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.

  • $\begingroup$ 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|. $\endgroup$
    – BBred
    Jul 10, 2012 at 7:36
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    $\begingroup$ I find this question slightly confusing, because in applying little $o$ (supposedly for "order") it suggests that $HK$ is a subgroup, which is not true without additional hypotheses. In fact one can have finite subgroups $H,K$ that span an infinite subgroup. I think the question should make the setting more clear. $\endgroup$ Jul 10, 2012 at 8:38
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    $\begingroup$ Such a formula is similar to $\mathrm{lcm}(x,y) = \dfrac{xy}{\mathrm{gcd}(x,y)}$ $\endgroup$
    – Watson
    Nov 30, 2016 at 20:08
  • $\begingroup$ right, HK is only a subgroup when one or both of the two groups H and K is normal. $\endgroup$
    – R_Squared
    Jan 18 at 15:37

6 Answers 6


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.

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    $\begingroup$ Beautiful, simple and elegant proof. +1 $\endgroup$
    – DonAntonio
    Jul 10, 2012 at 19:37
  • $\begingroup$ @DonAntonio Sorry to ask this late, but does the orbit-stabilizer theorem not imply that only $$|HK| = |(H \times K)/(H \times K)_1|$$ for which we only can conclude if $K$ and $H$ are finite? Do I overlook something? I am referring to the part that this also works when $H$ and $K$ are infinte. $\endgroup$ Feb 5, 2017 at 16:41
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    $\begingroup$ The oribt-stabilizer theorem states: Let $G$ be a group which acts on a set $X$. Let $x \in X$. Then $\# Gx \cdot \# G_x = \# G$. This is an equation of cardinals. $\endgroup$ Apr 12, 2017 at 11:22
  • $\begingroup$ Most beautiful proof I have seen of this. +1 $\endgroup$
    – Ryan S
    May 16, 2017 at 10:34
  • $\begingroup$ Follow post for this answer: math.stackexchange.com/questions/2282538/…. Beautiful proof!. If you don't mind, it is totally yours or have you seen it in any book? Thanks $\endgroup$ Apr 27, 2020 at 21:11

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 $t\in H\cap K$.)

  • $\begingroup$ Let me ask you one questuon: why did you write "by at least..."? I think here should be "exactly. Please explain this moment $\endgroup$
    – RFZ
    Nov 27, 2017 at 13:52
  • $\begingroup$ @RFZ Just to make sure, you noticed that the above was copied (and edited for better readability) from BBred's comment, right? $\endgroup$ Nov 27, 2017 at 13:55
  • $\begingroup$ @RFZ As far as I can say, the expression $hk=(ht)(t^-1k)$ shows that there are at least $|H\cap K|$ such products. And the argument in the following paragraph shows that this number is actually exact. (By show that there are at most $|H\cap K|$ such products.) $\endgroup$ Nov 27, 2017 at 14:06
  • $\begingroup$ BTW now I see that this is exactly what you asked in your question here: Theorem 2.5.1 from Herstein's book. So I guess you will get a more detailed explanation there. $\endgroup$ Nov 27, 2017 at 14:16
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    $\begingroup$ Doesn't this answer imply that the groups in questions are finite? $\endgroup$
    – user5826
    Sep 23, 2018 at 20:55

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.


This is similar to Martin Brandenburg's answer. It is from Steve Roman's Fundamentals Of Group Theory.

Define a function $f: H \times K \to HK$ which maps $(h, k) \mapsto hk$. This is a surjective function and the fibers $f^{-1}(y)$ partition $H \times K$ into $|HK|$ disjoint sets. Since $|H \times K| = |H| \cdot |K|$, all that remains is to show that each fiber $f^{-1}(y)$ has size $|H \cap K|$.

Fix $y = hk \in HK$. Every element in $H \times K$ can be written as $(hu, vk)$ for some $u \in H$ and $v \in K$. But $$f(hu, vk) = hk \text{ iff } huvk = hk \text{ iff } uv = 1 \text{ iff } v = u^{-1} $$ which means $$f^{-1}(y) = \{(hu, u^{-1}k) \: : \: u \in H \cap K\}$$ Thus, $f^{-1}(y)$ indeed has size $|H \cap K|$.

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    $\begingroup$ This answer is relatively concise and does not use any theorem, which is nice to beginner. Because the orbit-stabilizer theorem is usually introduced after this exercise in most of textbooks, a proof of this exercise without quoting the theorem can avoid the concern about "cyclic proof" (i.e. one uses the orbit-stabilizer theorem to prove this exercise while someone else may use this exercise to prove the orbit-stabilizer theorem). $\endgroup$
    – Sam Wong
    Dec 9, 2022 at 13:48

Lemma: $\color{blue}{ H\times K/\mathcal R}$ has $n$ elements that is, $\color{blue}{n= |H\times K/\mathcal R| }$ and we have, $$ \color{blue}{|HK|= |H\times K/\mathcal R| =\frac{|H\|K|}{|H\cap K|}} $$ This is a consequence of $E_1$ and $E_2$ see below for all the details.

Consider the map \begin{split} \phi :&& H\times K\to HK\\ && (h,k)\mapsto hk \end{split} Clearly, $\phi $ is onto (surjective ). Now we consider the relation,

$$\color{red}{(h,k)\mathcal R(h',k')\Longleftrightarrow hk=h'k'\Longleftrightarrow \phi(h,k)=\phi(h',k')}$$ It is easy to check that $\mathcal R$ is an equivalent relation on $H\times K$.

Fact.I. Let define by $[h,k]_\mathcal R$ the class of an element $(h,k) \in H\times K.$ that is $$[h,k]_\mathcal R =\left\{ (a,b) \in H\times K: (a,b)\mathcal R (h,k) \right\}$$ Fact. II. let denote $n\in \mathbb N$ the number of classes of $ H\times K$ with respect to the relation $\mathcal R.$ Also we denote by $\color{blue}{ H\times K/\mathcal R}$ the set of class. Precisely we have, $$\color{blue}{ H\times K/\mathcal R= \{[h_1,k_1]_\mathcal R, [h_2,k_2]_\mathcal R\cdots, [h_n,k_n]_\mathcal R\}} $$ Where, $(h_j,k_j)_j$ is a set of representative class of $H\times K/\mathcal R $.

For instance a representative class in $\Bbb Z_2 $ is $\{0,1\}$

First Equality: We consider the $\overline{\phi}$ the quotient map of $\phi$ w.r.t $\mathcal R.$ defines as follows:

\begin{split} \overline{\phi} :&& H\times K/\mathcal R \to HK\\ && [h,k]_\mathcal R \mapsto \phi(h,k) = hk \end{split}

  • $\overline{\phi} $ is well define since from the red line above we have,

$$\color{red}{(h',k') \in [h,k]_\mathcal R\Longleftrightarrow (h,k)\mathcal R(h',k')\Longleftrightarrow hk=h'k'\Longleftrightarrow \overline{\phi}([h,k]_\mathcal R)=\overline{\phi}([h',k']_\mathcal R)}\tag{Eq}.$$

  • $\overline{\phi}$ is onto (surjective): in fact for $g \in HK$ by definition of $HK$ there exist $h\in H$ and $k\in K $ such that $ g = hk = \overline{\phi}([h,k]_\mathcal R)$

  • $\overline{\phi}$ is one-to-one(injective): This is a direct consequence of (Eq) since we have
    $$\color{red}{[h',k']_\mathcal R = [h,k]_\mathcal R\Longleftrightarrow (h',k') \in [h,k]_\mathcal R \Longleftrightarrow \overline{\phi}([h,k]_\mathcal R)=\overline{\phi}([h',k']_\mathcal R)}\tag{Eq}.$$

conclusion $\overline{\phi}$ is a bijection and therfore, $$\color{blue}{n=|H\times K/\mathcal R| = |HK|}\tag{$E_1$} $$

we are jumping to the second way, starting from the following observation.

Fact. III Since $\mathcal R$ is an equivalent relation, we know that $\color{red}{([h_j,k_j]_\mathcal R)_{1\le j\le n}}$ is a partition of $H\times K$ that is, $$\color{red}{ |H\times K| = \sum_{j=1}^{n} |[h_j,k_j]_\mathcal R| }$$ Claim:(see the proof Below) $$\color{red}{|[h_j,k_j]_\mathcal R| = |H\cap K|}$$

Second Equality: Since for any finte sets A and B we have $|A\times B| =|A|\times|B|,$ using the claim and the foregoing relations, we get that $$\color{blue}{|H|\times|K| = |H\times K| = \sum_{j=1}^{n} |[h_j,k_j]_\mathcal R| = \sum_{j=1}^{n} |H\cap K| = n |H\cap K| \\=|H\times K/\mathcal R||H\cap K|}$$ Since $ n = |H\times K/\mathcal R|$.

Then $$ \color{blue}{n= |H\times K/\mathcal R| =\frac{|H\|K|}{|H\cap K|}}\tag{$E_2$} $$

Proof of the claim: Now we would like to investigate $|[h,k]_\mathcal R|$.

$$\color{blue}{(h',k')\in [h,k]_\mathcal R \Longleftrightarrow hk=h'k'\Longleftrightarrow h'^{-1}h=k'k^{-1}\in H\cap K .}$$

Consider the map \begin{split} f :&& [h,k]_\mathcal R \to H\cap K\\ && (h',k')\mapsto h'^{-1}h=k'k^{-1} \end{split} The above relation shows that $f$ is well defined as a map. We will show that $f$ is a bijective map to conclude.

  • $f$ is onto(surjective): Let $s\in H\cap K $. if we let $ k' = h s^{-1}~~~\text{and}~~~ k'=sk$ then $$\color{red}{h'k' = hs^{-1} sk =hk\Longleftrightarrow (h',k')\mathcal R (h,k) \implies (h',k') \in [h,k]_\mathcal R}$$ and $$\color{red}{f(h',k')= f(hs^{-1}, sk)} = s.$$

this prove that $f$ is onto.

  • $f$ is one-to-one(injective): let, $(a,b), (x,y)\in [h,k]_\mathcal R $ such that $f(a,b)=f(x,y)$. we have $$f(a,b) =a^{-1}h =bk^{-1} ~~~~\text{and}~~~f(x,y) =x^{-1}h =yk^{-1}$$

then, \begin{split} f(a,b)=f(x,y)&\implies& \color{blue}{a^{-1}h =bk^{-1}} =\color{red}{x^{-1}h =yk^{-1}}\\ &\implies& \color{blue}{a^{-1}h =x^{-1}h} ~~~~\text{and}~~~~\color{red}{ bk^{-1}=yk^{-1}}\\ &\implies& \color{blue}{a^{-1} =x^{-1}} ~~~~\text{and}~~~~\color{red}{ b=y}\\ &\implies& \color{red}{a =x} ~~~~\text{and}~~~~\color{red}{ b=y}\\ &\implies& (a,b)=(x,y) \end{split}

Hence $f$ is bijective then the claim follows


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 $|HK|= \frac {|H||K|}{|T|} = \frac {|H||K|}{|H \cap K|}$.

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    $\begingroup$ Please rewrite your answer using TeX. This link might be of use meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – Ludolila
    Jun 15, 2014 at 18:15
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    $\begingroup$ Thanks for the link Ludolila. This is the first time I'm using TeX. It is not as difficult as I feared. $\endgroup$ Jun 17, 2014 at 3:57
  • $\begingroup$ great! you are welcome! :) $\endgroup$
    – Ludolila
    Jun 18, 2014 at 7:50

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