# Thread: Prove that every element of G has finite order

1. ## Prove that every element of G has finite order

If $\displaystyle N$ is a normal subgroup of $\displaystyle G$ and if every element of $\displaystyle N$ and of $\displaystyle G/N$ has finite order, prove that every element of $\displaystyle G$ has finite order.

I started off doing this:

Let $\displaystyle g \in G$. By showing that $\displaystyle g^n = e$, $\displaystyle n \in \mathbb{Z}$, $\displaystyle |G| = n <\infty$.

$\displaystyle (Ng)^k = Ne$ for some $\displaystyle k \in \mathbb{Z}$
$\displaystyle \Rightarrow (eg)^k \in (Ng)^k = Ne = N$
$\displaystyle \Rightarrow (eg)^k \in N$
$\displaystyle \Rightarrow (e^k)(g^k) \in N$
$\displaystyle \Rightarrow eg^k \in N$
$\displaystyle \Rightarrow g^k \in N$
$\displaystyle \Rightarrow \exists$ $\displaystyle n \in \mathbb{Z}$ such that $\displaystyle (g^k)^n = e$

I don't know how to finish off this problem after point. Any help would be greatly appreciated.

2. Originally Posted by crushingyen
If $\displaystyle N$ is a normal subgroup of $\displaystyle G$ and if every element of $\displaystyle N$ and of $\displaystyle G/N$ has finite order, prove that every element of $\displaystyle G$ has finite order.

I started off doing this:

Let $\displaystyle g \in G$. By showing that $\displaystyle g^n = e$, $\displaystyle n \in \mathbb{Z}$

$\displaystyle (Ng)^k = Ne$ for some $\displaystyle k \in \mathbb{Z}$
$\displaystyle \Rightarrow (eg)^k \in (Ng)^k = Ne = N$
$\displaystyle \Rightarrow (eg)^k \in N$
$\displaystyle \Rightarrow (e^k)(g^k) \in N$
$\displaystyle \Rightarrow eg^k \in N$
$\displaystyle \Rightarrow g^k \in N$
$\displaystyle \Rightarrow \exists$ $\displaystyle n \in \mathbb{Z}$ such that $\displaystyle (g^k)^n = e$

I don't know how to finish off this problem after point. Any help would be greatly appreciated.
I think this is what you did...

Lemma: Let $\displaystyle N\lhd G$ and $\displaystyle m=\left[G:N\right]$, then $\displaystyle g^m\in N$ for every $\displaystyle g\in G$

Proof: Since $\displaystyle N\lhd G$ we know that $\displaystyle G/N$ is a group of order $\displaystyle m$ and so given any element $\displaystyle K$ in $\displaystyle G/N$ we have that, by basic group theory, that $\displaystyle K^m=e=N$. In particular, $\displaystyle \left(Ng\right)^m=Ng^m=N$ from where it follows that $\displaystyle g^m\in N$. $\displaystyle \blacksquare$.

Is that what you wre getting at??

3. Sorry, I edited the my post up above.

Pretty much, I'm trying to show that the group G is finite, not the lemma.

4. Never mind, I found out how to do it.

Since $\displaystyle g^k$ has finite order, $\displaystyle |g^k| = n$ for some $\displaystyle n \in \mathbb{Z}$

$\displaystyle \Rightarrow (g^k)^n = e$
$\displaystyle \Rightarrow g^{kn} = e$
$\displaystyle \Rightarrow |g| = kn < \infty$