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

  • Thread starter tonit
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  • #1
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Homework Statement


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?
Thanks
 

Answers and Replies

  • #2
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Think about the map

[tex]\varphi:G\rightarrow G:g\rightarrow g^m[/tex]

being injective or surjective.
 
  • #3
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Since [itex]G[/itex] is a group, it is closed under the operation. So for each [itex]g[/itex], there exists [itex]g^m[/itex] in [itex]G[/itex]. Since [itex]g^m_1 \neq g^{m}_2[/itex] for [itex]g_1 \neq g_2[/itex] the mapping must be bijective. So considering [itex]\varphi^{-1}[/itex], we get the desired result.

Please correct me if my reasoning line is not correct.
 
  • #4
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3,297
It's correct.
 
  • #5
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Thank you for the help.
 

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