Link between cyclic and abelian groups I know that every cyclic group is abelian. But what about the converse? Today in my lecture, the lecturer used the fact that a group was simple and abelian to conclude that it is cyclic of prime order. Why is this true?
 A: There are numbers $n$ such that every group of order $n$ is cyclic. These are called cyclic numbers and are characterized by $\gcd(n,\phi(n))=1$, where $\phi$ is Euler's function. See also this question.
In particular, every prime is a cyclic number, that is, every group of prime order is cyclic.
Not every cyclic number is prime. Examples are $15, 33, 35$. See A050384 .
There is also the notion of abelian number. See this question.
Finally, every abelian group of order $n$ is cyclic iff $n$ is squarefree. See this question.
A: 
But what about the converse?

Is false. Take $G=\mathbb{Z}_2\oplus\mathbb{Z}_2$. Or more explicitely $G=\{e, a, b, c\}$ with addition
$$e\mbox{ is neutral}$$
$$a+a=e$$
$$b+b=e$$
$$c+c=e$$
$$a+b=c$$
$$a+c=b$$
$$b+c=a$$
Also known as the Klien four-group. This group is abelian. It is not cyclic because it is of order $4$ while every (nontrivial) element is of order $2$.
Finite (or even finitely generated) abelian groups are all known. If $G$ is a finite abelian group then
$$G\simeq\mathbb{Z}_{p_1^{a_1}}\oplus\cdots\oplus \mathbb{Z}_{p_m^{a_m}}$$
for some (not necessarily distinct) primes $p_1,\ldots, p_m$ and naturals $a_1,\ldots,a_m$.
For the finitely generated case you have to add $\mathbb{Z}^k$ term.

a group was simple and abelian to conclude that it is cyclic of prime order. Why is this true?

Abelian groups have this neat property: every subgroup is normal. Now if $G$ is of order $n$ then by Cauchy's theorem it has a subgroup $H$ of prime order $p | n$. This subgroup is nontrivial. It is proper if $n$ is not prime. If additionally $G$ is abelian then $H$ is normal, hence $G$ is not simple.
So the only possibility for an abelian group to be simple is when it is of prime order. Again Cauchy's theorem implies that such group has to be cyclic and indeed every $\mathbb{Z}_p$ is simple.
A: "what additional conditions do you need to conclude that it's cyclic?" I think "cyclic" is the most describing criterion that is both necessary and sufficient. Or equivalent things, like "there is an element $a$ in the group such that each element of the group is a multiple of $a$", or "has an element of order equal to the order of the group" or something. I'm sorry if that seems disappointing.
If you have access to some decomposition of your group into products of cyclic groups, then your group is cyclic iff all the orders of the groups in the decomposition are pairwise coprime.
Simple and abelian, however, is equivalent to cyclic of prime order. You can see this by taking the subgroup generated by any non-identity element $a$. It must be the whole group, because the group is simple. And it must have prime order because if the order of $a$ is $nm$, with $m,n>1$, then $a^m$ generates a proper, non-trivial subgroup.
A: If your group is abelian and simple, then its subgroups are normal (as it is abelian) and its only subgroups are itself and {e} being the identity element (as it is simple). Let $G= [{e,a_1, a_2,...,a_n}]$. Show $G$ is cyclic of order $p$. 
So we must show that any element of $G$ must generate it to prove the order is prime, as the subgroup's order, generated by any element of $G$ must equal $G$ itself. 
Let $a_i$ $0<i<n$.  $H=<a_i>$ is the smallest non trivial subgroup of $G$. So $|H|$ must divide $|G|$ by Lagrange Theorem.  But G is simple, therefore for $|H|$ to divide $|G|$, $|H|$ must equal $|G|$, so $H=G$. So any element of $G$ generates $H$ which is equal to $G$, which means the order of each element of $G$ is $|G|$. 
The order of an element is equal to the order of the group divided by the $gcd$ of the element and the order $(|a_i|=|G|/|gcd(a_i,|G|)$ Lagrange). So if all elements orders are $G$, then the gcd between the elements an G is 1, so all elements are co-prime to $|G|$. So $|G|$ must be prime. So G is cyclic of order p.  
