TITLE: Is this correct? $rH=Hr^{-1}$
QUESTION [6 upvotes]: Is this reasoning correct ?
$g∈rH$
$⇔∃h∈H: g=rh$
$⇔h=r^{-1}g$
$⇔h^{-1}=g^{-1}r$
$⇔∃(h'=h^{-1}∈H): g=h' r^{-1}$
$⇔g∈Hr^{-1}$  
Therefore $rH=Hr^{-1}$

REPLY [4 votes]: As Andreas Caranti has pointed out, the statement itself is not correct.
Where you went wrong in your reasoning is the second-to-last line.  You are claiming $g = h^{-1}r^{-1}$.  But $g = rh$, so this is only true if $g = g^{-1}$, that is, if $g$ has order at most $2$.