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

When I pursued this line of reasoning, I was trying to show that $n|ord(a)$, and thus $n = ord(a)$. 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 $Na \in G/N$ have order n for some $a \in G$. Then $(Na)^n = Na^n = e$, so $a^n \in N$. Since N is finite, let $ord(N) = m$. Then $(a^n)^m = e$, so $ord(a)|nm$.

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

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

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

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

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

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