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Math Help - p-groups

  1. #1
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    p-groups

    I have this Lemma: If G is a finite abelian group whose order is divisible by a prime p, then G contains an element of order p.


    I have shown that for x \in G, x \ne \{{1}\} and order of x is pm, then x^{m} is an element in G of order p.
    Now, suppose that x has order t, where (p,t)=1. Since G is abelian, <x> is a normal subgroup of G and G/<x> is an abelian group of order |G|/t. It is easy to see that |G|/t is divisible by p (since |G| is divisible by p and (p,t)=1) and is less than |G|.
    Afterwards I know I need to show that there is an element in G/<x> whose order is p, but I'm stuck here. Please help me in completing the proof.


    I have another corollary: A finite group G is a p-group \Longleftrightarrow |G| is a power of p.

    I have shown ( \Longrightarrow ).
    For ( \Longleftarrow ), I used Lagrange's Theorem and shown that every subgroup of G has order of power of p. How to continue the proof from here?
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  2. #2
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    The corollary will based on the Lemma above and Cauchy's Theorem: If G is a finite group whose order is divisible by a prime p, then G contains an element of order p.
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  3. #3
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    Quote Originally Posted by deniselim17 View Post
    I have this Lemma: If G is a finite abelian group whose order is divisible by a prime p, then G contains an element of order p.
    do it by induction on the order of the group: if the order is 1, there's nothing to prove. so suppose |G| = n with n> 1, and the claim is true for any abelian group of order less than n.

    let p be a prime divisor of |G|. and choose 1 \neq x \in G. let o(x)=m and H = <x>. if \gcd(m,p)=1, then p will divide |G/H| and obviously |G/H| < |G|. so by induction, G/H has

    an element of order p, say Hy. so y^p \in H. hence y^{pm}=1, which gives you o(y^m)=p, because y^m \neq 1. why? if p \mid m, then m = kp and o(x^k)=p and you're done!


    I have another corollary: A finite group G is a p-group \Longleftrightarrow |G| is a power of p.

    For ( \Longleftarrow ), I used Lagrange's Theorem and shown that every subgroup of G has order of power of p. How to continue the proof from here?
    there's nothing left to continue! here's the definition of a p-group: a (finite or infinite) group is called a p-group if the order of every element of the group is a power of p.


    Proof of the corollary: do induction again: if G is abelian, we're done. if there exists x \in G - Z(G) such that p does not divide [G:C_G(x)], then since p divides |G|, we must have p \mid |C_G(x)|.

    since |C_G(x)| < |G|, we're done by induction. if p \mid [G:C_G(x)], for any x \in G - Z(G), then p \mid |Z(G)|, by the class equation, and we're done again because Z(G) is abelian. Q.E.D.


    for another proof of Cauchy's theorem see here.
    Last edited by NonCommAlg; November 3rd 2008 at 05:42 PM.
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  4. #4
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    In following the proof by NonCommAlg by the conjugacy class equation the only step that needs work to be proven is that if G is a finite abelian group and p divides |G| then there is an element of order p. Here is another way, though I perfer the one by NonCommAlg, involving the decomposition theorem. You can write G \simeq C_1 \times ... \times C_n where C_i are cyclic groups of prime order. Since p divides |G| it means p divides some C_i. Therefore, C_i \simeq \mathbb{Z}_{p^r}. Let H = p^{r-1}\mathbb{Z}_{p^r} . Then |H| = p and H is cyclic therefore it is generated by an element of order |H| = p. This shows there is an element of order p.
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