Let G be a group of order n, and let m be an integer such that gcd(m,n) = 1.
Prove that [itex]g^m = 1 => g = 1[/itex]
and show that each [itex]g \in G[/itex] has an mth root, that is [itex]g = a^m[/itex], for some [itex]a \in G[/itex]
The Attempt at a Solution
Now by Lagrange's theorem, [itex]g^n = 1[/itex].
Since gcd(m,n) = 1, we can write mx + yn = 1.
Now [itex]g = g^1 = g[/itex][itex]mx+yn[/itex][itex] = (g^m)^x (g^n)^y = 1^x 1^y = 1[/itex].
Also I found another way to solve the first part. Anyone tell me if it is correct:
[itex] g^m = 1\Rightarrow o(g) | m[/itex]
Since [itex] |G| = n \Rightarrow o(g) | n \Rightarrow o(g) = 1 [/itex] since [itex] gcd(m,n)=1[/itex] which implies that g = 1.
Now can anyone give me a hint for the second part?