Order of an element and its inverse [duplicate]

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, $\langle a\rangle = \langle a^{-1}\rangle$.

Solution 1:

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 $\langle a\rangle=\langle a^{-1}\rangle$. More generally, this proves that if $S,S'\subset G$ and if $S'$ is obtained formn $S$ 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.

Solution 2:

You can use the definition directly: $$ \langle a\rangle = \{ a^n : n \in \mathbb Z \} = \{ a^{-n} : n \in \mathbb Z \} = \{ (a^{-1})^{n} : n \in \mathbb Z \} = \langle a^{-1}\rangle $$

Solution 3:

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.