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

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data
    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]



    3. 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
     
  2. jcsd
  3. Jun 8, 2012 #2

    micromass

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    Think about the map

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

    being injective or surjective.
     
  4. Jun 9, 2012 #3
    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.
     
  5. Jun 9, 2012 #4

    micromass

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    It's correct.
     
  6. Jun 9, 2012 #5
    Thank you for the help.
     
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