## If the index and order of a normal subgroup and subgroup are relatively prime, then the subgroup is contained in the normal subgroup

Let $G$ be a group and let $H,N \leq G$ with $N$ normal in $G$. Prove that if $|H|$ and $[G:N]$ are relatively prime then $H \leq N$.

First we prove a lemma.

Lemma: Let $G$ be a group, $H \leq G$, and $x \in G$ an element of finite order $n$. If $k$ is the least positive integer such that $x^k \in H$, then $k|n$. Proof: If $k$ does not divide $n$, we have $n = qk+r$ for some $0 < r < k$ by the division algorithm. Now $1 = x^n = x^{qk}x^r \in H$, and $x^{qk} = (x^k)^q \in H$. Thus $x^r \in H$, which contradicts the minimality of $k$. Thus $k|n$. $\square$

Now to the main result.

Suppose $x \in H$, and let $k$ be the least positive integer such that $x^k \in N$. ($k$ exists since $H$ is finite.) By a previous exercise, as an element of $G/N$, $|xN| = k$, so that $k$ divides $[G:N]$. Moreover, we have $|x|$ divides $|H|$ by Lagrange, so that (by the lemma) $k$ divides $|x|$ and thus divides $|H|$. Because $|H|$ and $[G:N]$ are relatively prime, then, $k = 1$. But then $|xN| = 1$, so $xN = N$, and we have $x \in N$. So $H \subseteq N$. By a previous exercise $H \leq N$.

• Bobby Brown  On September 26, 2010 at 1:39 pm

I think you may want that k divides the order of x, why would k divide the index of N in G?

• nbloomf  On September 26, 2010 at 6:49 pm

We’re thinking of $G/N$ as a group of order $[G:N]$. There, $xN$ is an element of order $k$. So by Lagrange, $k$ divides $[G:N]$.

We need this in order to take advantage of the fact that $|H|$ and $[G:N]$ are relatively prime.

• Bobby Brown  On September 29, 2010 at 8:48 pm

I think I’ve found my error. So the order of xN is not the number of elements in that coset, rather the number of operations required in G/N to reach the identity coset? Sorry, and thanks so much for the explanations.

• nbloomf  On September 29, 2010 at 9:34 pm

That’s exactly right. No problem!

• inquisitive  On March 6, 2011 at 8:56 pm

but G can also be infinite?

• nbloomf  On March 7, 2011 at 7:40 am

We started out with the assumption that $G$ is finite. However, it looks like the proof still works for infinite $G$

• quadro  On March 9, 2011 at 4:12 am

in the solution parag no.2,|x| divides |H| clearly but why should k divide |H|?

• nbloomf  On March 9, 2011 at 9:26 am

This is using the lemma and Lagrange’s theorem. $k$ divides the order of $x$ (by the lemma) which divides $|H|$ (By Lagrange).

• quadro  On March 12, 2011 at 7:16 am

thanks..