If $a,b \in G$, show that $ab^{n}a^{-1}=(aba^{-1})^{n}$.

Question

If $a,b \in G$, show that $ab^{n}a^{-1}=(aba^{-1})^{n}$.

I have not found any question similar to this, and I am wondering how to prove this... My work so far seems very far from what needs to be shown.

I thought maybe this would work:

$ab^{n}a^{-1} = (a^{-1}a)^{n}ab^{n}a^{-1}(a^{-1}a)^{n}$

And this would somehow simplify? Can anyone show me the way?

Answer 1

Using associativity of the group operation, we have:$$(aba^{-1})^n=(aba^{-1})(aba^{-1})\cdots(aba^{-1})=ab(a^{-1}a)b(a^{-1}a)\cdots(a^{-1}a)ba^{-1}$$ so the $a^{-1}a$ terms all cancel out (i.e. they all yield the identity, and this does nothing to the product), and you're left with $b^n$ between the terminal $a$ and $a^{-1}$. You can show this more formally with induction on $n$.

Answer 2

Let $a \in G$ fixed. I claim that the map $c_a:G \to G, g \mapsto aga^{-1}$ satisfies $c_a(gh)=c_a(g)c_a(h)$. Indeed, we can compute $c_a(g)c_a(h)=(aga^{-1})(aha^{-1})=ag(a^{-1}a)ha^{-1}=a(gh)a^{-1}=c_a(gh)$. The map $c_a$ is called conjugation by $a$ and it is quite important in group theory.

Now using this property, we can show by induction that $c_a(b^n)=(c_a(b))^n$ which is a reformulation of the question. Indeed, this is trivially true for $n=1$ and if it holds for a fixed $n$, we get $c_a(b^{n+1})=c_a(b b^{n})=c_a(b)c_a(b^n)=c_a(b)(c_a(b))^n=(c_a(b))^{n+1}$, so it also holds for $n+1$.

Answer 3

Try induction.

It's is clearly true for $n=0$. Suppose it is true for some $n\geq 1$ and let's prove it for $n+1$:

$$(aba^{-1})^{n+1} = aba^{-1}(aba^{-1})^n = aba^{-1}ab^na^{-1} = ab^{n+1}a^{-1}.$$

Thus, it is true for every $n\in\mathbb N$.

To show it for every $n\in\mathbb Z$ you need just prove it for $n=-1$ and use that $x^{-n} = (x^n)^{-1}$.