1

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?

Jack M
  • 28,518
  • 7
  • 71
  • 136
lovw
  • 129
  • Near duplicates 1, 2. The exact dupe is closed so I'm a bit hesitant to use my dupehammer. – Jyrki Lahtonen Jul 08 '18 at 10:12
  • 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 :-) – Jyrki Lahtonen Jul 08 '18 at 10:15
  • 2
    @Jyrki Lahtonen I am disagree with your opinion regarding closing this question. – old Jul 08 '18 at 10:18
  • 1
    @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. – José Carlos Santos Jul 08 '18 at 10:19
  • @JyrkiLahtonen what about https://math.stackexchange.com/questions/253131/h-1-h-2-unlhd-g-with-h-1-cap-h-2-1-g-prove-every-two-elements – quid Jul 08 '18 at 13:07
  • A great catch @quid. Using that. Anyone wanting to add a new answer can still do so at the dupe target (or at one of the listed alternatives). – Jyrki Lahtonen Jul 08 '18 at 14:26
  • @JoséCarlosSantos Understood. I appreciate your concern. Also, I have occasionally been closing dupes rather militantly, so use your vote when/if you disagree. I do think that quid found an ideal dupe target. – Jyrki Lahtonen Jul 08 '18 at 14:29
  • @JyrkiLahtonen I fully agree: that's an ideal dupe target. – José Carlos Santos Jul 08 '18 at 14:31

3 Answers3

5

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 $e$.

mathworker21
  • 35,247
  • 1
  • 34
  • 88
2

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 $xyx^{-1}y^{-1}\in K$....

Angina Seng
  • 161,540
1

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.

Jack M
  • 28,518
  • 7
  • 71
  • 136