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Math Help - Algebra, Problems For Fun (10)

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    Algebra, Problems For Fun (10)

    Okay, this one is easy but I really like it: ( \varphi(n) here is the Euler totient function)

    Let G be a group of order n. Prove that if \gcd(n, \varphi(n))=1, then G is abelian.
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    by Counting by order of elements theorem since n is a divisor of n \implies\ number of elements in G of order n is a multiple of \varphi( n) \implies\ n=k*\varphi(n) for some k=positive integer.
    gcd( n,\varphi(n))=gcd( k*\varphi(n),\varphi(n))=1 for some k=positive integer.
    \implies\ \varphi( n)=1 \implies\ n=1 or n=2
    if ord(G)=1 then G={e} \implies\ G is abelian
    if ord(G)=2 then G is abelian since if a,b in G then a*b=b*a because (a=b) or (either a or b equal e).

    is this correct?
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    Quote Originally Posted by jaco View Post
    by Counting by order of elements theorem since n is a divisor of n \implies\ number of elements in G of order n is a multiple of \varphi( n) \implies\ n=k*\varphi(n) for some k=positive integer.
    gcd( n,\varphi(n))=gcd( k*\varphi(n),\varphi(n))=1 for some k=positive integer.
    \implies\ \varphi( n)=1 \implies\ n=1 or n=2
    if ord(G)=1 then G={e} \implies\ G is abelian
    if ord(G)=2 then G is abelian since if a,b in G then a*b=b*a because (a=b) or (either a or b equal e).

    is this correct?
    no it's not! there are infinitely many integers n with \gcd(n, \varphi(n))=1. for example if n is any prime number. you need Sylow theorems to solve the problem.
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    Quote Originally Posted by NonCommAlg View Post
    Okay, this one is easy but I really like it: ( \varphi(n) here is the Euler totient function)

    Let G be a group of order n. Prove that if \gcd(n, \varphi(n))=1, then G is abelian.
    here's a solution: excluding the trivial case n=1, we must have n=p_1p_2 \cdots p_k, where p_j are distinct primes, since \gcd(n,\varphi(n))=1. in this case \varphi(n)=(p_1-1)(p_2-1) \cdots (p_k - 1). now

    for any j \leq k, let n_j be the number of Sylow p_j subgroups of G. suppose n_j > 1. then since n_j \mid p_i, for some i \neq j, we must have n_j=p_i. but we also have that p_j \mid n_j - 1 = p_i - 1, which

    is impossible becasue then p_j \mid n and p_j \mid \varphi(n). thus n_j=1 for all j, i.e. every Sylow subgroup of G is normal. let P_j be the Sylow p_j subgroup of G. then:

    G \cong P_1 \times  \cdots \times P_k \cong \mathbb{Z}/p_1 \mathbb{Z} \times \cdots \times \mathbb{Z}/p_k \mathbb{Z} \cong \mathbb{Z}/p_1 \cdots p_k \mathbb{Z}=\mathbb{Z}/n \mathbb{Z}. so we actually proved that G is more than abelian, it's cyclic!
    Last edited by NonCommAlg; May 31st 2009 at 06:21 PM.
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