# Prove that every element of G has finite order

• February 9th 2010, 02:49 PM
crushingyen
Prove that every element of G has finite order
If $N$ is a normal subgroup of $G$ and if every element of $N$ and of $G/N$ has finite order, prove that every element of $G$ has finite order.

I started off doing this:

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

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

I don't know how to finish off this problem after point. Any help would be greatly appreciated.
• February 9th 2010, 03:10 PM
Drexel28
Quote:

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

I started off doing this:

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

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

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

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

Is that what you wre getting at??
• February 9th 2010, 04:25 PM
crushingyen
Sorry, I edited the my post up above.

Pretty much, I'm trying to show that the group G is finite, not the lemma.
• February 9th 2010, 07:21 PM
crushingyen
Never mind, I found out how to do it.

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

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