If you have a group $G$ with $|G|={\rm ord}(G)=n$, does that imply that there exists an element $g$ in $G$ such that ${\rm ord}(g)=n$? I am new to group theory this semester and have found myself getting confused with regards to the order of elements in groups.
My question is if you have a group $G$ with $|G|={\rm ord}(G)=n$, does that imply that there exists an element $g$ in $G$ such that ${\rm ord}(g)=n$?
I know that this would hold true for cyclic groups by definition, but I am having a hard time understanding if this applies to other kinds of groups as well.
 A: No.
Consider $\Bbb Z_2\times \Bbb Z_2$, which has order four. All its nontrivial elements have order two.
A: No if you take $$G = \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$$ Then you have $|G| = 4$, but for all $g = (a,b)\in G$ we have $g^2 = (a^2,b^2) = (e,e)$. So no elements of your group have order $4$.
A: Just to move away from the order $4$ (minimal) counterexample, consider $S_3$, of order $6$: all its nontrivial elements have order $2$ or $3$.
A: Despite all the counterexamples, the statement is true if and only if $G$ is cyclic of order $n$. That is $G \cong \mathbb{Z}/n\mathbb{Z}$.
A: What a cute question.
Well,  just to reiterate @NickyHekster, a group is cyclic if and only if it's singly generated.  That's if and only if it can be written $G=\langle g\rangle =\{g^n:n\in\Bbb Z\}$.
If $G$ is finite of order $n$, then $G\cong\Bbb Z_n$, and $g$ has order $n$.
So the answer to your question is no.  However,  if $p\mid n$, where $p$ is prime,  then there is an element of order $p$ (Cauchy's theorem).
There's one more possibility though:  there's an infinite cyclic group.   And you're surely familiar with it.  It's the integers, denoted $\Bbb Z$.

Actually there's a group,  called the circle group,  and I believe denoted $\Bbb T$, which contains all the finite cyclic groups (and lots of copies of $\Bbb Z$) as subgroups.   That's the aforementioned $\Bbb Z_n'$s.  They are also the so-called $n-$th roots of unity. They are evenly spaced around the unit circle.
And apparently that's where the term "cyclic" originates.

One more thing.  Cyclic groups are abelian.   So as soon as a group isn't abelian,  it can't be cyclic.
