I'm struggling to proof that if $H$ and $K$ are subgroups of finite index of a group $G$ such that $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$. I don't know why I can't answer it, because this question seems easy. I'm stuck maybe because I've studied so far just Lagrange's theorem and some of its consequences. But I think we don't need much more, because this is the material covered so far by the Hungerford's book.

I need help.


  • $\begingroup$ @CamiloArosemena no, maybe it's my problem, I found the proposition 4.8 and 4.9 very unintuitive and a little bit out of context. $\endgroup$
    – user42912
    Apr 25, 2013 at 23:10

3 Answers 3


We have the following fact:

  1. If $H$ and $K$ are subgroups of a group $G$, then $[H:H\cap K]\leq [G:K]$. If $[G:K]$ is finite, then $[H:H\cap K]=[G:K]$ if and only if $G=KH$. This is proposition $4.8$ of chapter I in Hungerford's Algebra.

It is easy to prove that $[G:H][H:H\cap K]=[G:K][K:H\cap K]$, from this conclude that $[G:H]\mid [K:H\cap K]$; since $[G:H]$ and $[G:K]$ are relatively prime, and combinig this with $(1)$ we get $[G:H]=[K:H\cap K]$, so...

  • $\begingroup$ But how to $G=KH$, you've only proved $G=KH$, remember $KH$ is not necessarily equal to $HK$, thank you for your answer :) $\endgroup$
    – user42912
    Apr 25, 2013 at 23:45
  • 1
    $\begingroup$ Interchange the roles of $H$ and $K$. You're welcome. $\endgroup$ Apr 25, 2013 at 23:46
  • $\begingroup$ I edited the question $\endgroup$ Apr 25, 2013 at 23:49
  • $\begingroup$ Thank you again, I'm writing down the solution, seeing if everything's all right :) $\endgroup$
    – user42912
    Apr 25, 2013 at 23:53
  • 2
    $\begingroup$ @user42912 In general it is not true that $HK = KH$. But it is true that if $H$ and $K$ are subgroups of $G$ then, $HK = KH$ if and only if $HK$ is a subgroup of G. So since CamiloArosemena has shown that $KH = G$ then in particular, $KH$ is a subgroup of $G$, so $KH = HK$ $\endgroup$
    – user58514
    Apr 25, 2013 at 23:58

Try proving that [$G$ : $H\cap K$] $=$ [$G$ : $H$][$G$ : $K$].

And then see how you can use the fact that $|HK| = \frac{|H||K|}{|H\cap K|}$

  • 8
    $\begingroup$ $G$ was not supposed to be necessarily finite, and $HK$ is not necessarily a subgroup.. $\endgroup$
    – Berci
    Apr 25, 2013 at 23:14
  • $\begingroup$ @Berci Just wondering,how can HK not be a subgroup when the question wants us to prove that G =HK which means HK is itself a group? $\endgroup$ Nov 24, 2014 at 12:25

If $[G:H]$ and $[G:K]$ are coprime, $[G:H∩K]=[G:H][G:K]$ (see here for a proof)

Observe that a coset of $H\cap K$ is just an intersection of a left coset of $H$ and a left coset of $K$.

Thus, every intersection of cosets of $H$ and $K$ are non-empty and make a coset of $H∩K$. In particular, $$xK∩H≠\emptyset $$ $∀x∈G$. So if $g\in G$, then there exists a $h\in H$ such that $gK=hK$

So for every there is an element of $H$ in every left coset of $K$. Hence $$G=\bigcup_{g\in G} gK=\bigcup_{h\in H} hK=HK$$

Notation: $[G:H]$ is the index of subgroup $H$ in group $G$


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