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

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.

• Are you sure this is true? It sounds like there's a counter-example to me. – Yanko Nov 18 '18 at 13:01
• If $K=H$, the property is equivalent to $H$ being a normal subgroup. – egreg Nov 18 '18 at 16:42

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 $$Kba \neq Kfg$$.

• See the comment I wrote to Yanko – user401516 Nov 18 '18 at 13:24

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 $$K$$.

• This is interesting. The reason I'm trying to prove it is actually an answer to this post: math.stackexchange.com/questions/730728/… (I'm trying to prove the function defined in the first answer is well defined). Does this mean the solution there is wrong? – user401516 Nov 18 '18 at 13:21
• The solution there is not wrong. You are attempting to generalize the result too much (allowing all $a,b,f,g$ to vary arbitrarily as opposed to within fixed transversals). – user10354138 Nov 18 '18 at 13:35
• @user10354138 How would you prove the function defined there is well defined? – user401516 Nov 18 '18 at 14:10
• @user401516 I suggest you ask that as another question. Put a link to that question and say that you don't understand why the function is well defined. – Yanko Nov 18 '18 at 15:04
• I actually did that: math.stackexchange.com/questions/3001398/… – user401516 Nov 18 '18 at 15:30