Every element of a finite abelian group with square free order equivalence I'm currently having some trouble with this problem:


Given $G$ a finite abelian group, prove the following are equivalent:
$1.$ Given any subgroup $H$, there exists a subgroup $K$ such that $HK = G$ and $H \cap K = \{e\}$
$2.$ Every element of $G$ has square-free order


I'm working on the $1 \to 2$ direction, but I'm not sure what I should do.  I first assumed that $1$ was true and assumed that there was a $x$ with order $p^2$ with the expectation that the case $p^kn$ would follow.  Clearly $<x>$ is a subgroup of $G$ so there exists such a $K$, and $<x>$ is isomorphic to $C_{p^2}$.  I showed that $C_{p^2}$ did not have the 1st property, and I know that $<x> \times K \cong C_{p^2} \times K \cong G$
However, I get stuck here.  Since $K$ is a finite abelian group, it can be decomposed into a product of cyclic groups, all of prime order.  If I could show that $|K|$ and $p^2$ (or $p^kn$) were relatively prime, I would be done, but I don't know if this is true.
We have the Chinese Remainder Theorem, decompositions of finite abelian groups (both cyclic and p-groups), and uniqueness.
 A: The Zhou's answer is perfect, but I want to describe more such groups.
A subgroup $K$ of a group $G$ is said to be complemented in $G$ if there exists a subgroup $H$ of $G$ such that $KH = G$ and $H\cap K= 1$. If every subgroup of $G$ has a proper supplement (complement), then $G$ is called an $aC$-group.
There are vast number papers about finite and infinite $aC$-group. In $1937$,  Hall characterized finite $aC$-groups for first time; however, he did not use term $aC$-group. He used Complemented groups instead of $aC$-group. In $2000$, Kappe and Kirtland published a paper which is a short review about $aC$-groups.
$1.$ P. Hall, Complemented groups, J. London Math. Soc. $12$ $(1937)$ $201–204$.
$2.$ L.-C. Kappe, J. Kirtland, Supplementaion in groups, Glasgow Math. J. $42$ $(2000)$ $37–50$.
A: Assume that $x\in G$ such that $|x|=p^2$. By 1, $G=<x^p> L$ for some $L \le G$ and $<x^p> \cap L=\{e\}$. Now $<x>=<x> \cap G=<x> \cap <x^p>L=<x^p> (<x> \cap L)$ (I hope you know Dedekind modular law), this is impossible. 
