Does every finite group have a finite cyclic subgroup? 
Does every finite group of size at least 2 have a finite cyclic subgroup of size at least 2?

Lagrange's theorem doesn't help, because it doesn't say anything about existence of subgroup of some order.
 A: It's not necessary to invoke Cauchy here, because you are not asking for a specific-order cyclic subgroup. 
It is trivial that if $g \in G$ then $\langle g \rangle \leq G$ (which is the smallest subgroup which contains $g$). 
And it's trivial that $\langle g \rangle \leq G$ is cyclic (in fact it contains elements of the form $g^m$ for some $m$.)
Then pick one non identity element of $G$ (it exists because $|G| \geq 2$), then it has order at least 2 (it contains at least $1_G$ and $g$) and so $\langle g \rangle \leq G$ is a cyclic subgroup of order at least 2.
A: Suppose that $G$ is a finite group of order $n$. Then there exists a prime $p$ with $p\mid n$. According to Cauchy's theorem, $G$ has an element $g$ of order $p$, and hence a cyclic subgroup $H=\langle g \rangle$ of order $p$.
Cauchy's theorem can be seen as a partial converse to Lagrange's theorem. There is also a class of groups, so called CLT groups, for which the converse of Lagrange's theorem holds. 
A finite group $G$ is a CLT-group if and only if for each positive divisor $d$ of $|G|$ there exists at least one subgroup
$H\le G$ with $|H|=d$. It turns out that every CLT-group is solvable, and every supersolvable group is a CLT-group. For example, every finite group $G$ with
$(G:Z(G))<12$ is a CLT-group. The result is best possible, because $A_4$ satisfies $(G:Z(G))=12$ and $A_4$ is not a CLT-group (it has no subgroup of order $6$, although $6\mid 12=|A_4|$).
