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Math Help - p-group

  1. #1
    Junior Member ibnashraf's Avatar
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    p-group

    Hi, can someone help me understand the following question ....

    "Let G be a finite group. Prove that G is a p-group iff every element of G has order a power of p."
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  2. #2
    Junior Member ibnashraf's Avatar
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    Re: p-group

    Quote Originally Posted by ibnashraf View Post
    "Let G be a finite group. Prove that G is a p-group iff every element of G has order a power of p."
    ok this was my attempt at the question:

    Suppose G is a p-group
    \Rightarrow |G|=p^n, where n\in Z
    Let g \in G \Rightarrow |G|/|g|
    \therefore p^n/|g|
    Hence, every element of G has order a power of p.

    Conversely, suppose that every element of G has order a power of p
    \Rightarrow |g|=p^n
    But |G|/|g|
    \therefore p^n divides |G|
    Hence, G is a p-group.


    can someone look over this and tell me what mistakes i made if any please ?
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  3. #3
    MHF Contributor

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    Re: p-group

    one implication is obvious:

    if G is a p-group, (so that |G| = pn) then the only possible divisors of |G| are powers of p, and since |g| divides |G|, |g| = pk, for some 0 ≤ k ≤ n.

    the other implication is not so obvious: if for any g in G,|g| = pk, for some non-negative integer k, we know that |G| = pnm, for some positive integer m with gcd(p,m) = 1, and some positive integer n.

    but it is not clear that we must have m = 1.

    however, by Cauchy's theorem, if q ≠ p is a prime that divides m, we must have an element x of G of order q. but there are no such elements, since every element of G has order a power of p. therefore, there are no prime numbers dividing m, so m = 1.
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