To Show that all ﬁbers of this map has exactly $|H \cap K|$ elements and $|HK| = \frac{|H|\times|K|}{|H\cap K|}$

Question

I want to solve :

I think the inclusion $H \subset HK$ and the projection $HK \to HK/K$ are both group homomorphisms and so is their composition $H \to HK/K$. May be we should find the kernel of this map. But I am not sure how to write the complete proof for this.

What makes you think $HK$ is a group though? This is not always true. — Mark, Sep 05 '19 at 12:25
@Mark Do we need to use the fact that HK is a Group. I mean can't we do this considering HK only as a Set. — Adam Warlock, Sep 05 '19 at 12:29
Yes, we can. The statement is true even if $HK$ is not a group. If you want I can write an answer and give a hint there. — Mark, Sep 05 '19 at 12:32

score 0 · Answer 1 · answered Sep 05 '19 at 12:37

Alright, let's call your map $f$. Note that if $x\in H\cap K$ then $f(h,k)=f(hx,x^{-1}k)$. Conversely, if $f(h_1,k_1)=f(h_2,k_2)$ then there is some $x\in H\cap K$ such that $h_2=h_1x$ and $k_2=x^{-1}k_1$. Indeed, we can take $x=k_1k_2^{-1}=h_1^{-1}h_2$. Can you finish from here?

To Show that all ﬁbers of this map has exactly $|H \cap K|$ elements and $|HK| = \frac{|H|\times|K|}{|H\cap K|}$

1 Answers1