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.

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

  • $\begingroup$ See the comment I wrote to Yanko $\endgroup$ – 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 $K$.

  • $\begingroup$ 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? $\endgroup$ – user401516 Nov 18 '18 at 13:21
  • $\begingroup$ 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). $\endgroup$ – user10354138 Nov 18 '18 at 13:35
  • $\begingroup$ @user10354138 How would you prove the function defined there is well defined? $\endgroup$ – user401516 Nov 18 '18 at 14:10
  • $\begingroup$ @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. $\endgroup$ – Yanko Nov 18 '18 at 15:04
  • $\begingroup$ I actually did that: math.stackexchange.com/questions/3001398/… $\endgroup$ – user401516 Nov 18 '18 at 15:30

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