An equality in a group Let $G$ be a non-abelian group of automorphisms, where composition of automorphisms is the group operator. We have the following notation
$$g = g_1 \quad x_0 = g_1^{-1}g_2 \quad x_i = g^{-i}x_0g^i$$
where $g_1,g_2 \in G$. Now I want to show that
$$g_1^{-j}g_2^j=x_{j-1} x_{j-2}\cdots x_0$$
I have looked at it for some time now, and I can not get it to work. It is part of a proof in "Algebraic Graph Theory" by Biggs 1974.
 A: Note that $g_1^{-1}x_kg_1 = x_{k+1}$ for all $k$: indeed, $g_1^{-1}x_kg_1 = g_1^{-1}g_1^{-k}x_0g_1^kg_1 = g_1^{-(k+1)} x_0 g_1^{k+1} = x_{k+1}$. 
We now proceed by induction on $j$. If $j=1$, then the left hand side is $g_1^{-1}g_2$, and the right hand side is $x_0$, so we have equality.
Assume the asserted equality holds for $j$. Then
$$\begin{align*}
g_1^{-j-1}g_2^{j+1} &= g_1^{-1}\Bigl( g_1^{-j}g_2^j\Bigr)g_2\\
&= g_1^{-1}\Bigl( x_{j-1}x_{j-2}\cdots x_0\Bigr) g_2\\
&= g_1^{-1}x_{j-1}1x_{j-2}1\cdots 1x_01g_2\\
&= g_1^{-1}x_{j-1}(g_1g_1^{-1})x_{j-2}(g_1g_1^{-1})\cdots (g_1g_1^{-1})x_0(g_1g_1^{-1})g_2\\
&= (g_1^{-1}x_{j-1}g_1) (g_1^{-1}x_{j-2}g_1) \cdots (g_1^{-1}x_0g_1) g_1^{-1}g_2\\
&= x_{j}x_{j-1}x_{j-2}\cdots x_1(g_1^{-1}g_2)\\
&= x_jx_{j-1}\cdots x_1x_0,
\end{align*}$$
as desired.
Note. It doesn't matter that the elements of $G$ are automorphisms, just that you have a group. The fact that $G$ is nonabelian is also irrelevant (though if $G$ is abelian then $x_i = x_0$ for all $i$, and the equation reduces to $(g_1^{-1}g_2)^j = x_0^j$, which is trivially true). And we don't actually need to rename $g_1$ (though it was probably done for some reason in the proof you were reading). 
