Existence of an element $g \in G$ of order $N$ in a finite group $G$ of order $N$

Question

Let $g \in G$

Where $G$ if finite and of order $N$

Does there exist an element $g\in G$ such that it is of order $N$ ?

If yes how do I prove it.

If no what would be an explanation.

Sorry if this question sounds trivial, I just started out in group theory and was pondering over this.

Any help or insight is deeply appreciated.

Answer 1

Counterexample : $G=D_3=\{1,x,x^2,y,xy,x^2y\}$ where $x^3=1,y^2=1,yx=x^2y$

Note that $G$ is of order $6$ and it can be easily verified that every element in $G$ is not of order 6.

If you look for explanation, the reason is if for a group $G$ of order $n$, if there exists an element $g\in G$ of order $n$ also, $g$ will generate $G$, which means that $G$ is cyclic. But since not all finite group is cyclic, your statement does not hold.

Answer 2

This is not always possible,

Take Symmetric group of degree $3$. It has order $6$ but there is no element of order $6$.

As pointed out in comments, you just have to find a counterexample to convince yourself.