# Thread: [SOLVED] Prove that G has an element of order n

1. ## [SOLVED] Prove that G has an element of order n

If N is a finite normal subgroup of a group G and if G/N contains an element of order n, prove that G contains an element of order n.
This is what I have so far:

Let $\displaystyle Na \in G/N$ have order n for some $\displaystyle a \in G$. Then $\displaystyle (Na)^n = Na^n = e$, so $\displaystyle a^n \in N$. Since N is finite, let $\displaystyle ord(N) = m$. Then $\displaystyle (a^n)^m = e$, so $\displaystyle ord(a)|nm$.

When I pursued this line of reasoning, I was trying to show that $\displaystyle n|ord(a)$, and thus $\displaystyle n = ord(a)$. However, I'm stuck here. Have I made a mistake somewhere, or am I missing something?

Thanks for any insights.

2. Originally Posted by Tachyon
This is what I have so far:

Let $\displaystyle Na \in G/N$ have order n for some $\displaystyle a \in G$. Then $\displaystyle (Na)^n = Na^n = e$, so $\displaystyle a^n \in N$. Since N is finite, let $\displaystyle ord(N) = m$. Then $\displaystyle (a^n)^m = e$, so $\displaystyle ord(a)|nm$.

When I pursued this line of reasoning, I was trying to show that $\displaystyle n|ord(a)$, and thus $\displaystyle n = ord(a)$. However, I'm stuck here. Have I made a mistake somewhere, or am I missing something?

Thanks for any insights.
$\displaystyle o(a^n) < \infty$ because $\displaystyle a^n \in N$ and $\displaystyle |N| < \infty.$ let $\displaystyle o(a^n)=k.$ then $\displaystyle (a^k)^n=e$ and thus $\displaystyle o(a^k) < \infty.$ let $\displaystyle o(a^k)=m.$ the claim is that $\displaystyle m=n.$ clearly $\displaystyle m \mid n.$ now $\displaystyle (Na)^{km}=N$ and hence $\displaystyle n \mid km.$

let $\displaystyle \frac{km}{n}=\ell.$ then $\displaystyle a^{\ell n}=e.$ thus $\displaystyle k \mid \ell$ and therefore $\displaystyle n \mid m.$

3. Originally Posted by NonCommAlg
now $\displaystyle (Na)^{km}=N$ and hence $\displaystyle n \mid km.$
I understand that $\displaystyle (Na)^{km} = N$, but could you clarify why $\displaystyle n \mid km$ necessarily follows? If $\displaystyle o(Na) = n$, wouldn't we need $\displaystyle (Na)^{km} = e$ in order to have $\displaystyle n \mid km$?

4. Originally Posted by Tachyon
I understand that $\displaystyle (Na)^{km} = N$, but could you clarify why $\displaystyle n \mid km$ necessarily follows? If $\displaystyle o(Na) = n$, wouldn't we need $\displaystyle (Na)^{km} = e$ in order to have $\displaystyle n \mid km$?
the identity element of $\displaystyle G/N$ is the coset $\displaystyle N.$ so $\displaystyle (Na)^{km}=N=1_{G/N}.$

5. Originally Posted by NonCommAlg
the identity element of $\displaystyle G/N$ is the coset $\displaystyle N.$ so $\displaystyle (Na)^{km}=N=1_{G/N}.$
D'oh. That's what I deserve for my notation confusing $\displaystyle 1_{G/N}$ and $\displaystyle 1_G$. Thanks for your help; I understand now.