In a group $G$, if $aba^{-1}=b^i$, show that $a^rba^{-r}=b^{(i^r)}$ for all positive integers $r$.

In a group $G$, if $aba^{-1}=b^i$, show that $a^rba^{-r}=b^{(i^r)}$ for all positive integers $r$.

If we interpret it as a rule then we will that $a(aba^{-1})a^{-1}={(b^{i})}^i=b^{i^2}$ and so on, then problem solved. But if it's a special case, we propably won't have $a(b^i)a^{-1}=(b^i)^i$, so is this sort of different perspectives different results?

• Is it $b^{\left(i^r\right)}$ or $\left(b^i\right)^r$? Apr 26 '14 at 10:02
• $b^{ii...i}$ so you get it Apr 26 '14 at 10:13
• $(b^i)^r$ would normally be written as $b^{ir}$. Apr 26 '14 at 10:14

This has to do with the fact that the mapping $\varphi_a\colon G\to G$ defined by $$\varphi_a(x)=axa^{-1}$$ is an automorphism of $G$ (that is, a bijective homomorphism, prove it). But there's more: the mapping $$a\mapsto \varphi_a$$ defines a group homomorphism $G\to\operatorname{Aut}(G)$, the codomain being the group of automorphisms of $G$ under map composition. Indeed, for $x\in G$, \begin{align} \varphi_{ab}(x)&=(ab)x(ab)^{-1}=(ab)x(b^{-1}a^{-1})=a(bxb^{-1})a^{-1}\\ &=\varphi_a(bxb^{-1})=\phi_a(\varphi_b(x))\\ &=\varphi_a\circ\varphi_b(x) \end{align} so we have proved that $\varphi_{ab}=\varphi_a\circ\varphi_{b}$.

In particular, $\varphi_{a^r}=\underbrace{\varphi_a\circ\dots\circ\varphi_a}_{\text{$r$times}}$.

Now, in your case, $\varphi_a$ acts on $b$ as $\varphi_a(b)=b^i$; by definition, $\varphi_{a^0}(b)=b=b^{i^0}$, which is the basis for the induction. Then, assuming the result true for $r$, we can do \begin{align} \varphi_{a^{r+1}}(b)=\varphi_{a\cdot a^r}&=\varphi_{a}\circ\varphi_{a^r}(b) &&\text{$a\mapsto\varphi_a$ is a homomorphism}\\ &=\varphi_a(\varphi_{a^r}(b)) &&\text{definition of $\circ$}\\ &=\varphi_a(b^{i^r}) &&\text{induction hypothesis}\\ &=\bigl(\varphi_a(b)\bigr)^{i^r} &&\text{$\varphi_a$ is a homomorphism}\\ &=(b^i)^{i^r} &&\varphi_a(b)=b^i\\ &=b^{i\cdot i^r}=b^{i^{r+1}} \end{align}

Notice that $x\cdot yz\cdot x^{-1}=xyx^{-1}\cdot xzx^{-1}$.

• Thank you sir, but sorry, I still don't get it Apr 26 '14 at 9:35

I'll do it for $r=2$, and then you can try the general case by induction.

$$a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^ia^{-1} = (aba^{-1})^i = (b^i)^i = b^{i^2}.$$

• So it's a rule not a special case. Apr 26 '14 at 10:16

You can prove this by induction on r:

$$a^rb a^{-r} = a(a^{r-1}ba^{-(r-1)})a^{-1}=ab^{(i^{r-1})}a^{-1}=(aba^{-1})^{(i^{r-1})}=(b^i)^{(i^{r-1})}=b^{(i^r)}$$

Induction is of course the easiest.

But you can also note how by repeatedly applying the given condition:
$b^{i^{r}}=\big(aba^{-1} \big)^{i^{r-1}} = \big( (aba^{-1})^{i}\big)^{i^{r-2}} = \big(a\cdot aba^{-1} \cdot a^{-1} \big)^{i^{r-2}} = \big( a^{2}ba^{-2} \big)^{i^{r-2}}$