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Given $H$ and $K$ are normal subgroups in $G$ such that $H \bigcap K=${$1_G$}

SHOW: $xy = yx$ for all $x \in H$ and $y \in K$

This is what I have so far:

$x \in H$ and $y \in K$

$\therefore g_{1}xg_{1}^{-1}=x$

and $g_{2}yg_{2}^{-1}=y$

for $g_{1}, g_{2} \in G$

$\therefore xy=g_{1}xg_{1}^{-1}g_{2}yg_{2}^{-1}$

$=g_{2}yg_{2}^{-1}g_{1}xg_{1}^{-1}$

$=yx$

is this correct?

  • 3
    No, that's not correct. You've just asserted that $g_1xg_1^{-1}g_2yg_2^{-1}=g_2yg_2^{-1}g_1xg_1^{-1}$ without giving any justification at all.2012-06-03
  • 2
    Where does $xyx^{-1}y^{-1}$ live?2012-06-03
  • 0
    Hint: You want to prove that $xy=yx$, i.e. that $xyx^{-1}y^{-1}=1_G$. One of your hypotheses is of the form "if $g$ satisfies , then $g=1_G$". So try to show that $xyx^{-1}y^{-1}$ satisfies those conditions.2012-06-03
  • 1
    See also this [question](http://math.stackexchange.com/questions/147530/normal-subgroups-that-intersect-trivially).2012-06-03

3 Answers 3