If an element $g \in G$ has $|g| = |G|$, then is $g$ automatically the cyclic generator of $G$? Is the following statement true for all groups in general?  If the order of an element $g \in G$ is equal to $|G|$ then $G = <g>$?
And thus, if a group does not have an element that is of the same order as the group itself then it cannot be cyclic.
 A: For each $g \in G$ we have $\langle g \rangle \le G$. But if $G$ is finite group and $|g|=|G|$ then we have that $\langle g \rangle$ is contained in a group of the same order as itself, so it must be the actual group itself. Hence $\langle g \rangle = G$.
On the other hand for $G$ to be cyclic it must have a generator, i.e. an element $g \in G$ s.t. for each $a \in G$ there exists a natural number $n$ s.t. $g^n = a$. This implies that order of $g$ is greater or equal than $|G|$. But as $|g|$ divide $|G|$ we must have $|g| = |G|$.
Therefore a finite group $G$ is cyclic iff there is an element $g \in G$ s.t. $|g| = |G|$
A: Yes. If $A\subseteq B$ and $A$ and $B$ are finite sets of equal cardinality, then clearly $A=B.$ The result now follows since the subgroup generated by $g,$ which is by definition cyclic, has the same number of elements as $G.$
A: Yes for finite groups (this may not apply for infinite cardinalities), if the order of $g$ is $|G|$, then it would be a contradiction to assume otherwise that $g$ is not the cyclical generator of $G$. $g^n$ produces $n$ different elements (including the identity). By definition, $<g>$ is a subgroup with the same cardinality as $G$, so it must be that $$<g> = G$$
