf{1}\times\mathbf{1}\neq G\times G=AB\cap CD$.

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Thank you very much for the counter example. And thanks for every one who tried to help. –  user28083 Apr 1 '12 at 16:29
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When $A=D$ is not trivial, $B=C=\{e\}$ is a counterexample.

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Thank you very much for the counter example. And thanks for every one who tried to help. –  user28083 Apr 1 '12 at 18:01
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I don't think (AB)∩(CD)=(A∩C)(B∩D) is true.Consider dihedral group D12 generated by a and t such that a^6=1, t^2=1, atat=1. The subgroups A,B,C,D of D12, in which A={1,t},B={1,ta^5},C={1,a^2,a^4},,D={1,a^3}, contradict to (AB)∩(CD)=(A∩C)(B∩D), because (AB)∩(CD)={1,a^5} and (A∩C)(B∩D)={1}.

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Thank you very much for the counter example. And thanks for every one who tried to help. –  user28083 Apr 1 '12 at 18:00
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