# Question about normal subgroup and relatively prime index

Suppose $G$ is finite, $K$ is a normal subgroup in $G$, $H$ is a subgroup of $G$, and $|K|$ is relatively prime to $[G:H]$. Show that $K$ is a subgroup of $H$.

I don't know where to begin...

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$[HK:H]=[K:K\cap H]$ divides both $[G:H]$ and $|K|$. – user641 May 23 '12 at 2:35
Yes, this is waaaaay easier...:) – DonAntonio May 23 '12 at 2:44
You don't really need $K$ normal, you just need $HK$ to be a subgroup. – Arturo Magidin May 23 '12 at 2:57
$G$ finite is not necessary. If $[G:H]$ finite and $|K|$ so is, then $K\leq H$. – Gastón Burrull May 23 '12 at 3:21
I got it. Thanks for the useful and free help! – Steven Li May 23 '12 at 3:33

You don't need $K$ to be normal, you just need $HK$ to be a subgroup (which it is when $K$ is normal). Note that $K\subseteq H$ if and only if $H\cap K=K$.

In general, we know that $|HK||H\cap K| = |H||K|$. Therefore, if all quantities are finite, we have $$\frac{|HK|}{|H|} = \frac{|K|}{|H\cap K|}.$$ If $HK$ is a subgroup and $G$ is finite, then $|HK|$ divides $|G|$, so $$\frac{|K|}{|H\cap K|} = \frac{|HK|}{|H|} \text{ divides }\frac{|G|}{|H|} = [G:H].$$

From this you should be able to finish.

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How do you understand the equality |HK||H∩K|=|H||K|? I can prove it, but I don't understand it well enough to apply it on my own. – Steven Li May 23 '12 at 4:24
@sli: It's a counting statement: if you take all products of elements of $H$ and elements of $K$ you get $|H\cap K|$ copies of each element of $HK$. Not sure what else you are supposed to "understand", except perhaps that as a special case you get the equality $|H|/|H\cap K| = |HK|/|K|$, which of course "should" follow from one of the Isomorphism Theorems. Perhaps you can view it as a generalization of that in the case where $K$ is not necessarily normal? – Arturo Magidin May 23 '12 at 4:33
I see the counting statement now. Thank you! – Steven Li May 23 '12 at 4:37
why do we need HK to be a subgroup? – user136266 Nov 24 '14 at 14:10
what is the difference between using second isomorphism theorem and the counting argument here? – user136266 Nov 24 '14 at 14:23

I'll prove a more general result:

If $K\triangleleft G$, $|K|$ finite, $H\leq G$, $[G:H]$ finite and $|K|$, $[G:H]$ are relatively prime then $H\leq K$

We need to prove the next Lemma for a very elemental proof.

Lemma: 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 $G=KH$.

Proof: If $A$ is collection of all left cosets of $H\cap K$ in $H$ and $B$ is the collection of all left cosets of $K$ in $G$. The map $\phi:A\to B$ given by $\phi(h(H\cap K))=hK$ (remember that $h$ must be in $H$) is well defined since $h(H\cap K)=h'(H\cap K)$ implies $h'h^{-1}\in (H\cap K)$ in particular $h'h^{-1}\in K$, then $h'K=hK$. Map is inyective since $h'K=hK$ with $h\in H$ implies $h'h^{-1}\in K$ and $h'h^{-1}\in H$ therefore $h'h^{-1}\in H\cap K$ and $h(H\cap K)=h'(H\cap K)$. Then $[H:H\cap K]=|A|\leq|B|=[G:K]$. Let $[G:K]=n<\infty$ then $[H:H\cap K]=[G:K]$ if and only if $\phi$ is surjective if and only if $G=KH$ $\square$.

Since $K$ is normal, $KH$ is a subgroup. By Lemma we have $[K:H\cap K]=[HK:K]$ but $|K|$ is finite then $|H\cap K|$ so is and we know that $[K:H\cap K]|H\cap K|=|K|$ then we have that $[K:H\cap K]$ is finite and $[G:H]=[G:HK][HK:H]$.

Note that $[HK:H]=[K: H\cap K]$ (second isomorphism theorem) and both are finite. Clearly $[HK:H]$ divides $[G:H]$ and $[K: H\cap K]$ divides $|K|$ but are relatively primes then $[K: H\cap K]=1$ implies $K\leq H$.

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Hi just wondering, isn't the lemma saying that $[K:H\cap K] \leq [HK: K]$? Why do you put equality? And why we need the lemma when we already have the second isomorphism theorem? – user136266 Nov 24 '14 at 13:48
does saying |K|,[G:H] relatively prime imply that they are finite? – user136266 Nov 24 '14 at 14:32

1) Suppose $\,H\triangleleft G\,$ , and let us look at $\,kH\in G/H\,\,,\,\,k\in K\,$ , so one hand we have $\,\left(kH\right)^{|K|}=k^{|K|}H=H\,$, and OTOH we also have $\,\left(kH\right)^{[G:H]}=k^{[G:H]}H=H\,$ , so both numbers $\,|K|\,,\,[G:H]\,$ have to be multiples of this element's order, which can be only if its order is 1, and we're done.

2) If H is not normal in G, just take $\,\displaystyle{\cap_{g\in G}H^g=}\,$ the core or H in G, instead of H.

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