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?

  • 1
    $\begingroup$ First part looks okay. The second follows directly from the first by setting $S = \{a\}$, as you suggested. Finally, for a group $G$ and $a \in G$, we can think of $o(a)$ as being the cardinality of the subgroup $\langle a \rangle \subset G$. Hence, by what you've shown in the first two parts, $o(a) = |\langle a \rangle| = |\langle a^{-1} \rangle| = o(a^{-1})$. $\endgroup$
    – jonan
    Jan 2, 2022 at 20:52
  • $\begingroup$ How do you define $o(x)$? Because some people define it to be $\lvert \langle x\rangle\rvert$. $\endgroup$
    – Shaun
    Jan 2, 2022 at 20:54
  • 2
    $\begingroup$ The first part is nice, but, if your definition of $\langle S\rangle$ is $\{x_1\cdots x_n:x_i\in S\cup S^{-1}\}$, then the equality $\langle S\rangle=\langle S^{-1}\rangle $ is trivial. $\endgroup$
    – moqui
    Jan 2, 2022 at 21:07

1 Answer 1


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

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

  • 1
    $\begingroup$ Okay great. I thought as much but I guess I'm still in the assuming everything is more complicated than it is phase of this course and not allowing myself to just see what's there. Thanks again for the comprehensive answer @Shaun $\endgroup$ Jan 3, 2022 at 17:35
  • 1
    $\begingroup$ You're welcome, @KyleO'Connell. Such a "phase" is routine and, for me, it helps to remind myself that things can seem easy once you figure them out and that, later on, when you're not immersed in the subject as you were when it made sense, things can seem complicated again; the thing is: it is that complicated, usually. Give yourself some credit. $\endgroup$
    – Shaun
    Jan 3, 2022 at 17:39
  • 1
    $\begingroup$ Great advice. I will try to internalize it and go forward with renewed confidence! As always, be safe and be well my friend. $\endgroup$ Jan 3, 2022 at 17:43

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