$\underline{\textbf{Answer:}\Rightarrow}\;(3)$
$\underline{\textbf{Explanation:}\Rightarrow}$
$\underline{\textbf{Proof for statement}\;\mathbf{S_1:}}$
This is the same as proving $\mathbf n$ is a multiple of $\mathbf{O(g)}$.
Let $\mathbf{g^n =e}$, so $\mathbf g$ has a finite order, say $\mathbf{m = O(g)}$.
By the division algorithm, there exist, unique integers $\mathbf {r,\;q}\in \mathbb Z$ such that:
$\mathrm {n = mq + r},\;\;\;\mathbf{0 \le r <m}$
Now, $\mathrm{e = g^n = g^{mq+r} = g^{mq}.g^r = \left (g^m \right )^q.g^r = e^q.g^r = eg^r = g^r}$
$\therefore \mathrm{g^r = e}$, but $\mathbf m$ is the smallest prositive integer such that: $\mathbf{g^m = e}$ and $\mathbf{r < m}$, this makes $\mathbf{r = 0}$
Then, $\mathbf{mq + 0}\Rightarrow \mathbf{n = mq}$
$\Rightarrow \mathbf{ m\mid n}$
$\therefore \mathbf{O(g)\mid n}$
$\underline{\textbf{Proof for statement}\;\mathbf{S_2:}}$
On pairing the each element of $\mathbf G$ with its inverse, and observing that:
$\mathrm{g^2 \neq e \iff g \neq g^{-1} \iff \text{there exists the pair} (g, g^{-1})}$
Now, $\because $ there is one element left which has no pairing, that is $\mathbf {e}\;\;,\text{($\because e = e^{-1} \iff e^2 = e$ )}$
Then, $\because$ the number of elements of $\mathbf{G}$ is even there $\color{red}{\textbf{must}}$ be at least one element more, let $\mathbf{e \neq a, a \in G}$, without pairing.
$\therefore \mathbf{a = a^{-1} \iff a^2 = e}$
$\therefore $ Both the statements are true and the correct option is $\mathbf{(3)}$
