Originally Posted by
jakncoke 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> $