Clarification on the proof of $|HK|=\displaystyle{\frac{|H||K|}{|H\cap K|}}$.

Question

Okay so I would like some clarification on a very elegant answer given by Martin Brandenburg on the following post:

I am well aware of the orbit stabiliser theorem and understand the direction he wants to take with the proof. There is just $2$ steps which I think should be easy to fill in in his proof but I cannot seem to do either.

1) Why is it obvious that the action is transitive?

That is given $hk$ and $h'k'$ in $HK$ why is it true that there exists $(h^*,k^*)$ such that $(h^*,k^*)hk=h'k'$.

I tried to reverse engineer an element to show transitivity but failed.

2) Why is it easy to see the stabiliser of $1 \in HK$ is isomorphic to $H \cap K$.

Could someone explain in more detail the missing parts of the argument here so I can understand the details of this proof.

Thanks!

https://math.stackexchange.com/questions/168942/order-of-a-product-of-subgroups?noredirect=1&lq=1 Apologies I forgot to include a link! — Ryan S, May 15 '17 at 21:22

score 3 · Accepted Answer · answered May 15 '17 at 21:39

3

The action given there is $(h,k)x=hxk^{-1}$, so your line here really looks like $$h^*hk{k^*}^{-1}=h'k'$$ Now take $h^*:=h'h^{-1}$ and $k^*:=k'^{-1}k$.
The stabilizer of $1$ consists of those $(h,k)\in H\times K$ for which $(h,k)1=1$, i.e. $h1k^{-1}=1$, which means $h=k$.
So, it will be $\{(x,x):x\in H\cap K\}$.

answered May 15 '17 at 21:39

Berci

90,745

Thank you. I understand clearly now! – Ryan S May 16 '17 at 10:33

Clarification on the proof of $|HK|=\displaystyle{\frac{|H||K|}{|H\cap K|}}$.

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