I'm trying to understand a proof in "Combinatorial Group Theory" by Lyndon & Schupp, page 26, Lemma 4.7.

Here's my problem:

$F$ is a free group with $X$ as basis. $x \in X$. $\alpha \in IA(F)$ - meaning it's an automorphism of $F$ which induces the identity automorphism on $F/[F,F]$. $k \in F_n$ - a term in the descending central series for F. $x\alpha$ and $xk$ are conjugates. In addition: for some $N\geq1$, $\alpha^N \in JA(F)$ - the group of inner automorphisms of F.

"Elementary Fact" used in the proof: $\alpha \in IA(F)$ and $k \in F_n$ implies $k \alpha = k$ (modulo $F_{n+1}$).

Right after mentioning this "elementary fact" they say it implies: $x\alpha^N = xk^N$ (modulo $F_{n+1}$). Why is that?

Edit: A little clarification: $x\alpha$ is the image of $x$ under the automorphism $\alpha$, following the notation in this book. $xk$ is just $x$ times $k$, multiplication in the group $F$.

  • $\begingroup$ @user3533: You are also missing the hypothesis that $N\geq 1$ is an integer such that $\alpha^N\in JA(F)$, the group of inner automorphisms of $F$. $\endgroup$ – Arturo Magidin Nov 23 '10 at 18:31
  • $\begingroup$ @Arturo Magidin: Thank you very much. It seems like this part of the hypothesis is only used later, but I'd better add it to the post. $\endgroup$ – user3533 Nov 23 '10 at 19:50
  • $\begingroup$ @user3533: are you having trouble with both statements, or just the latter? I can give you an induction proof for the former, but I'm still trying to track down the latter (to wit, I can post a proof that $k\alpha\equiv k\pmod{F_{n+1}}$, at least for the lower central series, but I'm still trying to figure out $x\alpha^N\equiv xk^N\pmod{F_{n+1}}$). $\endgroup$ – Arturo Magidin Nov 23 '10 at 19:52
  • $\begingroup$ @Arturu Magidin: I'm having trouble with both statements. A proof of any of them would be appreciated, if you can. $\endgroup$ – user3533 Nov 23 '10 at 19:55
  • $\begingroup$ @user3533: Don't go accepting yet! I haven't done the second assertion yet. (Unless this got you going enough to figure it out). $\endgroup$ – Arturo Magidin Nov 23 '10 at 20:51

For the first assertion, I will work with $F_n$ being the $n$th term of the lower central series; but the argument should be about the same for any central series.

We show that if $\alpha$ is the identity on $F/[F,F]$, then it induces the identity on $F_n/F_{n+1}$ for all $n$. The proof is by induction.

The assertion for $n=1$ is the original assumption. Assume now that the result holds for $k$, so that if $w\in F_k$, then $w\alpha\equiv w\pmod{F_{k+1}}$. We want to show that the result holds for $k+1$. The generators of $F_{k+1}$ are contained in the set $[F,F_{k+1}]$, so let $u\in F$ and $w\in F_k$; if we prove the result for $[u,w]$, then it will follow for $F_{k+1}$.

We will use the following commutator identity: in any group, for all $x,y,z,t$, we have: $$[xy,zt] = [x,t]^y[y,t][x,z]^{yt}[y,z]^t,$$ where $a^b = b^{-1}ab$ is conjugation.

Now, since $u\alpha\equiv u\pmod{F_2}$, then $u\alpha=uc$ with $c\in F_2$. And since $w\alpha\equiv w\pmod{F_{k+1}}$, then $w\alpha = wd$ with $d\in F_{k+1}$. Therefore, $$[u,w]\alpha = [u\alpha,w\alpha]= [uc,wd] = [u,d]^{c}[c,d][u,w]^{cd}[c,w]^{d} \equiv [u,w]\pmod{F_{k+2}},$$ because $[u,d]\in[F_1,F_{k+1}]$, $[c,d]\in[F_2,F_{k+1}]$, $[c,w]\in[F_2,F_k]$, and $[u,w]^{cw}=[u,w][[u,w],cw]\equiv [u,w]\pmod_{F_{k+2}}$. Therefore, $\alpha$ acts as the identity on $F_{k+1}/F_{k+2}$, as desired.

(Alternatively, the basic commutators of weight $k$ give a basis for the free abelian group $F_{k}/F_{k+1}$, and it is an easy matter to check that if $\alpha$ induces the identity on $F/F_{2}$, then it maps a basic commutator of weight $c$ to the product of itself times commutators of weight greater than $c$).

I'm still working on the second assertion, but for instance, for $n=1$, we have that $x\alpha= xk$ for some $k\in F_2$, hence $x\alpha\equiv xk\pmod{F_3}$, and since $k\alpha\equiv k\pmod{F_3}$, you get that $x\alpha^2 \equiv x\alpha k \equiv x k^2\pmod{F_3}$, and so inductively $x\alpha^N \equiv x k^{N}\pmod{F_3}$.

Added: I double-checked the book on my way out of the office and to the gym, and I noticed that the book did not have " $x\alpha^N\equiv xk^N\pmod{F_{n+1}}$, but only $x\alpha^N\equiv xk^N$. I can show that under the given assumptions, $x\alpha^N$ is conjugate, modulo $F_{n+1}$, to $xk^N$, and that should be enough for the purposes of the argument, since the next step of the proof is to note that since $\alpha^N\in JA(F)$, then this means that $uxu^{-1}\equiv xk^N\pmod{F_{n+1}}$ for some $u$, and this will also hold if all you have is that $x\alpha^N$ is conjugate to $xk^N$ modulo $F_{n+1}$.

To show this weaker claim, I proceed by induction on $m$. We know that $x\alpha$ is conjugate to $xk$, hence $x\alpha \equiv (xk)^u\pmod{F_{n+1}}$ for some $u$. Assume that $x\alpha^m\equiv (xk^m)^v\pmod{F_{n+1}}$ for some $v\in F$, where $a^b = b^{-1}ab$ is just conjugation. Applying $\alpha$ on both sides, we have $x\alpha^{m+1}$ on the left hand side, and \begin{align*} (xk^m)^v\alpha &= ((x\alpha)(k^m\alpha))^{v\alpha}\\ & \equiv \bigl((xk)^u k^m\bigr)^{v\alpha}\pmod{F_{n+1}}\\ &\equiv \bigl( (xk^{m+1})^u\bigr)^{v\alpha}\pmod{F_{n+1}} &\text{(as $k$ is central in $F_/F_{n+1}$)}\\ &\equiv (xk^{m+1})^{u(v\alpha)}\pmod{F_{n+1}}. \end{align*} Therefore, $x\alpha^{m+1}\equiv (xk^{m+1})^w\pmod{F_{n+1}}$ for some $w\in F$; that is, $x\alpha^{m+1}$ is conjugate, modulo $F_{n+1}$, to $xk^{m+1}$. In particular, $x\alpha^N$ is conjugate, modulo $F_{n+1}$ to $xk^{N}$, and since $x$ is conjugate to $x\alpha^N$, then $x$ is conjugate to $xk^N$ modulo $F_{n+1}$. Now I think the argument given in the book can be continued from this point.

So, continuing on with the proof of the proposition. We know that $x\alpha^N$ is conjugate to $xk^N$ modulo $F_{n+1}$, and that $\alpha^N$ is equal to conjugation by some element. So there exists $u$ such that $x\alpha^N = u^{-1}xu = x[x,u]$ is equal, modulo $F_{n+1}$, to $xk^N$, and therefore $[x,u]=k^N \pmod{F_{n+1}}$ by cancelling $x$.

Now, $F_n/F_{n+1}$ is a finitely generated free abelian group generated by the basic commutators; $k\in F_n$, so we can write $k$ as a product of basic commutators, and the same is true for $[x,u]$. Because $[x,u]$ is an $N$th power, commutator calculus tells us that in fact $u$ must be an $N$th power, so that in fact we can find a $v$ such that $[x,v]\equiv k \pmod{F_{n+1}}$ (essentially, every commutator that shows up is an $N$th power, so dividing the exponent by $N$ we can construct an appropriate $v$). So now we have $[x,v]\equiv k\pmod{F_{n+1}}$.

Now: $x\alpha$ is conjugate to $xk$, so $v^{-1}x\alpha v$ is conjugate to $vxkv^{-1}$ modulo $F_{n+1}$. But $vxkv^{-1} \equiv vxv^{-1}k\pmod{F_{n+1}}$ (since $k$ is central in $F/F_{n+1}$), so $$vxkv^{-1} \equiv vxv^{-1}k \equiv x[x,v^{-1}]k \equiv x[x,v]^{-1}k \equiv xk^{-1}k\pmod{F_{n+1}}$$ (the next to last congruence by using commutator identities, since we are in the lowest term of the lower central series of $F/F_{n+1}$). In summary, $x\alpha$ is conjugate to $x$ modulo $F_{n+1}$. That means that there exists some $k'\in F_{n+1}$ and some $w\in F$ such that $w^{-1}(x\alpha)w\equiv x \pmod{F_{n+1}}$. But this means that $w^{-1}(x\alpha)w =xk'$ for some $k'\in F_{n+1}$, which is precisely what we wanted to prove to complete the inductive step.

  • $\begingroup$ Thank you! I'll try to follow this idea for the second assertion too. $\endgroup$ – user3533 Nov 23 '10 at 20:51

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