Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $H$, $K$ be subgroups of $G$. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$.

I need this theorem to prove something.

share|cite|improve this question
I'll be needing this theorem to prove other theorems. Actually I used it already, I want to cite this theorem unfortunately I don't have a proof for this. – BBred Jul 10 '12 at 7:29
I saw this proof. – BBred Jul 10 '12 at 7:31
@BBred: I've added LaTeXed version of your formula. If you are satisfied with the result, you can edit the post ant leave the LaTeX-ed version there. (Or make further edits, if needed.) For more about writing math at this site see here or here. – Martin Sleziak Jul 10 '12 at 7:35
Certainly the set HKhas |H||K|symbols. However,not all symbols need represent distinct group elements. That is,we may have hk=k'k' although h not equal to h' and k ot equal to k' We must determine the extent to which this happens. For every t in HandK, hk =(ht)(t^-1 k),so each group element in HK is represented by at least |HandK|products in HK. But hk = h'k' implies t = h^-1 h' = k(k')^-1 element of HandK so that h'=ht and k' = t^-1 k. Thus each element in HKis represented by exactly |HandK|products. So,|HK|= |H||K|/|HandK|. – BBred Jul 10 '12 at 7:36
Unfortunately, i didnt get it. – BBred Jul 10 '12 at 7:38
up vote 10 down vote accepted

Here is LaTex-ed version of the proof posted in BBred's comment. I've tried to add details of one place of the proof. If the OP explains which part of the proof is the problem, perhaps that part can be explained in more detail. I've made this answer a CW - anyone, feel free to contribute.

Certainly the set $HK$ has $|H||K|$ symbols. However,not all symbols need represent distinct group elements. That is, we may have $hk=h'k'$ although $h\ne h'$ and $k\ne k'$. We must determine the extent to which this happens.

For every $t\in H\cap K$, $hk =(ht)(t^{-1} k)$, so each group element in $HK$ is represented by at least $|H\cap K|$ products in $HK$.

But $hk = h'k'$ implies $t = h^{-1} h' = k(k')^{-1}\in H\cap K$ so that $h'=ht$ and $k' = t^{-1} k$. Thus each element in $HK$ is represented by exactly $|H\cap K|$ products. So, $$|HK|= \frac{|H||K|}{|H\cap K|}.$$

If we have $hk=h'k'$ and we multiply this by $h^{-1}$ from left and by ${k'}^{-1}$ from right, we get $$k{k'}^{-1}=h^{-1}h.$$ Maybe it should be stressed that $t\in H$, since $t=h^{-1}h'$; and $t\in K$ since $t=k{k'}^{-1}$. (Which means $t\in H\cap K$.)

share|cite|improve this answer
Thanks so much Martin. – BBred Jul 10 '12 at 15:42

The group $H \times K$ acts on the set $HK \subseteq G$ via $(h,k) x := hxk^{-1}$. Cleary the action is transitive. The stabilizer of $1 \in HK$ is easily seen to be isomorphic to $H \cap K$. The orbit-stabilizer "theorem" implies $|HK| \cdot |H \cap K| = |H \times K| = |H| \cdot |K|$.

By the way, this proof also works when $H,K$ are infinite.

share|cite|improve this answer
Beautiful, simple and elegant proof. +1 – DonAntonio Jul 10 '12 at 19:37

We know that $$HK=\bigcup_{h\in H} hK$$ and each $hK$ has the same cardinality $|hK|=|K|$. (See ProofWiki.)

We also know that for any $h,h'\in G$ either $hK\cap h'K=\emptyset$ or $hK=h'K$.

So the only problem is to find out how many of the cosets $hK$, $h\in H$, are distinct.

Since $$hK=h'K \Leftrightarrow h^{-1}h'\in K$$ (see ProofWiki) we see that for each $k\in K$, the elements $h'=hk$ represent the same set. (We have $k=h^{-1}h'$.) We also see that if $k=h^{-1}h'$ then $k$ must belong to $H$.

Since the number of elements that represent the same coset is $|H\cap K|$, we have $|H|/|H\cap K|$ distinct cosets and $\frac{|H||K|}{|H\cap K|}$ elements in the union.

share|cite|improve this answer

Firstly, it can be shown that $HK \le G \iff H \unlhd HK$ or $K \unlhd HK$. Without loss of generality, we can assume that $K \unlhd HK$.

Let $T = H\cap K$. Then $T \unlhd H$.

Consider the function $f: H/T \to HK/K$ where $f(hT)=hK$ for each left coset $hT \in H/T$. Suppose $f(hT)=f(gT)$ for some $hT, gT \in H/T$. Then $hK=gK$. So $h^{-1}g \in K$. But since $h, g \in H, h^{-1}g \in H$. So $h^{-1}g \in T$. Then $hT=gT$. So $f$ is an injective function.

Now take $(hk)K \in HK/K$ where $h \in H$ and $k \in K$. Then $(hk)K=hK$. So there exists $hT \in H/T$ such that $f(hT)= (hk)K$. So $f$ is a surjective function.

Since $f$ is a bijective function, $|H/T|=|HK/K|$. Then $\frac {|H|}{|T|}= \frac {|HK|}{|K|}$. Thus $|HK|= \frac {|H||K|}{|T|} = \frac {|H||K|}{|H \cap K|}$.

share|cite|improve this answer
Please rewrite your answer using TeX. This link might be of use… – Ludolila Jun 15 '14 at 18:15
Thanks for the link Ludolila. This is the first time I'm using TeX. It is not as difficult as I feared. – Andrew Rajah Jun 17 '14 at 3:57
great! you are welcome! :) – Ludolila Jun 18 '14 at 7:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.