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Tricky group theory proof

Hi, I am trying to prove the following result:

Lemma:Let G be a group and let H,K be finite subgroups of G. Then∣HK∣=∣H∣∣K∣∣H∩K∣|HK|=\frac{|H||K|}{|H \cap K|}.

where HK={hk:h∈H,k∈K}HK=\{hk:h\in H, k\in K\}

I know |HK| corresponds to the number of ways I can do h*k (this corresponds to |H|*|K|) subtract all the repeated h*k. I can prove the lemma when {H∩K}\{H \cap K\} only has one element (the identity) but not when H and K share more elements. I have a feeling I am missing something really obvious but I can't figure it out!

Thanks in advance.
Reply 1
Avatar for Mark13
Mark13
Original post by Kelvinator
I can prove the lemma when {H∩K}\{H \cap K\} only has one element (the identity) but not when H and K share more elements.


That's a good start.

There might be better ways of doing this, but my first thought is to do define a map ϕ:H×K→HK\phi : H \times K \rightarrow HK by ϕ(h,k)=hk\phi(h,k)=hk. You can then argue that this map is surjective, and that for every element in HK, there are exactly |HnK| elements in HxK which are mapped to it by phi - see if you can prove this, and see if you can finish the proof off from here.
Reply 2
Avatar for Kelvinator
Kelvinator
OP
Thanks for your help. I managed to prove it with the method you suggested.

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