MHB Group Theory: Finite Group Has Prime Order Element

[alexmahone]

Show that if G is a finite group, then it contains at least one element g with |g| a prime number. (|g| is the order of g.)

Hints only as this is an assignment problem.

 

[Nono713]

Let $g$ be an arbitrary non-identity element of the group, with order $n > 1$. Suppose $n$ is not prime, and so $g^n = 1$ but $g^k \ne 1$ for all $1 \leq k < n$. If $n$ is not prime, it is divided by some prime $p$. Can you find an element of the group which has order $p$?

 

[Deveno]

It's not *quite* true- if $G$ has order 1...

 

[alexmahone]

Let $g$ be an arbitrary non-identity element of the group, with order $n > 1$. Suppose $n$ is not prime, and so $g^n = 1$ but $g^k \ne 1$ for all $1 \leq k < n$. If $n$ is not prime, it is divided by some prime $p$. Can you find an element of the group which has order $p$?

Let $n=ap$.

$g^n=g^{ap}=(g^a)^p$

So, $(g^a)^p=1$ ------------- (1)

Let us consider $(g^a)^m$ where $1\le m<p$.

Multiplying this inequality by $a$, we get $a\le am<ap$ or $a\le am<n$.

We know that $g^k\ne 1$ for all $1\le k<n$. Since $am<n$, $g^{am}\ne 1$.

So, $(g^a)^m\ne 1$ where $1\le m<p$ ------------- (2)

From (1) and (2), the order of $g^a$ is $p$.

Is this correct?

 

[Nono713]

Let $n=ap$.

$g^n=g^{ap}=(g^a)^p$

So, $(g^a)^p=1$ ------------- (1)

Let us consider $(g^a)^m$ where $1\le m<p$.

Multiplying this inequality by $a$, we get $a\le am<ap$ or $a\le am<n$.

We know that $g^k\ne 1$ for all $1\le k<n$. Since $am<n$, $g^{am}\ne 1$.

So, $(g^a)^m\ne 1$ where $1\le m<p$ ------------- (2)

From (1) and (2), the order of $g^a$ is $p$.

Is this correct?

That's right :) using this technique you can easily construct a group element that has the desired prime order, as you have shown. In general, if $g$ has order $nm$ then $g^n$ has order $m$ and symmetrically $g^m$ has order $n$.