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.

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You can't get it to work because it's false. – Harald Hanche-Olsen Jun 5 '12 at 21:00
Well, that truly is sad, since it is part of the proof I need :(. Do anybody know of an errata for the book, or have the second edition where this might be fixed? – utdiscant Jun 5 '12 at 21:01
@Harald: Are you sure? Because I thought I proved it.... – Arturo Magidin Jun 5 '12 at 21:02
@Arturo: Oops, no. I read the definition of $x_i$ as $x_i=g_1^{-i}x_0g_2^{i}$, and then it would be wrong. (I was also wondering what the point of $g$ (without any index) might be, since it wasn't referenced, or so I thought. Amazing how a single misreading sticks in the brain and refuses to let go …) – Harald Hanche-Olsen Jun 5 '12 at 21:08

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).

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Can you elaborate on your third equality in the induction step? – utdiscant Jun 5 '12 at 21:18
@utdiscant This follows from the fact that $1_G=g_1g_1^{-1}$, or equivalently, he's "sticking in" $1_G=g_1g_1^{-1}$ between each $x_i$ – M Turgeon Jun 5 '12 at 21:25
@utdiscant: I added two extra steps; I trust that will clarify what is happening. – Arturo Magidin Jun 5 '12 at 23:31
@utdiscant: I also want to add: if you aren't quite clear on the answer, don't accept it until you are! – Arturo Magidin Jun 5 '12 at 23:45