TITLE: $X$-permutable subgroup
QUESTION [5 upvotes]: Let $A$ and $B$ be subgroup of a group $G$ and $\phi \neq X \subseteq‎ G$. $A$ is $X$-permutable with $B$ if $AB^{x}=B^{x}A$ for some $x$.      Let $A, B ,X$ be subgroups of a group $G$ and $K\lhd G$. $\qquad$ let $K\leq A$ If  $A/K$ is $XK/K$-permutable with $BK/K$ then $A$ is $X$-permutable with $B$ in $G$

REPLY [2 votes]: This is mostly a straightforward verification:
Suppose $(A/K)(BK/K)^{xK} = (BK/K)^{xK} (A/K)$. Written out, $$(A/K)(BK/K)^{xK} = \{ aK (bK)^{xK} : a \in A,b \in B \} = \{ a b^x K : a \in A, b \in B \}$$ so we get $\{ a b^x K : a \in A,b \in B  \} = \{ b^x a K : a \in A,b \in B \}$.
Now $ab^x K = (b')^x a' K$ implies $ab^x \in B^x A K$. Since this is true for all $a \in A, b \in B$, we get $AB^x K \subseteq B^x AK$. Reversing roles, we get the opposite containment, so $AB^x K = B^xAK$.
However $K$ is normal, so $B^xK = KB^x$, and $K \leq A$ so $AK=A$. Combining this we get $AB^xK = AKB^x = AB^x$ and $B^xAK = B^xA$, so we are left with $AB^x = B^xA$, as was to be shown.