# Thread: prove that |HK| = |H||K| / |H ∩ K|

1. ## prove that |HK| = |H||K| / |H ∩ K|

Given that H and K are subgroups of G, prove that |HK| = |H||K| / |H ∩ K|. i have that if H and K share elements then there is some h1k1 = h2k2 in HK. but since H and K are subgroups, the inverse elements of each element exists and so $h_2^{-1}h_1 = k_2 k_1^{-1} = x \in H\intersect K$. so whenever a product equals another, there is an element x in the intersection that lets you go from one to another. in this example $h_2 = h_1 x^{-1}$ and $k_2 = xk_1$.

what i am having trouble showing is that every element in HK has multiple elements in H x K mapping to it. in fact each element in HK has n different products that multiply to it where n is |H ∩ K|. i know that there is some element in HK that is repeated as i showed above but i am having trouble seeing that for every single element. it seems to me like its possible to miss some elements so although some elements in HK are repeated some others are unique and are not repeated. but i know that for the formula i need to prove to be true, every element in HK has to be repeated the same number of times. but if i did not know the formula beforehand i would think some elements could be repeated while others are unique which is why this problem is confusing me so much. can someone help me out on the details of this proof? thanks!

2. ## Re: prove that |HK| = |H||K| / |H ∩ K|

suppose hk = h'k', where h ≠ h'. clearly k cannot be k' in this case, either, or we would have h = h', by the cancellation law.

well then h'^-1h = k'k^-1. this tells us that if we have a duplication, we also have an element in H∩K, namely h'^-1h.

so for each duplication, we have an element in the intersection. the question is, for a given hk in HK, how many possible ways

is there to write it as some other product h'k'? that is, does every element of H∩K gives a duplication in HK of hk, for a given hk?

well, if x is in H∩K, then x^-1 is in H∩K (since H∩K is a subgroup). so:

hk = h(xx^-1)k = (hx)(x^-1k). but hx is in H, since h is in H and x is in H∩K, and thus in H.

similarly, x^-1K is in K, since x^-1 is in H∩K, and thus in K, and k is in K.

taking h' = hx, and k' = x^-1k, we have hk = h'k'. so we see we get |H∩K| products h'k' all equal to hk,

one for each different x in H∩K (the identity e gives us the original hk).

3. ## Re: prove that |HK| = |H||K| / |H ∩ K|

If your curious, this can also be proved via the double-coset action as I discuss here on my blog.