Finite abelian p-group and an element of maximal order

I'm studying for an exam and I'm having trouble understanding the proof given for the following statement:

Suppose $G$ is a finite abelian $p$-group and $a \in G$ has maximum order, then there exists a subgroup $K \subseteq G$ such that:

$<a>\ast$ $K$ $= G$

$<a> \cap$ $K$ $= \{e\}$

What I have written down seems disjoint, so I probably missed a few details from lecture. Could anyone give me the proof, or a reference to one? For reference, this particular proof started with choosing $b \in G/<a>$ of minimal order and showing that $<a> \cap <b> = \{e\}$ and $|b|=p$, but it already has lost me by that point.

This was given near the beginning of the course: at that time, we only knew the Chinese Remainder theorem and that, given a finite abelian $G$ such that $\forall x \in G$, $x^{nm} = e$ with $\operatorname{gcd}(m,n) = 1$, if we define $G_n = \{x \in G : x^n = e\}$ and $G_m = \{x \in G : x^m = e\}$, we have that $G \cong G_n \times G_m$

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If you are supposed to know clasification of finite abelian group,it is trivial. –  mesel Mar 9 '14 at 23:01
@mesel This proof precedes that theorem. I'll clarify what I meant by splitting into a product of subgroups. –  Lost Mar 9 '14 at 23:02
@mesel Question has been edited to make what I meant more precise –  Lost Mar 9 '14 at 23:06
well, I guess I should sleep :) I can not think properly, I can say you can try induction, it seems to me work. –  mesel Mar 9 '14 at 23:25
@mesel No problem, I wasn't that clear anyway. However, I'm not sure how induction applies here. –  Lost Mar 9 '14 at 23:28

Actually, I want to write it as a comment

If you showed that there is an element $b$ of $G$ with oreder $p$ not contained in $<a>$,

Use induction on $|G|$,

$G/<b>$ satisfes the hypotesis,and $\bar{a}$ has maximum possible order in $G/<b>$ since $<a>\cap <b>=1$.(as $\bar{<a>}\cong <a>/(<a>\cap <b>)$)

Thus,$\bar{G}=\bar{<a>}\bar{K}$ such that $\bar{<a>}\cap \bar{K}=1$.By using isomorphism theorem, you can conclude that $G=<a>K$ and $<a>\cap K =1$.

Thus, key step is showing such a $b$ exists,After that you can use induction.

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