Group theory - inverses: $(a^{-1})^n=(a^n)^{-1}$

Question

Given group $A$ with operation $*$ and $a \in A$, prove that $$(a^{-1})^n=(a^n)^{-1} \in A$$

To be perfectly honest, I don't quite know where to start. I think that I will need to use the fact that inverses are unique and so I have computed $$ (a^n)^{-1}*a^n=e=a^{-n}*a^{n}$$ Inverses are are unique so $(a^n)^{-1}=a^{-n}$. But how do I show $(a^{-1})^n=a^{-n}= (a^n)^{-1}$?

Answer 1

To perhaps remove any confusion that may result with induction in your case...let us attempt to take a different approach.

Given any $g \in G$ and $n \in \mathbb{N}$, by closure in $G$ it follows that $g^n \in G$.

Claim: $\left(g^{-1}\right)^n$ is the inverse element in $G$ of $g^n$.

Thus consider: $$\left(g^{-1}\right)^n * g^n = \underbrace{g^{-1} * ...* g^{-1}}_{n} * \underbrace{g * ... * g}_{n} = \underbrace{g^{-1}* ... * g^{-1}}_{n-1} * e * \underbrace{g* ... * g}_{n-1} = e$$

Similarly, $$g^n * \left(g^{-1}\right)^n = e$$

Hence $\left(g^{-1}\right)^n$ is the inverse element in $G$ of $g^n$, i.e. $\left(g^{-1}\right)^n = \left(g^n\right)^{-1}$.

Answer 2

Hint:

use induction showing that:

$ (A^2)^{-1}=(A^{-1})^2$

and

$(A^n)^{-1}=(A^{-1})^n \quad \Rightarrow \quad(A^{n+1})^{-1}=(A^{-1})^{n+1} $

Answer 3

For $n=1$ the claim is obvious. Proceed inductively:

$$(a^{-1})^n\cdot a^n = (a^{-1})^{n-1}\cdot \underbrace{a^{-1} \cdot a}_{=e} \cdot a^{n-1} = (a^{-1})^{n-1} \cdot a^{n-1} = \ldots = e$$ $$a^n\cdot (a^{-1})^n = a^{n-1}\cdot \underbrace{a \cdot a^{-1}}_{=e} \cdot (a^{-1})^{n-1} = a^{n-1}\cdot (a^{-1})^{n-1} = \ldots =e$$

Answer 4

Consider the assertion$$(\forall n\in\mathbb{N}):(a^{-1})^n*a^n=e.\tag1$$If $n=1$, this is trivial. Suppose that is true for a certain $n\in\mathbb N$. Then$$(a^{-1})^{n+1}*a^{n+1}=(a^{-1})^n*a^{-1}*a*a^n=(a^{-1})^n*e*a^n=(a^{-1})^n*a^n=e.$$So, the assertion $(1)$ holds and therefore, by the definition of inverse, for each $n\in\mathbb N$, $(a^{-1})^n$ is the inverse of $a^n$.