# If $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$

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.

Thanks.

• @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. Apr 25, 2013 at 23:10

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...

• But how to $G=KH$, you've only proved $G=KH$, remember $KH$ is not necessarily equal to $HK$, thank you for your answer :) Apr 25, 2013 at 23:45
• Interchange the roles of $H$ and $K$. You're welcome. Apr 25, 2013 at 23:46
• I edited the question Apr 25, 2013 at 23:49
• Thank you again, I'm writing down the solution, seeing if everything's all right :) Apr 25, 2013 at 23:53
• @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 $KH = HK$
– 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 $|HK| = \frac{|H||K|}{|H\cap K|}$

• $G$ was not supposed to be necessarily finite, and $HK$ is not necessarily a subgroup.. Apr 25, 2013 at 23:14
• @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? 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 $$G$$