Let $G$ a group and $a\in G$ such that $a$ is an element of finite order, then $\lvert \langle a \rangle\rvert=o(a)$. I have a theorem to establish:

Let $G$ a group and $a\in G$ such that $a$ is an element of finite order, then $\lvert \langle a \rangle\rvert=o(a)$.

We denote $o(a)$ as the order of $a$.
So my question is...

Is it true that $G= \langle a \rangle$ if and only if $\lvert G \rvert=o(a)$?

I am getting confused if it follows by definition or it requires a formal proof.
THANKS, for the help :D

You are correct. If for some $a \in G$ we have $|G| = o(a) < \infty$, then we must have
$G = \langle a \rangle$.
To see why, we can use containment and finiteness.
Can you show $\langle a \rangle \subseteq G$? This will use the fact that $G$ is closed under its multiplication, and $a \in G$. If you want to be extra formal, you might show each $a^n \in G$ by induction on $n$.
Next, we use a crucial fact about finite sets. If $|X| = |Y| < \infty$ and $X \subseteq Y$, then $X = Y$. That is, when we're in the finite world, you cannot pull any hilbert's hotel type tricks. So if $X \subseteq Y$ and they are the same size, they must actually be the same.
But we showed earlier that $\langle a \rangle \subseteq G$, and we are assuming that $|G| = | \langle a \rangle | < \infty$. So $G = \langle a \rangle$.

I hope this helps! ^_^
