Let $G$ be a group and $a$ belongs to $G$. Prove that $\langle a^{-1}\rangle = \langle a\rangle$. Is this a correct approach?
I have used the result which states that $\langle a^i\rangle= \langle a^j\rangle$ if and only if $\gcd(n,i)=\gcd(n,j)$, where $n=|a|$
Let $|a|=n$
Therefore, $|a^{-1}|=n$
Now, $\gcd(n,1)=1$ and $\gcd(n,-1)=1$
Hence, $\langle a\rangle = \langle a^{-1}\rangle$.
 A: If $H$ is a subgroup of $G$ such that $a\in H$, then $a^{-1}\in H$ too. Since $\langle a\rangle$ is the smallest subgroup containing $a$, this proves that $\langle a\rangle=\left\langle a^{-1}\right\rangle$. More generally, this proves that if $S,S'\subset G$ and if $S'$ is obtained from $S$ replacing some of its elements by their inverses, then $\langle S\rangle=\langle S'\rangle$.
Your approach assumes the extra hypothesis that the order of $a$ is finite.
A: You can use the definition directly:
$$
\langle a\rangle
= \{ a^n : n \in \mathbb Z \}
= \{ a^{-n} : n \in \mathbb Z \}
= \{ (a^{-1})^{n} : n \in \mathbb Z \}
= \langle a^{-1}\rangle
$$
A: I found this to be fairly straightforward and short:
($\subseteq$)  Let $b \in \langle a^{-1} \rangle$.  Then $b = (a^{-1})^n$ where $n \in \mathbb{Z}$ .  Consider that $b = (a^{-1})^n = a^{-n}$.  Since $-n \in \mathbb{Z}$ , $b \in \langle a \rangle$.
($\supseteq$)  Let $d \in \langle a \rangle $.  Then $d = a^n$ where $n \in \mathbb{Z}$ .  Note that $d = a^n = (a^{-1})^{-n}$.  Because $-n \in \mathbb{Z}$ , $d \in \langle a^{-1} \rangle$.
Therefore, $\langle a^{-1} \rangle = \langle a \rangle$.
A: For a morphism perspective: For $a\in G$ the subgroup $\langle a\rangle$ is the image of the unique homomorphism $f_a\colon \mathbb Z \to \mathbb Z$ determined by $1\mapsto a$. This is basically the definition of the generated subgroup. So, to show $\langle a\rangle=\langle a^{-1}\rangle$ we must show that the images of $f_a$ and $f_{a^{-1}}$ are the same. This is true since $f_{a^{-1}}=  f_a\circ \psi$, where $\psi \colon \mathbb Z \to \mathbb Z$ is the isomorphism $n\mapsto -n$. Pre-composing with a bijection does not affect the image, hence the desired equality of the images. 
