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.