# Suppose G is a group, H a subgroup of G, and a and b elements of G. If a in bH then b in aH.

Suppose G is a group, H a subgroup of G, and a and b elements of G. If $a\in bH$ then $b\in aH$.

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Suppose
$aH\cap bH\ne \mathrm{\varnothing }$
Let $x\in aH\cap bH$, then there exist ${h}_{1},{h}_{2}\in H$
Such that,
$x=a{h}_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x=b{h}_{2}$
$a=x{h}_{1}^{-1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b=x{h}_{2}^{-1}$
$a=b{h}_{2}{h}_{1}^{-1}$
$aH=b{h}_{2}{h}_{1}^{-}1H=bH$
(since $aH=H\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}a\in H$)
Now, if $a\in bH$, using above theorem
Since $b\in bH$, then $aH=bH⇒b\in aH$
$a\in H\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}a\in aH⇒a\in aH\cap bH$
$⇒aH\cap bH\ne 0$

$⇒b\in aH$
Hence G be a group, H a subgroup of G, and a and b elements of G. If $a\in bH$ then $b\in aH$.

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