Two Normal Subgroups Intersecting Trivially Commute Each Other Problem 196

Let $G$ be a group. Assume that $H$ and $K$ are both normal subgroups of $G$ and $H \cap K=1$. Then for any elements $h \in H$ and $k\in K$, show that $hk=kh$. Add to solve later

Proof.

It suffices to show that $h^{-1}k^{-1}hk \in H \cap K$.
In fact, if this it true then we have $h^{-1}k^{-1}hk=1$, and thus $hk=kh$.

Since $h\in H$ and $H$ is a normal subgroup of $G$, we see that the conjugate $k^{-1}hk\in H$.
Thus we have
$h^{-1}k^{-1}hk =h^{-1}(k^{-1}hk)\in H. \tag{*}$

Also, since $k^{-1}\in K$ and $K$ is a normal subgroup of $G$, we have the conjugate $h^{-1}k^{-1}h\in K$.
Hence, we see that
$h^{-1}k^{-1}hk =(h^{-1}k^{-1}h)k\in K. \tag{**}$

From (*) and (**), we see that the element $h^{-1}k^{-1}hk$ is in both $H$ and $K$, hence in $H\cap K$ as claimed. Add to solve later

More from my site

You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory Abelian Normal Subgroup, Intersection, and Product of Groups

Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$. (That is,...

Close