1

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?

Asaf Karagila
  • 405,794
sarah jamal
  • 1,473

3 Answers3

2

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 $\,[x,y]\in [H,K]\,$ ...

DonAntonio
  • 214,715
2

Try proving that the term $(xyx^{-1})y^{-1}= x (yx^{-1}y^{-1}) \in H \cap K =\{e\}$

0

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 :)

N8tron
  • 2,537

  • "Since your g is a generic element you've assume the group G is commutative."

    Actually, it's not quite correct. x can belong to the center of G and G may not be equal to its center.

     – Nikita Hismatov Jan 22 '21 at 09:23