If H and K are finite subgroups of G (another proof )

Question

I have a question and it's solution , but I want another proof if there exist .

Thanks

Answer 1

Write $HK = \bigcup_{h \in H} hK.$ So we need to find the number of distinct left cosets of the form $hK, h \in H.$ Note that $h_1K = h_2K \Leftrightarrow h_1( H \cap K) = h_2(H \cap K) \Rightarrow $ number of distinct cosets of the form $hK, h \in H$ is $\dfrac{|H|}{|H \cap K|}.$

Answer 2

If $G$ is a group, a $G$-set is a set equipped with a group action of $G$. Suppose $X$ and $Y$ are $G$-sets. Then a function $f:X\to Y$ is called a morphism (of $G$-sets) if $f(gx)=gf(x)$ for all $x\in X,g\in G$; we also say it intertwines with the action, and call it $G$-equivariant.

If $H,K\le G$ are arbitrary subgroups, then $HK=\{hk:h\in H,k\in K\}$ is a union of left cosets of $K$, and so $HK/K$ makes sense as a coset space. There is an action of $H$ on $HK/K$ and $H/(H\cap K)$ both by left multiplication, and an $H$-equivariant bijection given by $hK\mapsto h(H\cap K)$ (which I leave the reader to verify). Therefore $HK/K\cong H/(H\cap K)$ are isomorphic as $H$-sets.