Chapter6: AutomorphismsofFreeGroups

Chapter 6: Automorphisms of Free Groups

LÉTOFFÉ Olivier

Université Toulouse 3 – Faculté des sciences et ingénierie

November 13, 2020

LÉTOFFÉ (UT3 – FSI) Chapter 6: Automorphisms of Free Groups November 13, 2020 1 / 11

I Generalities and Tools



Definition (Automorphism):Let G be a group, we call Automorphism a morphism ϕ : G → G .Aut(G) = {ϕ : G → G | ϕ Automorphism} group (for composition).

Definition (Inner Automorphism):

∀g ∈ G , we call ρg :{

G → Gh 7→ ghg−1 ∈ Aut(G) an Inner Automorphism.

In(G) = {ρg | g ∈ G}C Aut(G) normal subgroup.

Examples:

←→ϕ :{

F2 → F2 =< a; b >(wi)i∈J1;KK 7→ (←→wi )i∈J1;KK

∈ Aut(F2).∣∣∣∣∣ ←→a = b

←→b = a

←→a−1 = b−1

←→b−1 = a−1

ρabc :{

F3 → F3 =< a; b; c >w 7→ abcwc−1b−1a−1 ∈ In(F3).




Definition (Inner Automorphism / Outer Automorphism):



In(G) = {ρg | g ∈ G}C Aut(G) normal subgroup.∃!Out(G) < Aut(G) s.t. Aut(G) ' In(G)× Out(G).Out(G) subgroup of Outer Automorphisms.

Fact:In(G)× Z (G) ' G with Z (G) = {g ∈ G | ∀z ∈ G , gz = zg}.

In particular: In(Zn) ' {Id} ' In(F1) and ∀n > 2, In(Fn) ' Fn.




Definition (Inner Automorphism / Outer Automorphism):



In(G) = {ρg | g ∈ G}C Aut(G) normal subgroup.∃!Out(G) < Aut(G) s.t. Aut(G) ' In(G)× Out(G).Out(G) subgroup of Outer Automorphisms.

Fact:In(G)× Z (G) ' G with Z (G) = {g ∈ G | ∀z ∈ G , gz = zg}.In particular: In(Zn) ' {Id} ' In(F1) and ∀n > 2, In(Fn) ' Fn.


II Basis of Zn and Fn



Proposition:∀n ∈ N∗,Aut(Zn) ' GLn(Z) = {M = (mk;l)k;l∈J1;nK ∈ Mn(Z) | Det(M)2 = 1}.




Proof:Let (vk)k∈J1;nK be the canonical basis of Zn.

∀ϕ ∈ Aut(Zn),with Mϕ = (mk;l)k;l∈J1;nK ∈ Mn(Z) s.t. ∀l ∈ J1; nK, (mk;l)k∈J1;nK = ϕ(vl).Det(Mϕ) ∈ Z. ϕ−1 ∈ Aut(Zn), so Det(Mϕ−1) ∈ Z. But Mϕ−1 = Mϕ

−1:

ϕ−1

(ϕ

( n∑k=1

(xkvk)))

= ϕ

(ϕ−1

( n∑k=1

(xkvk)))

=n∑

k=1(xkvk)

so Mϕ−1Mϕ = MϕMϕ−1 = Id , so Det(Mϕ); Det(Mϕ)−1 ∈ Z: Mϕ ∈ GLn(Z).∀M ∈ GLn(Z),∃ϕ ∈ Aut(Zn) s.t. M = Mϕ is obvious so Aut(Zn) ' GLn(Z) �


II Basis of Zn and FnProposition:∀n ∈ N∗,Aut(Zn) ' GLn(Z) = {M = (mk;l)k;l∈J1;nK ∈ Mn(Z) | Det(M)2 = 1}.

Proof:Let (vk)k∈J1;nK be the canonical basis of Zn.∀ϕ ∈ Aut(Zn),

ϕ

( n∑k=1

(xkvk))

=n∑

k=1(ϕ(xkvk)) =

n∑k=1

(xkϕ(vk))

with Mϕ = (mk;l)k;l∈J1;nK ∈ Mn(Z) s.t. ∀l ∈ J1; nK, (mk;l)k∈J1;nK = ϕ(vl).Det(Mϕ) ∈ Z. ϕ−1 ∈ Aut(Zn), so Det(Mϕ−1) ∈ Z. But Mϕ−1 = Mϕ

−1:

ϕ−1

(ϕ

( n∑k=1

(xkvk)))

= ϕ

(ϕ−1

( n∑k=1

(xkvk)))

=n∑

k=1(xkvk)





ϕ

( n∑k=1

(xkvk))

=n∑

k=1(ϕ(xkvk)) =

n∑k=1

(xkϕ(vk)) = Mϕ

n∑k=1

(xkvk)

with Mϕ = (mk;l)k;l∈J1;nK ∈ Mn(Z) s.t. ∀l ∈ J1; nK, (mk;l)k∈J1;nK = ϕ(vl).

Det(Mϕ) ∈ Z. ϕ−1 ∈ Aut(Zn), so Det(Mϕ−1) ∈ Z. But Mϕ−1 = Mϕ−1:

ϕ−1

(ϕ

( n∑k=1

(xkvk)))

= ϕ

(ϕ−1

( n∑k=1

(xkvk)))

=n∑

k=1(xkvk)





ϕ

( n∑k=1

(xkvk))

=n∑

k=1(ϕ(xkvk)) =

n∑k=1

(xkϕ(vk)) = Mϕ

n∑k=1

(xkvk)

with Mϕ = (mk;l)k;l∈J1;nK ∈ Mn(Z) s.t. ∀l ∈ J1; nK, (mk;l)k∈J1;nK = ϕ(vl).Det(Mϕ) ∈ Z. ϕ−1 ∈ Aut(Zn), so Det(Mϕ−1) ∈ Z.

But Mϕ−1 = Mϕ−1:

ϕ−1

(ϕ

( n∑k=1

(xkvk)))

= ϕ

(ϕ−1

( n∑k=1

(xkvk)))

=n∑

k=1(xkvk)





ϕ

( n∑k=1

(xkvk))

=n∑

k=1(ϕ(xkvk)) =

n∑k=1

(xkϕ(vk)) = Mϕ

n∑k=1

(xkvk)


−1:

ϕ−1

(ϕ

( n∑k=1

(xkvk)))

= ϕ

(ϕ−1

( n∑k=1

(xkvk)))

=n∑

k=1(xkvk)

so Mϕ−1Mϕ = MϕMϕ−1 = Id ,

so Det(Mϕ); Det(Mϕ)−1 ∈ Z: Mϕ ∈ GLn(Z).∀M ∈ GLn(Z),∃ϕ ∈ Aut(Zn) s.t. M = Mϕ is obvious so Aut(Zn) ' GLn(Z) �




ϕ

( n∑k=1

(xkvk))

=n∑

k=1(ϕ(xkvk)) =

n∑k=1

(xkϕ(vk)) = Mϕ

n∑k=1

(xkvk)


−1:

ϕ−1

(ϕ

( n∑k=1

(xkvk)))

= ϕ

(ϕ−1

( n∑k=1

(xkvk)))

=n∑

k=1(xkvk)

so Mϕ−1Mϕ = MϕMϕ−1 = Id , so Det(Mϕ); Det(Mϕ)−1 ∈ Z: Mϕ ∈ GLn(Z).

∀M ∈ GLn(Z),∃ϕ ∈ Aut(Zn) s.t. M = Mϕ is obvious so Aut(Zn) ' GLn(Z) �




ϕ

( n∑k=1

(xkvk))

=n∑

k=1(ϕ(xkvk)) =

n∑k=1

(xkϕ(vk)) = Mϕ

n∑k=1

(xkvk)


−1:

ϕ−1

(ϕ

( n∑k=1

(xkvk)))

= ϕ

(ϕ−1

( n∑k=1

(xkvk)))

=n∑

k=1(xkvk)





Remark:Aut(F1) ' GL1(Z) = {−1; 1} but ∀n > 2,Aut(Fn) 6= GLn(Z).

We describe ϕ ∈ Aut(Fn) by ϕ(ak)k∈J1;nK. (Fn =< ak >k∈J1;nK).

Examples:

←→ϕ is{

a 7→ bb 7→ a ;

ρabc is

{a 7→ abcac−1b−1a−1b 7→ abcbc−1b−1a−1c 7→ abcb−1a−1

; IdAut(Fn) is{

ak 7→ akk ∈ J1; nK




Remark:Aut(F1) ' GL1(Z) = {−1; 1} but ∀n > 2,Aut(Fn) 6= GLn(Z).We describe ϕ ∈ Aut(Fn) by ϕ(ak)k∈J1;nK. (Fn =< ak >k∈J1;nK).

Examples:

←→ϕ is{

a 7→ bb 7→ a ;

ρabc is


; IdAut(Fn) is{






Examples:

←→ϕ is{

a 7→ bb 7→ a ; ρabc is


;

IdAut(Fn) is{






Examples:

←→ϕ is{

a 7→ bb 7→ a ; ρabc is


; IdAut(Fn) is{






Fact:

ϕ ={

ak 7→ ϕ(ak)k ∈ J1; nK Automorphism ⇐⇒ ∃ϕ−1 =

{ak 7→ ϕ−1(ak)k ∈ J1; nK s.t.

ϕ ◦ ϕ−1 = ϕ−1 ◦ ϕ = Id



Remark:We describe ϕ ∈ Aut(Fn) by ϕ(ak)k∈J1;nK. (Fn =< ak >k∈J1;nK).

Fact:

ϕ ={


{ak 7→ ϕ−1(ak)k ∈ J1; nK s.t.

ϕ ◦ ϕ−1 = ϕ−1 ◦ ϕ = Id

Examples:

←→ϕ ∈ Aut(F2) with ←→ϕ −1 =←→ϕ :

←→ϕ ←→ϕ

a 7→ b 7→ ab 7→ a 7→ b

ρabc ∈ Aut(F3) with ρ−1abc = ρc−1b−1a−1 (w = abc), we have:ρabc ρc−1b−1a−1

c 7→ (abc)c(c−1b−1a−1) 7→ w−1aww−1bww−1cww−1b−1ww−1a−1w= abcb−1a−1 = w−1abcb−1a−1w

= c−1b−1a−1abcb−1a−1abc = c




Fact:

ϕ ={


{ak 7→ ϕ−1(ak)k ∈ J1; nK s.t.

ϕ ◦ ϕ−1 = ϕ−1 ◦ ϕ = Id

Examples:

←→ϕ ∈ Aut(F2) with ←→ϕ −1 =←→ϕ :

←→ϕ ←→ϕ

a 7→ b 7→ ab 7→ a 7→ b

ρabc ∈ Aut(F3) with ρ−1abc = ρc−1b−1a−1

(w = abc), we have:ρabc ρc−1b−1a−1






Fact:

ϕ ={


{ak 7→ ϕ−1(ak)k ∈ J1; nK s.t.

ϕ ◦ ϕ−1 = ϕ−1 ◦ ϕ = Id

Examples:

←→ϕ ∈ Aut(F2) with ←→ϕ −1 =←→ϕ :

←→ϕ ←→ϕ

a 7→ b 7→ ab 7→ a 7→ b

ρabc ∈ Aut(F3) with ρ−1abc = ρc−1b−1a−1 (w = abc), we have:ρabc ρc−1b−1a−1






Fact:

ϕ ={


{ak 7→ ϕ−1(ak)k ∈ J1; nK s.t.

ϕ ◦ ϕ−1 = ϕ−1 ◦ ϕ = Id

Examples:{a 7→ ab2

b 7→ a−1 /∈ Aut(F2):The inverse doesn’t exist because:{

a−1 7→ bab2 7→ a =⇒

{a 7→ b−1

ab2 7→ a =⇒{

a 7→ b−1b2 7→ b−1a


II Basis of Zn and FnRemark:We describe ϕ ∈ Aut(Fn) by ϕ(ak)k∈J1;nK. (Fn =< ak >k∈J1;nK).

Fact:

ϕ ={


{ak 7→ ϕ−1(ak)k ∈ J1; nK s.t.

ϕ ◦ ϕ−1 = ϕ−1 ◦ ϕ = Id

Complicate Automorphism (F9):

ϕ∗ =

a 7→ aba−1b 7→ ac 7→ dg−1ecdc−1e−1gd−1d 7→ dg−1ecec−1e−1gd−1e 7→ dg−1ece−1gd−1f 7→ dg−1f 7g4d−1g 7→ dg−1f 2g2d−1h 7→ hi−2i 7→ hi−1

Is an Automorphism of F9.



Fact:

ϕ ={


{ak 7→ ϕ−1(ak)k ∈ J1; nK s.t.

ϕ ◦ ϕ−1 = ϕ−1 ◦ ϕ = Id


with ϕ−1∗ =

a 7→ bb 7→ b−1abc 7→ dg−1f 2gece−1d−1c−2e−1gd−1edg−1ec2dec−1e−1g−1f −2gd−1d 7→ dg−1f 2gece−1d−1c−2e−1gd−1cdg−1ec2dec−1e−1g−1f −2gd−1e 7→ dg−1f 2gece−1d−1c−2e−1gdg−1ec2dec−1e−1g−1f −2gd−1f 7→ dg−1f 2gece−1c−1e−1gd−1fg−3dg−1ecec−1e−1g−1f −2gd−1g 7→ dg−1f 2gece−1c−1e−1gd−1f −2g7dg−1ecec−1e−1g−1f −2gd−1h 7→ h−1i2i 7→ h−1i




Fact:

ϕ ={


{ak 7→ ϕ−1(ak)k ∈ J1; nK s.t.

ϕ ◦ ϕ−1 = ϕ−1 ◦ ϕ = Id


ϕ∗ =




III Abelianization of Free Groups



Fact:Let (vk)k∈J1;nK be the canonical basis of Zn and Fn =< ak >k∈J1;nK.χn: Aut(Fn)� Aut(Zn) ' GLn(Z).



Fact:Let (vk)k∈J1;nK be the canonical basis of Zn and Fn =< ak >k∈J1;nK.

gn:{

Fn � Zn

ak 7→ vkinduce χn: Aut(Fn)� Aut(Zn) ' GLn(Z).




gn:{

Fn � Zn


Remark:∀n ∈ N∗, In(Fn) ⊂ Ker(χn) .




gn:{

Fn � Zn


Remark:∀n ∈ N∗, In(Fn) ⊂ Ker(χn) : ∀w ∈ Fn, ∀k ∈ J1; nK,(χn ◦ ρw )(ak) = χn(wakw−1) = χn(w)vkχn(w)−1 = vk = (χn ◦ IdAut(Fn))(ak)




gn:{

Fn � Zn



Case n = 1:F1 ' Z so χ1 isomorphism and In(F1) = Ker(χ1) = {IdAut(F1)}.




gn:{

Fn � Zn



Proposition (Case n = 2):In(F2) = Ker(χ2) ( ⇐⇒ Aut(F2) ' GL2(Z)× In(F2) )




gn:{

Fn � Zn


Proposition (Case n = 2):In(F2) = Ker(χ2) ( ⇐⇒ Aut(F2) ' GL2(Z)× In(F2) ⇐⇒ GL2(Z) ' Out(F2) )



Fact:∃ χn: Aut(Fn)� Aut(Zn) ' GLn(Z).

Proposition (Case n = 2):In(F2) = Ker(χ2) ( ⇐⇒ GL2(Z) ' Out(F2) )

Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1

b 7→ am1;2bm2;2 ∈ Out(F2).





Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1

b 7→ am1;2bm2;2 ∈ Out(F2).

ϕM−1 ◦ ϕM ={

a 7→ am1;1bm2;1 7→ (am2;2b−m2;1)m1;1(a−m1;2bm1;1)m2;1 = ab 7→ am1;2bm2;2 7→ (am2;2b−m2;1)m1;2(a−m1;2bm1;1)m2;2 = b





Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1

b 7→ am1;2bm2;2 ∈ Out(F2).

ϕM−1 ◦ ϕM ={

a 7→ am1;1bm2;1 7→ (am2;2b−m2;1)m1;1(a−m1;2bm1;1)m2;1 = ab 7→ am1;2bm2;2 7→ (am2;2b−m2;1)m1;2(a−m1;2bm1;1)m2;2 = b

(am2;2b−m2;1)m1;1(a−m1;2bm1;1)m2;1 = (am2;2b−m2;1)m1;1(b−m1;1am1;2)−m2;1 = a





Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1

b 7→ am1;2bm2;2 ∈ Out(F2).∀ϕ ∈ Aut(F2), we have to prove: ∃!w ∈ F2,M ∈ GL2(Z) s.t. ϕ = ρw ◦ ϕM .





Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1

b 7→ am1;2bm2;2 ∈ Out(F2).∀ϕ ∈ Aut(F2), we have to prove: ∃!w ∈ F2,M ∈ GL2(Z) s.t. ϕ = ρw ◦ ϕM .∃!w ∈ F2,M ∈ Out(F2) s.t. ϕ = ρw ◦ ϕM and ϕM ∈ Out(F2). =⇒ Unicity.


III Abelianization of Free GroupsFact:∃ χn: Aut(Fn)� Aut(Zn) ' GLn(Z).


Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1

b 7→ am1;2bm2;2 ∈ Out(F2).∀ϕ ∈ Aut(F2), we have to prove: ∃!w ∈ F2,M ∈ GL2(Z) s.t. ϕ = ρw ◦ ϕM .∃!w ∈ F2,M ∈ Out(F2) s.t. ϕ = ρw ◦ ϕM and ϕM ∈ Out(F2). =⇒ Unicity.

WARNING:ϕ−1 6= ρw−1 ◦ ϕM−1

ϕ−1 = ϕM−1 ◦ ρw−1 = ρϕ(w)−1 ◦ ϕM−1





Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1

b 7→ am1;2bm2;2 ∈ Out(F2).∀ϕ ∈ Aut(F2), we have to prove: ∃!w ∈ F2,M ∈ GL2(Z) s.t. ϕ = ρw ◦ ϕM .∃!w ∈ F2,M ∈ Out(F2) s.t. ϕ = ρw ◦ ϕM and ϕM ∈ Out(F2). =⇒ Unicity.ϕ =

{a 7→ ϕ(a) = wabia aja w ′ab 7→ ϕ(b) = wbbib ajb w ′b

={

a 7→ ϕ(a) = wabia aja bka

b 7→ ϕ(b) = wbbib ajb bkb (w.l.g.)





Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1



={


b 7→ ϕ(b) = wbbib ajb bkb (w.l.g.)la = |wa|; lb = |wb| and L = la + lb + ia + ib.


III Abelianization of Free GroupsFact:∃ χn: Aut(Fn)� Aut(Zn) ' GLn(Z).


Proof (Project 3):

∀M = (mk;l)k;l∈{1;2} ∈ GL2(Z), ϕM ={

a 7→ am1;1bm2;1



={


b 7→ ϕ(b) = wbbib ajb bkb (w.l.g.)la = |wa|; lb = |wb| and L = la + lb + ia + ib. Induction on L =⇒ Existance:•L = 0 =⇒ OK.•L > 0 and ia × ib = 0 =⇒ wazajbk = waj+zbk =⇒ reduction of L.•Else, wbiajbk 7→ ajbk−iw−1 7→ wajbk−i =⇒ reduction of L. �




Remark:∀n ∈ N∗, In(Fn) ⊂ Ker(χn)

Case n > 3:

For w 6= Id ∈ F2 =< a; b >, ρw ={

a 7→ waw−1b 7→ wbw−1 ∈ In(F2) induce

ψw =

a 7→ waw−1b 7→ wbw−1c 7→ c

∈ Ker(χ3)

but /∈ In(F3) so Ker(χ3) 6= In(F3).

No contradition with Ker(χ2) = In(F2) because In(F1) = {Id}.




Remark:∀n ∈ N∗, In(Fn) ⊂ Ker(χn)

Case n > 3:

For w 6= Id ∈ F2 =< a; b >, ρw ={

a 7→ waw−1b 7→ wbw−1 ∈ In(F2) induce

ψw =

a 7→ waw−1b 7→ wbw−1c 7→ c

∈ Ker(χ3) but /∈ In(F3) so Ker(χ3) 6= In(F3).

No contradition with Ker(χ2) = In(F2) because In(F1) = {Id}.


IV An Easier Form for ϕ∗


IV An Easier Form for ϕ∗Basis Form:

ϕ∗ =


=

a 7→ a b a−1b 7→ a a a−1c 7→ dg−1 ec d c−1e−1 gd−1d 7→ dg−1 ec e c−1e−1 gd−1e 7→ dg−1 ec c c−1e−1 gd−1f 7→ dg−1 f 7g3 gd−1g 7→ dg−1 f 2g gd−1h 7→ hi−2i 7→ hi−1

Semi-Matrix Form:

ϕ∗ :

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1


IV An Easier Form for ϕ∗

Semi-Matrix Form:

ϕ∗ :

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1

Proposition:∀ϕ ∈ Aut(Fn), ∃σ ∈ Sn; M ∈ GLn(Z) diagonal by blocs and ρ composition ofinner automorphisms of these blocs s.t. ϕ = ρ ◦ ϕM ◦ σ. (No Unicity!)

Idea of Proof:Project 3 give n = 2 and a recurrence is given by the same proof as Project 3.


IV An Easier Form for ϕ∗Semi-Matrix Form:

ϕ∗ :

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1

Proposition:∀ϕ ∈ Aut(Fn), ∃σ ∈ Sn; M ∈ GLn(Z) diagonal by blocs and ρ composition ofinner automorphisms of these blocs s.t. ϕ = ρ ◦ ϕM ◦ σ. (No Unicity!)

Example (Case n = 2):

ρw ◦ ϕ( m1;1 m1;2m2;1 m2;2

) :(

w)◦(

m1;1 m1;2m2;1 m2;2

)


V Dynamics of Free Groups



Notation:Let ϕ ∈ Aut(Fn) and w ∈ Fn, ∀m ∈ N, we denote Lϕ;w (m) the length of ϕm(w).

Question:For k ∈ J1; nK, what are the properties of Lϕ;ak (m) when m is big?





Definition (Order):∀ϕ ∈ Aut(Fn), we call the order of ϕ the number O(ϕ) define by:•If ∃m ∈ N∗ s.t. ϕm = IdAut(Fn) then O(ϕ) = min(m ∈ N∗|ϕm = IdAut(Fn))•Else O(ϕ) = +∞

Example:

O(←→ϕ ) = 2 because ←→ϕ 6= IdAut(F2) and ←→ϕ 2 = IdAut(F2).(←→ϕ =

{a 7→ bb 7→ a

).






Example:

O(←→ϕ ) = 2 because ←→ϕ 6= IdAut(F2) and ←→ϕ 2 = IdAut(F2).(←→ϕ =

(0 11 0

)).



Recall ϕ∗:

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1

abcdefghi

Dynamics of ϕ∗:

•In < h; i >, ϕ∗ is(

1 −21 −1

)so ϕ4

∗ is(

1 −21 −1

)4=(

1 00 1

)∀w ∈< h; i >, Lϕ∗;w (4m) = LId ;w (m) = |w |.



Recall ϕ∗:

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1

abcdefghi

Dynamics of ϕ∗:• ∀w ∈< h; i >, Lϕ∗;w (4m) = LId ;w (m) = |w |.

•In < a; b >, ϕ∗ is(

a)◦(

0 11 0

)ϕ∗(ab) = aba−1a = ab (Fix point).



Recall ϕ∗:

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1

abcdefghi

Dynamics of ϕ∗:• ∀w ∈< h; i >, Lϕ∗;w (4m) = LId ;w (m) = |w |.

•In < a; b >, ϕ∗ is(

a)◦(

0 11 0

)ϕ∗(ab) = aba−1a = ab (Fix point). ϕ∗(b) = a and ϕ2

∗(a) = (ab)a(ab)−1so ∀m ∈ N, ϕ2m+1

∗ (b) = ϕ2m∗ (a) = (ab)ma(ab)−m.


V Dynamics of Free GroupsRecall ϕ∗:

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1

abcdefghi

Dynamics of ϕ∗:• ∀w ∈< h; i >, Lϕ∗;w (4m) = LId ;w (m) = |w |.• ∀m ∈ N, Lϕ∗;ab(m) = 2 and Lϕ∗;b(2m + 1) = Lϕ∗;a(2m) = 4m + 1.

Remark:

With the basis{ a = a

b = ab , we have:{∃a ∈ w =⇒ Lϕ∗;w (m) −→

m→+∞+∞

w = bk =⇒ ∀m ∈ N, Lϕ∗;w (m) = k.




WARNING:ϕ = ρw ◦ ϕM =⇒ ∀m ∈ N∗, ϕm = ρϕm−1(w) ◦ ϕMm





THEOREM:∀n ∈ N, ∀ϕ ∈ Aut(Fn) s.t. O(ϕ) < +∞, ∀p ∈ P, p > n =⇒ p - O(ϕ).

Example:@ϕ ∈ Aut(F3) s.t. O(ϕ) = 5.


V Dynamics of Free GroupsDefinition (Order):∀ϕ ∈ Aut(Fn), we call the order of ϕ the number O(ϕ) define by:•If ∃m ∈ N∗ s.t. ϕm = IdAut(Fn) then O(ϕ) = min(m ∈ N∗|ϕm = IdAut(Fn))•Else O(ϕ) = +∞


THEOREM:∀n ∈ N, ∀ϕ ∈ Aut(Fn) s.t. O(ϕ) < +∞, ∀p ∈ P, p > n =⇒ p - O(ϕ).

Idea of Proof:We use the Semi-Matrix Form:•If ϕ = ρw ◦ ϕM , the warning give obviously the result for all letters =⇒ OK.•In the general case, the order divide the product of the orders of blocksthis doesn’t contain a p > n because case 1 =⇒ OK.



Recall ϕ∗:

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1

abcdefghi

Dynamics of ϕ∗:• ∀w ∈< h; i >, Lϕ∗;w (4m) = LId ;w (m) = |w |.• ∀m ∈ N, Lϕ∗;ab(m) = 2 and Lϕ∗;b(2m + 1) = Lϕ∗;a(2m) = 4m + 1.•∀w ∈< f ; g >, Lϕ∗;w (m) −→

m→+∞+∞.



Recall ϕ∗:

(a)

dg−1

(

ec

) ◦

0 11 0

0 1 00 0 11 0 0

7 32 1

1 −21 −1

abcdefghi

Dynamics of ϕ∗:• ∀w ∈< h; i >, Lϕ∗;w (4m) = LId ;w (m) = |w |.• ∀m ∈ N, Lϕ∗;ab(m) = 2 and Lϕ∗;b(2m + 1) = Lϕ∗;a(2m) = 4m + 1.•∀w ∈< f ; g >, Lϕ∗;w (m) −→

m→+∞+∞.

=⇒ ∀w ∈< c; d ; e; f ; g >, Lϕ∗;w (m) −→m→+∞

+∞.



ψ∗

ψ∗ =

ec

◦ 0 1 0

0 0 11 0 0

=

c 7→ ecdc−1e−1d 7→ ecec−1e−1e 7→ ece−1

.

Question:Can we see more precisely the length of ψm

∗ (w)?



ψ∗

ψ∗ =

ec

◦ 0 1 0

0 0 11 0 0

=


.

Question:Can we see more precisely the length of ψm

∗ (w)?

Perron-Frobenius method:We give different length at letters: |c| = α; |d | = β and |e| = 1 and we search thefactor λ between w and ψ∗(w): ∀l ∈ {c; d ; e}, |ψ∗(l)| = |λl |. 2|c| + |d | + 2|e| = λ|c|

2|c| + + 3|e| = λ|d ||c| + + 2|e| = λ|e|

:

2 1 22 0 31 0 2

αβ1

= λ

αβ1


V Dynamics of Free Groupsψ∗

ψ∗ =

ec

◦ 0 1 0

0 0 11 0 0

=


.

Perron-Frobenius method:We give different length at letters: |c| = α; |d | = β and |e| = 1 and we search thefactor λ between w and ψ∗(w): ∀l ∈ {c; d ; e}, |ψ∗(l)| = |λl |.

N =

2 1 22 0 31 0 2

; v =

αβ1

; Mv = λv .

Fact (Perron-Frobenius Theorem):Let N be a nonnegative integer irreductible matrix. Then M has a unique (up toscalar multiplication) eigenvector v with positive coordinate, the associatedeigenvalue λ is the largest real root of det(N − xId) and λ > 1.


V Dynamics of Free Groupsψ∗

ψ∗ =

ec

◦ 0 1 0

0 0 11 0 0

=


.

Perron-Frobenius method:We give different length at letters: |c| = α; |d | = β and |e| = 1 and we search thefactor λ between w and ψ∗(w): ∀l ∈ {c; d ; e}, |ψ∗(l)| = |λl |.

N =

2 1 22 0 31 0 2

; v =

αβ1

; Mv = λv .

det(N − xId) = −x3 + 4x2 − 1 so λ ' 4.

Fact (Perron-Frobenius Theorem):Let N be a nonnegative integer irreductible matrix. Then M has a unique (up toscalar multiplication) eigenvector v with positive coordinate, the associatedeigenvalue λ is the largest real root of det(N − xId) and λ > 1.


Chapter6: AutomorphismsofFreeGroups

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Transcript of Chapter6: AutomorphismsofFreeGroups