G, group of order n, and m such that (m,n)=1, if g^m = 1 show that g = 1

[tonit]

Homework Statement

Let G be a group of order n, and let m be an integer such that gcd(m,n) = 1.
Prove that $g^m = 1 => g = 1$
and show that each $g \in G$ has an mth root, that is $g = a^m$, for some $a \in G$

The Attempt at a Solution

Now by Lagrange's theorem, $g^n = 1$.
Since gcd(m,n) = 1, we can write mx + yn = 1.

Now $g = g^1 = g$^$mx+yn$$= (g^m)^x (g^n)^y = 1^x 1^y = 1$.

Also I found another way to solve the first part. Anyone tell me if it is correct:

$g^m = 1\Rightarrow o(g) | m$

Since $|G| = n \Rightarrow o(g) | n \Rightarrow o(g) = 1$ since $gcd(m,n)=1$ which implies that g = 1.


Now can anyone give me a hint for the second part?
Thanks

 

[micromass]

Think about the map

$$\varphi:G\rightarrow G:g\rightarrow g^m$$

being injective or surjective.

 

[tonit]

Since $G$ is a group, it is closed under the operation. So for each $g$, there exists $g^m$ in $G$. Since $g^m_1 \neq g^{m}_2$ for $g_1 \neq g_2$ the mapping must be bijective. So considering $\varphi^{-1}$, we get the desired result.

Please correct me if my reasoning line is not correct.

 

[micromass]

It's correct.

 

[tonit]

Thank you for the help.