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?
Best Answer
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
$$\begin{align} o(a)&=\lvert \langle a\rangle\rvert\\ &=\lvert \langle a^{-1}\rangle\rvert\\ &=o(a^{-1}). \end{align}$$