# Order and index of a normal subgroup $N$ are relatively prime

Let $N$ be a normal subgroup of a finite group $G$. Assume that the order of $N$ and the index of $N$ in $G$ are relatively prime. Prove that if $g\in G$ satisfies $o(g)\mid o(N)$, then $g\in N$.

I tried using some properties from number theory, but didn't get anywhere. Any idea?

Note that $(gN)^{o(g)}=(gN)^{|G/N|}=eN$, and $o(g)$ and $|G/N|$ are relatively prime. There exist integers $m$ and $n$ such that $1=m\cdot o(g)+n|G/N|$. So $gN=(gN)^{m\cdot o(g)+n|G/N|}=eN$ and thus $g\in N$.

• Why is $$(gN)^{o(g)} = (gN)^{|G/N|}$$ true? – LKSR Sep 20 '15 at 16:03
• Sorry to open this up but I have a question, I understand that $gN=(gN)^{m\cdot o(g)+n|G/N|}=eN$ But don't quite follow why this then implies $g \in N$. Could anyone explain? – harry55 Apr 2 '17 at 13:24
• @harry55 this is a fundamental property of the left cosets. Note that: $$gN = N \Longleftrightarrow \forall n \in N \exists n' \in N \text{ such that } gn = n' \therefore g = n'n^{-1} \in N$$ – Danilo Gregorin Afonso Sep 8 '17 at 22:25
• @sephylnix Note that: $$(gN)^{o(g)} = g^{o(g)}N = N = (gN)^{|G/N|}$$ Remember that if n is the order of some group G, then $g^n = e \ \ \forall g \in G$ – Danilo Gregorin Afonso Sep 8 '17 at 22:32

Hint: If $A$ and $B$ are finite groups and $\phi:A\to B$ is a group homomorphism, then for any $a\in A$, we have $\mathrm{ord}(\phi(a))\mid \mathrm{ord}(a)$ because $$a^k=1_A\implies 1_B=\phi(1_A)=\phi(a^k)=\phi(a)^k$$ and we also have $\mathrm{ord}(\phi(a))\mid \lvert B\rvert$ because $\langle \phi(a)\rangle$ is a subgroup of $B$.

Thus, $\mathrm{ord}(\phi(a))\mid\gcd(\mathrm{ord}(a),|B|)$.

I leave it to you to figure out what $A$, $B$, and $\phi$ might be in your situation.

We have a homomorphism $\pi:G \to G/N$ (regular projection). Let's look at $\pi(<g>) \cong <g>/{<g>\cap N}$. $\vert \pi(<g>) \vert$ divides $|<g>|$ and hence divides $|N|$. And $|\pi(<g>)|$ divides $[G:N]$ so $|\pi(<g>)|$ divides the gcd of $N$ and of its index hence is 1. So $\pi(<g>)$ is the identity, hence $<g> \subseteq N$.

• < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. – Zev Chonoles Jun 1 '15 at 3:31

Note that $|G| = |N|[G:N]$.

Suppose $p$ is a prime dividing $|N|$ (I leave it to you to decide what happens when NO prime divides $|N|$). Then since $\gcd(|N|,[G:N]) = 1$, we have that $p\not\mid [G:N]$, that is: $p$ is not a factor of the index of $N$.

Now if $p|o(g)$, then by transitivity of divisibility, we have $p$ divides $|N|$ (since $o(g)$ divides $|N|$ by assumption). So none of the prime factors of $o(g)$ are in $[G:N]$.

Now consider the homomorphism $\pi: G \to G/N$ where $\pi(g) = gN$. We have $|G/N| = [G:N]$, and since $o(gN)|o(g)$ (since if $o(g) = k$, then $(gN)^k = g^kN = eN = N$), a prime factor of $o(gN)$ must divides $o(g)$ and $[G:N]$, and thus $N$ and $[G:N]$.

But by assumption these are co-prime, so we must have $o(gN) = 1$, that is: $gN = N$, whence $g \in N$.

(In all fairness to mich95, this is the same argument spelled out in slightly more elementary terms).