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Let $H$ and $K$ be subgroups of a group $G$. Suppose $H=A\times B$.

Does it follow that $H\cap K=(A\cap K)\times (B\cap K)$?

I'm having a hard time trying to prove that $H\cap K\le (A\cap K).(B\cap K)$.

Thanks, Robert.

EDIT: The question stated in this form has negative answer. I should have added the assumption that $A\cap K$ in non-trivial!

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Why not try and think of a small counterexample? – Derek Holt Nov 2 '11 at 23:42
There are counterexamples using vector spaces and subspaces, for example, which you can even draw! – Mariano Suárez-Alvarez Nov 2 '11 at 23:45
up vote 2 down vote accepted

$\DeclareMathOperator{\Z}{\mathbf{Z}}$Say $H = \Z/4\Z \times \Z/2\Z$. Let $K$ be the subgroup of this generated by $(1, 1)$, i.e. $$ K = \{(1, 1), (2, 0), (3, 1), (0, 0)\}. $$ Then $K \cap (\Z/4\Z \times \{0\}$) is a cyclic group of order $2$, but the desired statement clearly fails here.

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Shame on me! Thanks. Is there a simple example with $A\cap K$ non-trivial? This is actually the situation I have. – Robert Nov 3 '11 at 0:01
@Robert I'll think about it. Will remove this, in the meantime. – Dylan Moreland Nov 3 '11 at 0:34
@Robert There, I think this works. – Dylan Moreland Nov 3 '11 at 0:51
It does! Thanks. – Robert Nov 3 '11 at 0:55

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