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Claim : $H$ and $K$ are two normal subgroups of group $G$ such that $H\cap K = \{e\}$ then $ xy = yx $ for $x \in H$ and $y \in K$.

Proof : Let $x \in H$ and $y\in K$ then I need to prove that $xy =yx$. Let us assume that

$$y^{-1}xy \neq x$$ $y^{-1}xy = z$ where $z \in H$

Question : I am not getting how to proceed further?

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marked as duplicate by Jyrki Lahtonen group-theory Jul 8 '18 at 14:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Near duplicates 1, 2. The exact dupe is closed so I'm a bit hesitant to use my dupehammer. $\endgroup$ – Jyrki Lahtonen Jul 8 '18 at 10:12
  • $\begingroup$ We have used this result so many time in this site that it is surprising if it has not been handled explicitly many times. Therefore there is the urge to close it as a dupe of something :-) $\endgroup$ – Jyrki Lahtonen Jul 8 '18 at 10:15
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    $\begingroup$ @Jyrki Lahtonen I am disagree with your opinion regarding closing this question. $\endgroup$ – old Jul 8 '18 at 10:18
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    $\begingroup$ @JyrkiLahtonen Why? If it is not a duplicate of a non-closed and answered question, then the fact that it has been used a lot here is another reason not to close it as a duplicate. Then, if this question reappears again in the future (and I suppose it will), that new question will be closed as a duplicate of this one. $\endgroup$ – José Carlos Santos Jul 8 '18 at 10:19
  • $\begingroup$ @JyrkiLahtonen what about math.stackexchange.com/questions/253131/… $\endgroup$ – quid Jul 8 '18 at 13:07
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Note $x^{-1}(y^{-1}xy) = (x^{-1}y^{-1}x)y$. The LHS is in $H$ since $H$ is normal and is a subgroup. The RHS is in $K$ since $K$ is normal and is a subgroup. So, the common value must be $e$.

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  • $\begingroup$ We never used the finiteness of $G$, right? $\endgroup$ – Jack M Jul 8 '18 at 10:34
  • $\begingroup$ I don't think anyone said $G$ was finite. But yea we didn't need to assume $G$ is finite for this argument. $\endgroup$ – mathworker21 Jul 8 '18 at 11:08
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Let $\pi:G\to G/H$ be the projection. Then $\pi(x)=\pi(x^{-1})=e$ so $\pi(xyx^{-1}y^{-1})=\pi(y)\pi(y^{-1})=e$, so $xyx^{-1}y^{-1}\in H$. Similarly $xyx^{-1}y^{-1}\in K$....

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You started out well. From

$$yxy^{-1}=x$$

notice that on the left you have a conjugation, which smells like a good idea to me, since one of the definitions of a normal subgroup is that it's closed under conjugation. So the left hand side is in $H$. Now continue:

$$(yxy^{-1})x^{-1}=e$$

The left hand side is a product of two elements of $H$ and so is in $H$. If we could show it's in $K$ as well, then we'd be done. But in fact, all we have to do is shuffle the parentheses around to get $y(xy^{-1}x^{-1})$ and see that the left hand side is indeed in $K$, for exactly the same reason it's in $H$. Now you can check that each of the steps in our reasoning is reversible to get back to the original equation you wanted to prove, $yx=xy$.

Another, maybe more direct way is to just calculate the "difference" between $xy$ and $yx$ and check that it's "zero" (identity). So we try to calculate $xy(yx)^{-1}$ and show it equals $e$. You should find that the reasoning is almost exactly the same.

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