# Prove that $\langle a \rangle = \langle a^{-1} \rangle$

Let $G$ be a group and let $a \in G$. Prove that $\langle a \rangle = \langle a^{-1} \rangle$.

(where $\langle a \rangle = \{\ldots,-2,-1,0,1,2,\ldots\} = \{a^k : k \in \mathbb{Z}\}$)

My attempted work:

If $k=0$, then $a^k = e$ ($e$ being the identity element).

Then $(a^{-1})^k = a^{-1\cdot k}=(a^{k})^{-1}=e^{-1}=e$

Since $a \in G$, by inverse property there exists $a^{-1} \in G$ s.t. $a * a^{-1} = e = a^{-1} * a$. And G can be a subgroup of itself. So I think Lagrange's theorem can be used here, but not too sure about that. I believe that if $a^k = e$ and $(a^{-1})^k = e$, then by Lagrange's theorem, order of $a$ divides order of $a^{-1}$. Does this mean $\langle a \rangle$ divides the order of $\langle a^{-1} \rangle$? Nonetheless, I'm unable to establish the relation $\langle a \rangle = \langle a^{-1} \rangle$.

You cannot argue with order because nothing here need be finite. But: By definition, $\langle a\rangle$ is the smallest subgroup of $G$ that contains $a$. As a subgroup, it also contains $a^{-1}$ and with that also the smallest subgroup containing $a^{-1}$. In other words, $\langle a^{-1}\rangle\le \langle a\rangle$. By the same argument $\langle a\rangle\le \langle a^{-1}\rangle$, hence equality.
$\langle a^{-1} \rangle = \{a^{-k} : k \in \mathbb{Z}\} = \{a^{k} : k \in \mathbb{Z}\} = \langle a^k \rangle$
• If I may ask, how are $a^{-k}$ and $a^{k}$ equal in this context? Feb 2 '14 at 22:00
• They are not equal, it's just that $\mathbb{Z}$ is read "in the opposite order". $\{-k:k\in\mathbb{Z}\} = \{k:k\in\mathbb{Z}\}$ Feb 2 '14 at 22:01
$$a^{-1}\in\langle a\rangle\quad\land\quad a\in\langle a^{-1}\rangle$$ and notice that if an element $g$ is in a subgroup $H$ then $\langle g\rangle\subset H$.