Two subgroups $H$ and $K$ of a finite group $G$ are $4$-permutable if and only if $HKHK= \langle H,K\rangle$ (where $\langle H,K\rangle$ is the smallest subgroup of $G$ containing both $H$ and $K$).
Is it true that if $H$ $4$-perm $K$, then $H$ $4$-perm $K^g$, for every $g\in G$ ?
What are the conditions under which $H$ $4$-perm $K$ implies $H$ $4$-perm $K^g$.
As for the first part, the answer is no. As a counter example we may take $H=\langle (1,2)\rangle$ and $K=\langle(1,2,3)\rangle$ as subgroups of the symmetric group $S_4$. Then: $HKHK=\langle H,K\rangle=S_3$. But if we take $g=(2,4)\in S_4$. Then $HK^gHK^g\neq\langle H,K^g\rangle$.
What could be deduced from this example.
Thank you.
