If $H\cap K=e$ then $H$ and $K$ are normal For $|G|=p^km$ for $p$ is prime and $p$ does not divide $m$. Let $H=[x\in G \mid x^{p^k}=e]$ for $H<G$ and let $K=[x\mid x^m=e] $ for $K<G$. We are not assuming G is abelian
Show that $H\cap K=\{e\}.$ Then show $H$ and $K$ are normal.
Note that $H\cap K$ is a subgroup of $H$ and $K$. Hence, by Lagrange's theorem, the size of $H\cap K$ divides $p^k$ and m. $(p^k,m)=1$ thus $H\cap K=\{e\}$
Then how does this mean that H and K are normal?
 A: Note that, in general, $H$ and $K$ are not subgroups of $G$.
If you assume they are, then $H$ is a $p$-group and the order of any element in $K$ divides $m$. Since no power of $p$ divides $m$, an element in $H\cap K$ must have order $1$.
Recall that $(xyx^{-1})^r=xy^rx^{-1}$.

 If $h\in H$ and $g\in G$, then you have $h^{p^k}=1$ for some $k$ and so$$(ghg^{-1})^{p^k}=g(h^{p^k})g^{-1}=1$$so $ghg^{-1}\in H$.Can you do similarly for $K$?

A: First, note that $G$ is generated by $H$ and $K$: there are $p^km$ ways to write a product $hk$ for $h\in H, k\in K$. If $h_1k_1=h_2k_2$, then $k_1k_2^{-1}=h_1^{-1}h_2$. Since this element is in both $H$ and $K$, it equals $1$, so $h_1=h_2$ and $k_1=k_2$. Thus, the $p^km$ products are distinct and thus give all of $G$.
Now, let $k'\in K$, $g\in G$. We seek to show that $g^{-1}kg\in K$, which will prove $K$ is normal. We can write $g=hk, h\in H, k\in K$, by the above reasoning. Then $g^{-1}k'g=k^{-1}h^{-1}k'hk$. Note that $(k^{-1}h^{-1}k'hk)^m=k^{-1}h^{-1}k'^mhk=k^{-1}h^{-1}1hk$, since $k'\in K\implies k'^m=1$. But $k^{-1}h^{-1}1hk=1$. We have thus shown that $(g^{-1}k'g)^m=1$, so this element is in $K$, as desired. Thus, $K$ is normal.
By the same argument as above, $H$ is also normal, and in fact, $G\cong H\times K$.
