$Kb=Kf$, $Ha=Hg$ implies $Kba=Kfg$?

by user401516   Last Updated November 18, 2018 13:20 PM

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.

Answers 2

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

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

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 $a-g=a\in H$.

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

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

November 18, 2018 13:05 PM

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 $Kb=\{(123),(23)\}$.

$ba=(1423)$ and $fg=(1432)$, so now $Kba=\{(1423),(14)(23)\}$ and $Kfg=\{(1432),(143)\}$

November 18, 2018 13:19 PM

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