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Let $G$ be a group, $H<G$ a subgroup of $G$ and $K<H$ a subgroup of $H$. I'm trying to prove the following:

If $Ha=Hg$, $Kb=Kf$ for $a,g in G$, $b,f in H$ then $Kba=Kfg$.

I tried to show that given $ag^{-1} in H $, $bf^{-1} in K $ we have $ba(fg)^{-1} in K$, that is $ bag^{-1}f^{-1} in K$. But $ag^{-1} in H $ is "stuck in the middle" and I'm not sure how to continue.

I think this is not true:
Here is a counter example:
$G=S_4$, $K=S_2={e,(12)}$ and $H=S_3={e,(12),(13),(23),(1 2 3),(1 3 2)} $

and choose
$b=(123), f=(23), a=(14),g=(142)$.

Now
$Ha={(14),(142),(1423),(1432),(14)(23),(143)}=Hg$ and $Kb={(123),(23)}=Kf$.

$ba=(1423)$ and $fg=(1432)$, so now $Kba={(1423),(14)(23)}$ and $Kfg={(1432),(143)}$ so we can see that $Kba neq Kfg$.

You can't prove this because it's wrong, here's a counter-example:

Take $G=mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z} oplus mathbb{Z}/5mathbb{Z}$ (additive). $H$ be the subgroup corresponding to $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/3mathbb{Z}$ and $K$ to $mathbb{Z}/2mathbb{Z}$.

Now choose $b,f,g = 0_G$ and $a = (0,1,0)$.

Clearly, $K+b=K+f$ because $b-f=0_G$ and $H+a=H+g$ because $a-g=ain H$.

However $K+b+a = K+a not = K = K+f +g$. Because $anotin K$.

Note: You can construct infinitely many counter-examples by simply taking any $G$ and $H$ and any proper subgroup $K$ of $H$ (i.e. $Knot = H$) then take $b,f,g=0$ and $a$ be any element in $H$ that is not in $K$.