Order of elements within a group If $G$ is a finite group of order (size) $n$ then, for any $g \in G$, the order (period) of $g$ is a divisor of $n$.
Proof: $g$ must have finite order since $G$ is finite. If the order (period) of $g$ is $m$ then the order (size) of the cyclic subgroup $\left<g\right>$ it generates is also $m$. Since $G$ has a subgroup of size $m$, Lagrange's Theorem tells us that $m$ is a divisor of $n$.
What does this proof mean by "$g$ must have finite order since $G$ is finite". Firstly, are we talking about the order (period) or order (size) and either way why does $G$ being finite mean $g$ has finite order?
 A: If $G$ is finite, and $g \in G$, look at the sequence $g^i$ for integer $i \ge 1$, i.e. $g^i = \{ g, g^2, . . . , g^k, g^{k + 1}, . . . , \text{etc.} \}$  Since $G$ is finite, this sequence of elements of $G$ must repeat itself at some point; thus we must have $g^l = g^k$ for $l > k \ge 1$.  If we then choose $k$ to be the least integer for which such repetition occurs for some $l$, and $l$ to be the first integer after $k$ for which it does occur, we must have $g^{l - k} = e$, the identity of $G$, and $g^m \ne e$ for any $m < l - k$.  The order of $g$ is thus $l - k$.  Thus $G$ finite implies the order of any $g \in G$ finite, and now the argument based on Lagrange's theorem shows that the order of $g$ must divide that of $G$.
Hope this helps.  Happy Holidays,
and as always,
Fiat Lux!!!
A: For your first question: The period of $g$.
For your second question:
Suppose for a contradiction that g does not have a finite period. Then it will generate an infinite number of elements, resulting in an group with infinite number of elements which is a contradiction to the supposition that $G$ is finite.
A: It depends on the definition of order of an element.
My preferred definition is

Let $G$ be a group and $g\in G$. The order of $g$ is the number of elements of $\langle g\rangle$, if finite; otherwise we say the order is infinite.

Given $g\in G$, we can define $\varphi_g\colon\mathbb{Z}\to G$ by
$$
\varphi_g(n)=g^n \qquad(n\in\mathbb{Z})
$$
and $\varphi_g$ is easily seen to be a homomorphism; the image of $\varphi_g$ is exactly $\langle g\rangle$.
The homomorphism theorem then says that $\varphi_g$ induces an isomorphism between $\mathbb{Z}/\ker\varphi_g$ and $\langle g\rangle$. A known fact about subgroups of $\mathbb{Z}$ gives that $\ker\varphi_g=k\mathbb{Z}$, for a unique $k\ge0$.
If $k=0$, $\varphi_g$ is injective and so $g$ has infinite order.
Otherwise $k$ is exactly the order of $g$. In this case, $g^k=\varphi_g(k)=1$ and, for $0<n<k$, $g^n\ne1$, because $n\notin k\mathbb{Z}=\ker\varphi_g$. Another consequence, in this case, is that if $g^n=1$, then $k\mid n$.
Of course, if $G$ is finite, also $\langle g\rangle$ is finite and so the order of $g$ divides $|G|$ by Lagrange's theorem.

Defining the order of $g$ as the least positive integer $k$, if it exists, such that $g^k=1$, forces proving again, but in a complicated way, that the subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$. I see no reason for not using the homomorphism theorem, which is one of the most basic tools in group theory.
However, as already said in other answers or comments, when $G$ is finite there are two distinct positive integers $m<n$ such that $g^m=g^n$, so $g^{n-m}=1$. Therefore the order of $g$ is finite, according to the other definition.
A: Suppose $g$ is an element of the group $G$ with infinite order. What is the order of the group $G$ then? Can it be finite?
