Show that $\langle S^{-1}\rangle=\langle S\rangle$. In particular, $\langle a^{-1}\rangle=\langle a\rangle$, so also $o(a) = o(a^{-1})$ Problem Statement
Let $S$ be a subset of a group $G$, and let $S^{-1}$ denote $ \{ s^{-1} : s \in S \}$. Show that $\langle S^{-1} \rangle = \langle S \rangle$. In particular, $\langle a^{-1} \rangle = \langle a \rangle$, so also $o(a) = o(a^{-1})$

My attempt at a solution
The group $\langle S \rangle$ is closed under inverses thus $s_{i}^{-1} \in \langle S \rangle$. As well, it is closed under multiplication since $s_{i}s_{i}^{-1} = e \in\langle S \rangle$

*

*We can see that: $\langle S^{-1} \rangle \subseteq \langle S \rangle$
On the other hand, since inverses are unique and $\langle S^{-1} \rangle$ is closed under inverses  we can say that: $(s_{i}^{-1})^{-1} = s_{i} \in \langle S^{-1} \rangle$. As well, it is closed under multiplication as $s_{i}^{-1}s_{i} = e \in\langle S^{-1} \rangle$

*

*Hence  $\langle S \rangle \subseteq \langle S^{-1} \rangle$
Thus $\langle S^{-1} \rangle = \langle S \rangle$

Where I require guidance
I believe I have a sufficient proof for the first part of the problem statement - establishing that the elements of one group are contained in the other and that they are closed under multiplication and then using that to define a subgroup relation between the two groups then using that to make a statement about the equivalence of the two groups. Is this indeed sufficient? Am I missing any steps? Is my approach incorrect?
The next part I'm having a little uncertainty about: So $\langle a \rangle$ is the subgroup generated by the singleton $S = \{a\}$. I.e., it is the smallest subgroup of $G$ that contains the element $a$. My intuition is to just use the same argument that I used for $\langle S \rangle$ =  $\langle S^{-1} \rangle$ since this is essentially the same kind of object (a group), just a potentially different subgroup.
Could I then go on to say that $o(a) = o(a^{-1})$ by the simple fact that two equivalent groups necessarily have the same number of elements and thus the same order?

Hold on a sec here I may just be dense
Alternatively, am I misunderstanding the problem statement? The way I read it initially it seemed like there were $3$ parts to prove in this problem: $\langle S \rangle=  \langle S^{-1} \rangle$, $\langle a \rangle = \langle a^{-1} \rangle$ and $o(a) = o(a^{-1})$. Upon rereading it, it seems like maybe the $2^{nd}$ and $3^{rd}$ parts of the problem statement are just sort of nudges in the right direction of an argument. How would you interpret this problem statement?
 A: Your attempt is fine. It seems like you understand the question.

Since, for $S=\{ s_i\mid i\in I\}$ for some index set $I$,
$$\langle S\rangle =\left\{ s_{i_1}^{\varepsilon_{i_1}}\dots s_{i_k}^{\varepsilon_{i_k}}\,\middle|\, i_j\in I, k\in\Bbb N, \varepsilon_{i_j}\in\{1,-1\}\right\}\cup\{ e\},$$
we have
$$\langle S^{-1}\rangle =\left\{ s_{i_1}^{\delta_{i_1}}\dots s_{i_k}^{\delta_{i_k}}\,\middle|\, i_j\in I, k\in\Bbb N, \delta_{i_j}\in\{1,-1\}\right\}\cup\{ e\},$$
where $\delta_{i_j}=-\varepsilon_{i_j}$ implies $\langle S\rangle=\langle S^{-1}\rangle$; so if we let $S=\{a\}$, then by definition we get
$$\begin{align}
\langle a\rangle &=\langle \{ a\}\rangle \\
&=\langle \{ a\}^{-1}\rangle\\
&=\langle \{ a^{-1}\}\rangle \\
&=\langle a^{-1}\rangle.
\end{align}$$
Moreover, it is fairly simple (if not tautological by definition) that $o(x)=\lvert\langle x\rangle\rvert$ for all $x\in G$. Thus
$$\begin{align}
o(a)&=\lvert \langle a\rangle\rvert\\
&=\lvert \langle a^{-1}\rangle\rvert\\
&=o(a^{-1}).
\end{align}$$
