Prove that a finite abelian group is simple if and only if its order is prime. So I'm having trouble with this problem. I know that the definition of a simple group means that the group has no nontrivial subgroups. I know that this can be proven somehow with the help of the converse of Lagrange's Theorem for Abelian groups: If G is abelian of order n, and d is a divisor of n, then G has a subgroup of order d. 
My attempt:
(=>)Assume that G is a finite abelian group and is simple, then G has no nontrivial normal 
subgroups. 
(Now I don't know how to show that this implies that G has order p, where p is prime.
(<=)Assume that G is a finite abelian group with order p, where p is a prime. (Since the order of p is prime then what does this mean?)
Edit: Can someone check my new attempt at the proof? 
(=>) Suppose G is a simple finite abelian group. Suppose for the sake of contradiction that G does not have prime order, then |G|=p*k where p is a prime number and k is an integer such that k>1. Then G has an element of order p. Let the element of order p be called x. Then , the subgroup generated by x, is of order p and  is not all of G. Since G is abelian, this subgroup is normal, which leads us to a contradiction. Therefore, G must have prime order. 
(<=) Suppose that G is a finite abelian group and it’s order is p, a prime. Since G has prime order, then the only two subgroups of G are the trivial subgroup and the group G. Then, by definition the group G is simple since there are no nontrivial proper subgroups, and thus no nontrivial normal subgroups. 
 A: Edit: Can someone check my new attempt at the proof?
(=>) Suppose G is a simple finite abelian group. Suppose for the sake of contradiction that G does not have prime order, then |G|=p*k where p is a prime number and k is an integer such that k>1. Then G has an element of order p. Let the element of order p be called x. Then , the subgroup generated by x, is of order p and is not all of G. Since G is abelian, this subgroup is normal, which leads us to a contradiction. Therefore, G must have prime order.
(<=) Suppose that G is a finite abelian group and it’s order is p, a prime. Since G has prime order, then the only two subgroups of G are the trivial subgroup and the group G. Then, by definition the group G is simple since there are no nontrivial proper subgroups, and thus no nontrivial normal subgroups.
A: If $G$ is a simple finite abelian group, then it has no non-trivial subgroups, since all subgroups are normal because $G$ is abelian. This means that $G$ is cyclic, generated by any non-trivial element $g$. Let $n$ be the order of $G$. If $d$ divides $n$, then $\langle g^d \rangle$ is a subgroup of $G$. Thus, $d$ can only be $1$ or $n$, and so $n$ is prime.
If $G$ is a finite group whose order is prime, then $G$ is cyclic, generated by any non-trivial element. This implies that $G$ has no non-trivial subgroups.
A: Hint :
For $\Rightarrow $ : 
Suppose group has order $|G|=pqk$ where $p,q$ are distinct primes.
Now, what doe Cauchy theorem tells  for finite abelian groups?
For $\Leftarrow $ :
For a group to be simple, It should have some non trivial subgroup.
Order of a subgroup divides order of group.
A: Your proof seems good.  Here's a different approach without using cauchy's theorem.  
$(\Longrightarrow)$ If $G$ is a finite simple abelian group, pick any $g \in G$ non trivial.  Then $1  \not =<g> \trianglelefteq G$ since $G$ is abelian.  Since $G$ is simple, $G = <g>$.   That is, $G$ is cyclic.  Since $G$ is finite and cyclic, for every divisor $d$ of $|G|$ ,$G$ has a unique cyclic subgroup of order $d$.  $G$ is simple so that this cyclic subgroup is either $G$ or 1.  Thus the divisor $d$ is either 1 or $|G|$. This show $|G|$ must be prime.  
For fun, you can modify the problem a little bit more. 
Let $G$ be an simple abelian group, not necessarily finite.  Prove that $G$ must be finite and of prime order. 
A: I have noticed a few issues with the definition of 'simple' in a number of answers and comments. A finite group $G$ is simple it it has no non-trivial $normal$ subgroups. In the case where $G$ is abelian, every subgroup is normal by simply using the definition of normal. Just something to take note of and be carful of. 
