If $A \le G$ is s.t. $ \forall H \le G$, $AH = HA$ (i.e. $A$ permutes in $G$), then show $AB$ permutes in $G$ where $A^g=B \not= A$ for some $g\in G$.

Question

If $A \le G$ is such that $ \forall H \le G$, $AH = HA$ (i.e. $A$ permutes in $G$), then show $AB$ permutes in $G$ where $A^g=B \not= A$ for some $g\in G$.

What I have tried so far is:

$(AB)H = ((AB)H)^{g^{-1}g} = ((AA^g)H)^{g^{-1}g} = ((A^gA)H)^{g^{-1}g} = (A^g(AH))^{g^{-1}g} = (A(AH)^{g^{-1}})^g$

$=((AH)^{g^{-1}}A)^g=(AH)A^g=(HA)A^g=H(AA^g) = H(AB)$.

Is this right?

Answer 1

The solutions seems OK, but it could be made even shorter. For example:

$$(AB)H = A(BH) = (BH)A = gAg^{-1}HA = gA(g^{-1}Hg)g^{-1}A = g(g^{-1}Hg)Ag^{-1}A = H(gAa^{-1})A = HBA = H(AB)$$

In fact the shortest way is to prove that $B$ commutes too, which can be done by:

$$BH = gAg^{-1}H = gA(g^{-1}Hg)g^{-1} = gg^{-1}H(gAg^{-1}) = HB$$

And the result follows immediatelly.