3
$\begingroup$

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.

  • 1
    Just some clarification: are all of these groups subgroups of some larger group? Any assumption about normality?2012-04-01

3 Answers 3

6

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 $(A\cap C)(B\cap D)= \mathbf{1}\times\mathbf{1}\neq G\times G=AB\cap CD$.

  • 0
    Thank you very much for the counter example. And thanks for every one who tried to help.2012-04-01
3

When $A=D$ is not trivial, $B=C=\{e\}$ is a counterexample.

  • 0
    Thank you very much for the counter example. And thanks for every one $w$ho tried to help.2012-04-01
0

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}.

  • 0
    Thank you very much for the counter example. And thanks for every one who tried to help.2012-04-01