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Math Help - HELP: A finite abelian p-group generated by its elements of maximal order?

  1. #1
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    HELP: A finite abelian p-group generated by its elements of maximal order?

    We say that a group in which every element has order a power of a fixed prime p is called a "p-group." Then how do I prove that a finite abelian p-group is generated by its elements of maximal order?

    Algebra is so hard to understand. 囧rz
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    Re: HELP: A finite abelian p-group generated by its elements of maximal order?

    Assume G is a finite abelian p-group. Since G is finite, there exists  a \in G such that  |a| \geq |x| where  x \in G . This is a maximal element of G. Now if G = <a> =  p^n then we are done. Assume G  \not = <a> . Then  \exists a_1 \in G such that  a_1 \not \in <a> . Now since G is a group, closed under operation,  a_1 \dot a \in G so a_1 \dot a has power  p^k \leq p^n, Now abelian groups has a property that for  (a_1 \dot a)^n = (a_1)^n \dot a^n . So using this property,  ( a_1 \dot a )^{p^k}  = (a_1)^{p^k} \dot a^{p^k} = e . If (a_1)^{p^k} \not = e then it means (a_1)^{p^k} is the inverse of  a^{p^k} , which would mean, since <a> is a subgroup, it contains its own inverses, so  (a_1)^{p^k}  \in <a> which would mean  a_1 \in <a> which we said cannot be, so that means both  (a_1)^{p^k} = e and  a^{p^k} = e since  p^k \leq p^n , this means  p^k = p^n and  a_1 is also a maximal element. Now you can see how this would proceed further, take an element not in  <a_1> or  <a> and u would come to the same conclusion about the new element and thus u can see that since G is finite, you get  {a, a_1, a_2, ..., a_k } are maximal elements which when intersected give u {e}, and which  <a> \cup <a_1> \cup ... <a_k> = G . Since all subgroups of an abelian group G, are normal in G. then indeed G =  <a> + <a_1> + <a_2> ... + <a_k>
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    Re: HELP: A finite abelian p-group generated by its elements of maximal order?

    Quote Originally Posted by jakncoke View Post
    Assume G is a finite abelian p-group. Since G is finite, there exists  a \in G such that  |a| \geq |x| where  x \in G . This is a maximal element of G. Now if G = <a> =  p^n then we are done. Assume G  \not = <a> . Then  \exists a_1 \in G such that  a_1 \not \in <a> . Now since G is a group, closed under operation,  a_1 \dot a \in G so a_1 \dot a has power  p^k \leq p^n, Now abelian groups has a property that for  (a_1 \dot a)^n = (a_1)^n \dot a^n . So using this property,  ( a_1 \dot a )^{p^k}  = (a_1)^{p^k} \dot a^{p^k} = e . If (a_1)^{p^k} \not = e then it means (a_1)^{p^k} is the inverse of  a^{p^k} , which would mean, since <a> is a subgroup, it contains its own inverses, so  (a_1)^{p^k}  \in <a> which would mean  a_1 \in <a> which we said cannot be, so that means both  (a_1)^{p^k} = e and  a^{p^k} = e since  p^k \leq p^n , this means  p^k = p^n and  a_1 is also a maximal element. Now you can see how this would proceed further, take an element not in  <a_1> or  <a> and u would come to the same conclusion about the new element and thus u can see that since G is finite, you get  {a, a_1, a_2, ..., a_k } are maximal elements which when intersected give u {e}, and which  <a> \cup <a_1> \cup ... <a_k> = G . Since all subgroups of an abelian group G, are normal in G. then indeed G =  <a> + <a_1> + <a_2> ... + <a_k>
    i think your argument has a flaw in it.

    consider the abelian group Z4xZ8, which is a p-group with p = 2.

    the element (1,1) is of order 8, which is the maximal possible order. note that:

    <(1,1)> = {(0,0),(1,1),(2,2),(3,3),(0,4),(1,5),(2,6),(3,7)}

    note that (3,1) is not in <(1,1)>. we have however that 4[(3,1)+(1,1)] = 4(0,2) = (0,0), with neither element being the identity, and 4 < 8.

    two elements of order 8 may generate cyclic subgroups that intersect in a subgroup of order 4 (such as (1,1) and (3,1)) or order 2 (such as (1,1) and (0,1)). it is this difficulty i encountered when i started to post my own solution.

    note that <(1,1),(3,1)> has 16 elements:

    <(1,1),(3,1)> = {(0,0),(1,1),(2,2),(3,3),(0,4),(1,5),(2,6),(3,7),( 3,1),(1,3),(3,5),(1,7),(2,4),(0,2),(0,6),(2,0)}

    if we consider <(1,1),(3,1),(0,1)>, we see that <(1,1),(3,1)> ∩ <(0,1)> has order 4: <(1,1),(3,1)> ∩ <(0,1)> = {(0,0),(0,2),(0,4),(0,6)}, which means that:

    Z4xZ8 is generated by 3 elements of order 8: Z4xZ8 = <(1,1),(3,1),(0,1)>.

    i think perhaps a counting argument (counting the number of elements of maximal order first) might be the way to proceed, but i haven't hit upon the right tactic yet.
    Thanks from jakncoke
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