Page 128 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.
P. 128
3.5 Nilpotent groups and p -groups
Thus α fixes every generator of FratGr and we are done.
Lemma 3.5.30 Let N1 and N2 be subgroups of FratGr . Then Gr /N1 ∼= Gr /N2 if and only if
there exists α ∈ AutGr such that N1α = N2.
PROOF. It is obvious that if there exists α ∈ AutGr such that N1α = N2, then it induces an
isomorphism Gr /N1 → Gr /N2. Conversely, suppose there is an isomorphism α : Gr /N1 →
Gr /N2. Let y1, . . . , yr ∈ G be such that (N1xi )α = N2yi . By Lemma 3.5.28 there exists a ho-
momorphism α: Gr → Gr with x α = yi . Since α is an isomorphism, Gr = 〈y1, . . . , yr 〉N2.
i
But N2 ≤ FratGr , therefore Gr = 〈y1, . . . , yr 〉. Thus α is surjective. Since Gr is finite,
this implies that α is an isomorphism. It remains to show that N α = N2. By definition,
1
N2x α = (N1x )α for all x ∈ Gr , and the result follows easily from here.
A lower bound
A similar argument as in the proof of 3.2.9 shows the following:
Lemma 3.5.31 Let V be a vector space over GF(q ) of dimension d . For 0 ≤ k ≤ d , let n k ,d
be the number of subspaces of V of dimension k . Then
(q d − 1)(q d − q ) · · · (q d − q k −1)
n k ,d = (q k − 1)(q k − q ) · · · (q k − q k −1) .
In particular, q k (d −k ) ≤ n k ,d ≤ q k (d −k +1).
Proposition 3.5.32 Let r be a positive integer, and s an integer such that 1 ≤ s ≤ r (r +1)/2.
Then there are at least p r s (r +1)/2−r 2−s 2 isomorphism classes of groups of order p r +s .
PROOF. Let Gr be as above. Let X be the set of subgroups N ≤ FratGr of index p s in FratGr .
Each N ∈ X gives rise to a group Gr /N of order p r +s . Furthermore, Lemma 3.5.30 implies
that the set of isomorphism classes of these groups is in 1-1 correspondence with the set
of orbits of AutGr acting on X .
Let θ : AutGr → Aut(Gr / FratGr ) be the natural homomorphism. By Lemma 3.5.29
every α ∈ ker θ fixes FratGr pointwise and so acts trivially on X . Therefore ker θ is con-
tained in the stabilizer of every element of X , and so the length of any orbit of AutGr
acting on X is at most
| Aut G |/| ker θ | ≤ | Aut(G / Frat G )| = | Aut C r | = | GL(r, p )| ≤ p r 2 .
p
r r r
From Lemma 3.5.31 we conclude that |X | ≥ p s (r (r +1)/2−s , therefore there are at least
p s (r (r +1)/2−s /p r 2
Thus α fixes every generator of FratGr and we are done.
Lemma 3.5.30 Let N1 and N2 be subgroups of FratGr . Then Gr /N1 ∼= Gr /N2 if and only if
there exists α ∈ AutGr such that N1α = N2.
PROOF. It is obvious that if there exists α ∈ AutGr such that N1α = N2, then it induces an
isomorphism Gr /N1 → Gr /N2. Conversely, suppose there is an isomorphism α : Gr /N1 →
Gr /N2. Let y1, . . . , yr ∈ G be such that (N1xi )α = N2yi . By Lemma 3.5.28 there exists a ho-
momorphism α: Gr → Gr with x α = yi . Since α is an isomorphism, Gr = 〈y1, . . . , yr 〉N2.
i
But N2 ≤ FratGr , therefore Gr = 〈y1, . . . , yr 〉. Thus α is surjective. Since Gr is finite,
this implies that α is an isomorphism. It remains to show that N α = N2. By definition,
1
N2x α = (N1x )α for all x ∈ Gr , and the result follows easily from here.
A lower bound
A similar argument as in the proof of 3.2.9 shows the following:
Lemma 3.5.31 Let V be a vector space over GF(q ) of dimension d . For 0 ≤ k ≤ d , let n k ,d
be the number of subspaces of V of dimension k . Then
(q d − 1)(q d − q ) · · · (q d − q k −1)
n k ,d = (q k − 1)(q k − q ) · · · (q k − q k −1) .
In particular, q k (d −k ) ≤ n k ,d ≤ q k (d −k +1).
Proposition 3.5.32 Let r be a positive integer, and s an integer such that 1 ≤ s ≤ r (r +1)/2.
Then there are at least p r s (r +1)/2−r 2−s 2 isomorphism classes of groups of order p r +s .
PROOF. Let Gr be as above. Let X be the set of subgroups N ≤ FratGr of index p s in FratGr .
Each N ∈ X gives rise to a group Gr /N of order p r +s . Furthermore, Lemma 3.5.30 implies
that the set of isomorphism classes of these groups is in 1-1 correspondence with the set
of orbits of AutGr acting on X .
Let θ : AutGr → Aut(Gr / FratGr ) be the natural homomorphism. By Lemma 3.5.29
every α ∈ ker θ fixes FratGr pointwise and so acts trivially on X . Therefore ker θ is con-
tained in the stabilizer of every element of X , and so the length of any orbit of AutGr
acting on X is at most
| Aut G |/| ker θ | ≤ | Aut(G / Frat G )| = | Aut C r | = | GL(r, p )| ≤ p r 2 .
p
r r r
From Lemma 3.5.31 we conclude that |X | ≥ p s (r (r +1)/2−s , therefore there are at least
p s (r (r +1)/2−s /p r 2
