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

- Nov 12th 2012, 10:38 PMRitaHELP: 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 - Nov 19th 2012, 03:59 AMjakncokeRe: 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 $\displaystyle a \in G $ such that $\displaystyle |a| \geq |x| $ where $\displaystyle x \in G $ . This is a maximal element of G. Now if G = <a> = $\displaystyle p^n$ then we are done. Assume G $\displaystyle \not = $ <a> . Then $\displaystyle \exists a_1 \in G$ such that $\displaystyle a_1 \not \in <a> $. Now since G is a group, closed under operation, $\displaystyle a_1 \dot a \in G$ so $\displaystyle a_1 \dot a $ has power $\displaystyle p^k \leq p^n$, Now abelian groups has a property that for $\displaystyle (a_1 \dot a)^n = (a_1)^n \dot a^n $. So using this property, $\displaystyle ( a_1 \dot a )^{p^k} = (a_1)^{p^k} \dot a^{p^k} = e $. If $\displaystyle (a_1)^{p^k} \not = e $ then it means $\displaystyle (a_1)^{p^k} $ is the inverse of $\displaystyle a^{p^k} $, which would mean, since <a> is a subgroup, it contains its own inverses, so $\displaystyle (a_1)^{p^k} \in <a> $ which would mean $\displaystyle a_1 \in <a> $ which we said cannot be, so that means both $\displaystyle (a_1)^{p^k} = e $ and $\displaystyle a^{p^k} = e $ since $\displaystyle p^k \leq p^n $, this means $\displaystyle p^k = p^n $ and $\displaystyle a_1 $ is also a maximal element. Now you can see how this would proceed further, take an element not in $\displaystyle <a_1> $ or $\displaystyle <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 $\displaystyle {a, a_1, a_2, ..., a_k } $ are maximal elements which when intersected give u {e}, and which $\displaystyle <a> \cup <a_1> \cup ... <a_k> = G $. Since all subgroups of an abelian group G, are normal in G. then indeed G = $\displaystyle <a> + <a_1> + <a_2> ... + <a_k> $

- Nov 19th 2012, 12:21 PMDevenoRe: HELP: A finite abelian p-group generated by its elements of maximal order?
i think your argument has a flaw in it.

consider the abelian group Z_{4}xZ_{8}, 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:

Z_{4}xZ_{8}is generated by 3 elements of order 8: Z_{4}xZ_{8}= <(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.