1. 26.8.2012
l:split-el-ab
Lemma 1.1. Let G = A qC (E × U ) amalgamated free pro-p product
of finite p-groups. Assume that E is elementary abelian and is not
contained in C. Then d(G) > d(A).
Proof. Since E is elementary abelian it decomposes in a form E =
E ∩ C × E0 with E0 ∩ C = 1. Next observe that G = AG (E0 × U )G and
G/(AG ∩ (E0 × U )G ) ∼
= G/AG × G/(E0 ∩ U )G ∼
= (E0 × U ) × A.
Hence indeed d(G) > d(A).
l:socle
2. 30.8.2011
`
Lemma 2.1. Let G = G1 H G2 be an amalgamated free pro-p product
of finite p-groups. Suppose that socle(G1 ) ∪ socle(G2 ) ⊆ H. Then
socle(G1 ) ∩ socle(G2 ) 6= 1.
Proof. Let N /o G be maximal such that G1 ∩N = G2 ∩N = 1. There is
S /o G containing N with S/N ∼
= Cp . By our maximality assumption on
N w.l.o.g. we can assume that Q := G1 ∩ S ∼
= Cp . From Q ≤ H ≤ G2
deduce that Q = G2 ∩S /G2 . Hence 1 6= Q ≤ socle(G1 )∩socle(G2 ). p:dG-estimate-new
Proposition 2.2. Let G := G1 qH G2 be an amalgamated free pro-p
product of finite p-groups. Then d(G) ≥ min{d(G1 ), d(G2 )}.
Proof. We can assume that G is non-fictitious. Suppose there is a
counter-example. Then there is one with |G1 | + |G2 | minimal. We first
claim that socle(Gi ) ≤ H must hold for i = 1, 2.
If this is not the case then
there is i ∈ {1, 2} and N ≤ socle(Gi ) so
l:split-el-ab
that N ∩ H = 1. Lemma 1.1 implies that N ≤ Φ(Gi ) since otherwise
d(G) > d(Gj ) for the unique j ∈ {1, 2} with j 6= i.
We let “bar” denote passing to the quotient modulo N G . So we have
Ḡ = Ḡi qH Gj .
By our minimality assumption we must have that
d(Ḡ) ≥ min{d(Ḡi ), d(Gj )}.
Since d(Ḡi ) = d(Gi ) and d(G) = d(Ḡ) we can see that G cannot be a
counter-example, a contradiction.
1
2
Example 2.3. Let G := hx, y, u, v | x4 = y 4 = [y 2 , x] = [x2 , y] = 1i an
abstract group. Then G has the following properties:
• There is a short-exact sequence C2 × C2 ,→ G →
→ C2 ∗ C2 with
kernel hx2 , y 2 i = Z(G).
• G is residually 2.
• With G1 := hx, y 2 i, G2 := hy, x2 i and H := hx2 , y 2 i we find
that G = G1 ∗H G2 .
Note that d(G) = d(G1 ) = d(G2 ) = 2.
• The assertions of the previous item hold for the pro-2 completion.
Note that socle(G1 ) = socle(G2 ) = Z(G) = H = hx2 , y 2 i.
Moreover Φ(G1 ) = hx2 i and Φ(G2 ) = hy 2 i.
The following table contains conditions on N ≤ socle(Gi ) and then
the differences d(G) − d(Ḡ), d(G1 ) − d(Ḡ1 ) and d(G2 ) − d(Ḡ2 ).
N ∩ H = 1, N
N
N
N
N ≤ H,
N
N
N
N
∩ Φ(G1 ) = 1
∩ Φ(G2 ) = 1
≤ Φ(G1 )
≤ Φ(G2 )
≤ Φ(G1 ) ∩ Φ(G2 )
∩ Φ(G1 ) = N ∩ Φ(G2 ) = 1
∩ Φ(G1 ) = 1, N ≤ Φ(G2 ) = 1
≤ Φ(G1 ), N ∩ Φ(G2 ) = 1
1
1
0
0
0
1
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
1
0
1
3
l:G1HG2
Lemma 2.4. Let G := G1 qH G2 be proper. Let N ≤ socle(G1 ) and
N ∩ Φ(G1 ) = N ∩ H = 1. Then N ∩ Φ(G) = 1.
Proof. Suppose that N ≤ Φ(G). Then G = hH, N, G2 i = hH, G2 i =
G2 , a contradiction.
p:dG-estimate
Proposition 2.5. Let G := G1 qH G2 be proper and Z(G) = 1. Then
d(G) > min{d(G1 ), d(G2 )}.
Proof. Suppose, on the contrary, that there is a counter-example with
|G1 | + |G2 | minimal. We may assume that d(G1 ) ≥ d(G2 ). We must
have
H 6= 1 else by the pro-p version of the Grushko-Neumann theorem
RZ:00b
[?, Proposition 9.1.15] find d(G) = d(G1 ) + d(G2 ) > d(G2 ).
Claim 1: Whenever N ≤ socle(Gi ) and N ∩ H = 1 then N ≤ Φ(Gi ).
When i = 1 then Nl:G1HG2
∩ Φ(G1 ) = 1 implies that G1 = A × N with
H ≤ A. By Lemma 2.4 N ∩ Φ(G) = 1. If M ≤ socle(A qH G2 )
then M ≤ A ∩ G2 and then [M, N ] = 1 contradicting Z(G) = 1.
Hence Z(Ḡ) = 1 and therefore d(G) > d(Ḡ) ≥ d(Ḡ2 ) = d(G2 ), a
contradiction.
Now i = 2 and G2 = B × N with H ≤ B.
Similarly as before one
l:G1HG2
shows that Z(G1 qH Ḡ2 ) = 1. Then Lemma 2.4 implies d(G) = d(Ḡ)+1
and since d(Ḡ) > d(Ḡ2 ) = d(G2 ) − 1 find d(G) > d(G2 ), again a
contradiction.
Claim 2:l:socle
socle(Gi ) ≤ H must hold for i = 1, 2, i.e., the assumption of
Lemma 2.1 hold.
In fact, if N ∩ H = 1 for N ≤ socle(Gi ) then N ≤ Φ(Gi ). Suppose
that Z(Ḡ) 6= 1. Then there is M ≤ socle(Ḡ1 ) ∩ socle(Ḡ2 ) = socle(Ḡi ) ∩
socle(Gj ) with j 6= i. There is a finite subgroup M̃ ≤ Gi that maps
onto M and since M ∼
= Cp we must have M̃ ∼
= M × N . and so
d(G) = d(Ḡ) > d(Ḡ2 ) = d(G2 ), a contradiction.
Applying the Lemma shows the existence of N ≤ socle(G1 ) ∩
socle(G2 ). We can assume that G1 6= H 6= G2 .
We cannot have N ≤ Φ(G1 ) ∩ Φ(G2 ) else, d(G) = d(Ḡ) > d(Ḡ2 ) =
d(G2 ), a contradiction. If N ∩ Φ(G1 ) = N ∩ Φ(G2 ) = 1 then we find
factorizations G1 = A × N , G2 = B × N and H = C × N so that
G = (A qC B) × N . Therefore d(G) = d(Ḡ) + 1 > 1 + d(B) = d(G2 ),
a contradiction.
We are left with the cases
a) G1 = A × N , H = C × N and N ≤ Φ(G2 ).
b) G2 = B × N , H = C × N and N ≤ Φ(G1 ).
4
HNN1
Lemma 2.6. For a finitely generated pro-p HNN-extension G =
HNN(K, A, B, φ, t) there is an estimate d(K) ≤ d(G) + d(A) − 1.
Proof. Let “bar” denote passing to the quotient modulo Φ(G) and
employ additive notation. Then, from the pro-p presentation G =
hK, t | φ(a)−1 at | a ∈ Ai, one can read off Ḡ = K̄/((1 − φ̄)Ā) ⊕ Cp .
Hence dim(Ḡ) + dim((1 − φ̄)Ā)) = dim(K̄) + 1. Since dim((1 − φ̄)Ā) ≤
d(A) the result follows.
l:HNNandamalgrank
Lemma 2.7. Let (G, Γ) be a finite graph of finite p-groups and T a
spanning tree of Γ. Suppose that d := d(G) is finite and δ ≥ d(Gv ) for
all vertex groups Gv . Then G can be considered as an HNN-extension
G = HNN(G0 , Ai , Bi , φi , ti ) with i in a set I of cardinality at most d
and d(G0 ) ≤ d(G) + d(δ − 1).
Proof. Consider the standard tree S for G. Since d = d(G) is finite we
find that d ≥ d(G/hGm | m ∈ Si) = |Γ − T |. Therefore we can consider
G to be the result of at most d times forming an HNN-extension
Gk := HNN(Gk−1 , Ak , Bk , φk , tk ) starting from G0 := Π1 (G|T , T ) for
k = 1, . . . , d – leading to the description of G as an HNN-extension
G = HNN(G0 , HNN1
Ai , Bi , φi , ti ) with i in a set I of cardinality at most d.
Using Lemma 2.6 yields d(Gk−1 ) ≤ d(Gk ) + δ − 1 and summation for
k = 1, . . . , d yields the desired estimate.
Our first example shows that in a proper HNN-extension G =
HNN(K, A, B, φ, t) we can have d(K) arbitrarily large whereas d(G) =
2.
d(HNN)
Example 2.8. Let K be a finite vector space of dimension k := ps
(s > 0) over the field Fp and φ the automorphism defined by cyclically
permuting a basis B := {ei | i = 1 . . . k}. Let G := HNN(K, K, K, φ, t).
Then d(K) = k by our choice. On the other hand, d(G) = 2 since on
G/Φ(G) ∼
= K/(1 − φ)K ⊕ Cp ∼
= Cp ⊕ Cp .
5
3. An example
For i = 0, . . . , n+1 consider Ai := hai iohbi i and set zi := ai bi`
. Define
a pro-2 group Gn to be the quotient of the free pro-2 product n+1
i=0 Ai
modulo the pro-2 relations holding for i = 1, . . . , n:
a0 = bn+1 = 1
bi−1 = zi2 = ai+1 .
Let there be a sequence Bn := (han i o hbn i) × (hcn i o hdn i) and let
Cn−1n ∼
= C2 × C2 × C2 × C2 . There are embeddings φn−1 : Cn−1n → Bn
and ψn : Cn−1n → Bn that send the generators of Cn−1n to elements in
respectively Bn−1 and Bn according to the following table:
φn−1 (an−1 bn−1 )2 an−1 (cn−1 dn−1 )2 cn−1
ψn
bn
(an bn )2
dn
(cn dn )2
Using this we can construct the following finite tree of pro-2, 3 groups:
B0 Ca
CC
CC
C
φ0 CC
C01
=
{{
{
{
{{
{{
ψ1
B1 Ca
CC
CC
C
φ1 CC
@
...
ψ2
C12
...
Ba B
BB
BB
B
φn−1 BB
x;
xx
x
xx
xx
Bn
ψn
Cn−1n
Let Gn denote the corresponding tree product. We first prove by
induction that {b0 , d0 , an , cn } is a minimal generating set of Gn . When
n = 0 this is obvious. Suppose it holds for Gn−1 . Then
Gn = Gn−1 qCn−1n Bn
and we can read off the above table that
• b0 , d0 stay generators
• cn−1 , an−1 become identified with elements in Φ(Bn )
• dn , bn become identified with elements in Φ(Gn−1 ).
• an , cn become adjoined to be new generators.
Remarks:
acz
(1) Example 3.1 implies that
S Gn cannot be pro-2.
(2) We can form G∞ := n Gn (as an abstract group). Then
d(G∞ ) = ∞ although d(Gn ) = 4 for all n ≥ 1. Moreover, the
normal closure of any torsion element in G∞ has finite index.
(3) In the discrete situation (and if we consider C to contain all
finite groups whose order is divided by either 2 or 3) then the
graphs Γn can be arbitrary large while d(Gn ) is equal to 4.
acz
Example 3.1. Let G = ha, c, z, a0 , c0 , z 0 | a2 = c2 = z 2 = [a, z] = a0 2 =
0
c02 = z 0 2 = [a0 , z 0 ] = 1, ac = az, a0 c = a0 z 0 , z = c0 , c = z 0 i. Then, setting
A := ha, ci ∼
= C2 o C2 , B := ha0 , c0 i ∼
= C2 o C2 and C := hz, ci = hc0 , z 0 i it
turns out that G = A ∗C B. Moreover, G = C×(hai∗ha0 i) with a acting
6
1 1
1 0
0
like
and a like
. Evidently G is therefore residually
0 1
1 1
0 1
0 −1
acts like
(this automorphism has
finite. However, as aa
1 1
order 3) it cannot be residually 2. More precisely, the kernel K of the
action turns out to be K = h(aa0 −1 )3 i and hai ∗ ha0 i/K ∼
= S3 .
References
H64
[1] Higman G., Amalgams of p-groups. J. Algebra 1, 1964, 301–305.