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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?

  • 1
    See also this [question](http://math.stackexchange.com/questions/147530/normal-subgroups-that-intersect-trivially).2012-06-03

3 Answers 3

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Prove: A subgroup $\,H\leq G\,$ is normal iff $\,[H,G]:=\langle g^{-1}h^{-1}gh\,\,;\,\,g\in G\,,\,h\in H\rangle \leq H\,$ .

Apply this now to an element $\,[x,y]\in [H,K]\,$ ...

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    +1 for the answer, however, it would be better to elaborate it more for him.2012-06-03
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Try proving that the term $(xyx^{-1})y^{-1}= x (yx^{-1}y^{-1}) \in H \cap K =\{e\}$

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Sorry your proof is not correct. Just because $gHg^{-1}=H$ does not mean that it happens pointwise! Note $gxg^{-1}=x$ will be equivalent to $x$ and $g$ commuting. Since your $g$ is a generic element you've assume the group $G$ is commutative.

Several of the others have suggested better approaches :)