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

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

Let $$G$$ be a group, $$H a subgroup of $$G$$ and $$K 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.

Tags :

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 $$K$$.

Yanko
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 $$Kfg=\{(1432),(143)\}$$

mathnoob
November 18, 2018 13:19 PM