Permutable subgroups are subnormal Let $G$ be a finite group. A subgroup $H \le G$ is called permutable, if for each other subgroup $U \le G$ we have $UH = HU$. Show that if $H \le G$ is permutable, then $H$ must be subnormal.
Does anybody knows an elementary proof of this fact?
 A: This should be completely elementary. I got the idea for this answer by looking in the book "Products of Finite Groups" (link to publisher page), pg. 12, Lemma 1.2.9. This might be a good reference for more material about permutable subgroups.
First, two small results.

(1) If $H$ is a subgroup of $G$ and $G = H^g H$, then $H = G$.

Proof: Now $g^{-1} \in H^g H = g^{-1}HgH$, so $1 \in HgH$ and thus $g \in H$.

(2) Suppose that $H \leq G$ is permutable. If $H \leq M$ for a maximal subgroup $M$, then $H \leq M^g$ for all $g \in G$.

Proof: If $H \not\leq M^g$, then $HM^g = M^gH = G$ since $H$ is permutable and $M^g$ is maximal. But this implies $G = M^gM$, which is a contradiction by (1).

Now for your result, proceed by induction on the order of $G$. Assume that $H$ is a proper permutable subgroup of $G$.  Now $H$ is contained in a maximal subgroup $M$, and by (2) the subgroup $H$ is contained in the proper normal subgroup $\cap M^g$. Apply induction to $\cap M^g$ and we are done.
A: Assume the contrary and let $G$ be minimal counter example,
Let $N=H^G$, if $N$ is proper in $G  $ by our assumtion $H$ is subnormal in $N$ and $N$ is normal in $G$ we are done.
So, assume $N=G$ ,
Let $M$ be a maximal subgroup containing $H$.
Since, $H^G=G$ there exist a $M^x$ such that $H$ is not contained in $M^x$. 
We have $HM^x=G\implies MM^x=G$ which is impossible.
In Mikko's answer, it seems to be better explained. 
