This document discusses three theorems about Sylow subgroups in finite groups. Theorem 1 proves the existence of a group with q^e Sylow p-subgroups, where q and p are primes and q^e ≡ 1 (mod p). Theorem 2 shows that if p and q are "mod-1 related", meaning q ≡ 1 (mod p), then there exists a group with q^n Sylow p-subgroups for any n. Theorem 3 deals specifically with the 2-case, proving there exists a group with n Sylow 2-subgroups for any positive odd integer n. The document provides constructions of groups to satisfy the conditions of each theorem and proofs of subsidiary lemmas about the properties of

1. SPECIFIC FINITE GROUPS
SHANE NICKLAS
Introduction
The goal of this paper is too provide foundation to Sylows Theorem in regards
to its utilization. Starting with any general primes, p and q, how should a group be
constructed in order for it to have q Sylow p − subgroups? What about q2
Sylow
p − subgroups? qn
? Moreover, are there any special cases that provide the best
conditions in order to allow the most flexibility when constructing these groups?
The first two theorems are related to the general case; the first one proves existence
while the latter provides/proves a methodology in which to fully exploit the first.
The last theorem deals with the smoothest case when constructing: the 2 case. The
reasoning is because of the uniqueness of 2 compared to other primes, specially:
pn
≡ 1 · (mod2)∀ primes p(= 2) and n ∈ N>0. All three theorems provide intuition
on the existence of Sylow subgroups throughout all primes and their combinations
and moreover how to these construct groups to fit specific Sylow constraints.
Prerequisites to Theorem 1
• Let ’q’ and ’p’ both be distinct primes
• q > p ( p = 2. If p = 2, see 2-case)
• qei
≡ 1(modp) for some set of positive integers: E = {e1, e2, ..., en}
———————————————————————————————————
Theorem 1
For every e ∈ E , there exists a group with qe
Sylow p-subgroups.
———————————————————————————————————
Notations(1)
• First note that since |F∗
qei | = (qei
− 1) ≡ 0(modp) ⇒ (By Cauchy’s Theo-
rem) p | |F∗
qei |, therefore ∃λ ∈ Fqei such that o(λ) = p . Let λ denote this
such element.
• Denote the set:
(1) G = {(x, a)|x ∈ Fqei , a ∈< (λ) >}
Moreover since x has qei
possibillities and a has p possibillities ⇒ |G| =
qei
· p
1

2. 2 SHANE NICKLAS
• Denote the subset P ⊂ G, such that
(2) P = {0, l)|l ∈< (λ) >}
Moreover since 0 is fixed and l has p possibillities ⇒ |P| = p
———————————————————————————————————
Lemma 1
(G, * ) is a group via the operation: (x, b) ∗ (y, b) = (x + by, ab)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Proof of Lemma 1. Want to show G is a group by satisfaction of the 4 axioms:
i.) Closure
ii.) Associativity
iii.) Existence of Identity
iv.) Existence of Inverses
[Closure] For g1, g2 ∈ G where
g1 = (x, a)
g2 = (y, b)
g1 ∗ g2 = (x + ay, ab)
Clearly each (x + a · y) ∈ Fqei and (a · b) ∈< (λ) >⇒ (g1 ∗ g2) ∈ G
[Associativity] For the same g1, g2 ∈ G and g3 ∈ G such that g3 = (z, c) we have:
(g1 ∗ g2) ∗ g3 =
= (x + by, ab) ∗ g3 = (x + Ay + abz, abc)
= (x + a(y + bz), a(bc)) = (g1 ∗ (y + bz, bc)
= g1 ∗ (g2 ∗ g3)
[Existence of Identity] Consider e ∈ G such that e = (0, 1),
e ∗ g1 = (0 + (1 · x), 1 · a) = (0 + x, a) = (x, a) = g1
g1 ∗ e = (x + (a · 0), a · 1) = (x + 0, a) = (x, a) = g1
[Existence of Inverses] For g1, h ∈ G , where h = (−a−1
· x, a−1
) , we have:
g1∗h = (x, a)∗(−a−1
·x, a−1
) = (x−a·a−1
·x, a·a−1
) = (x−(1·x), 1) = (x−x, 1) = (0, 1) = e
h ∗ g1 = (−a−1
· x, a−1
) ∗ (x, a) = (−a−1
· x + a−1
· x, a−1
· a) = (0, 1) = e
⇒ h = g−1
1 ∈ G
The above axioms are satisfied thus (G, *) is a group.

3. SPECIFIC FINITE GROUPS 3
[Proof of Lemma 1]
———————————————————————————————————
Lemma 2
P is a Sylow p-subgroup of G
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Proof of Lemma 2. First to prove P < G by showing ∀t1, t2 ∈ P, (t1 ∗ t−1
2 ) ∈ P.
Given t1 = (0, r), t2 = (0, s) ⇒ t−1
2 = (0, s−1
). Thus (t1 ∗ t−1
2 ) =
(0, r) ∗ (0, s−1
) = (0 + r · 0, r · s−1
) = (0 + 0, r · s−1
) = (0, r · s−1
)
Clearly then (r · s−1
) ∈< (λ) > therfore (t1 ∗ t−1
2 ) ∈ P ⇒ P < G.
Moreover since |G| = qei
· p and since p qei
(p and q are distinct primes) , p1
is
the highest power of p for |G| and given |P| = p ⇒ P is a Sylow p − subgroup of G.
[Proof of Lemma 2]
———————————————————————————————————
Lemma 3
|SylP (G)| = qe
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Proof of Lemma 3. Want to show ∀g ∈ NG(P), g ∈ P. Proceed by contridiction,
assume g /∈ P. Consider (g ∗ k ∗ g−1
) ∈ P, ∀k ∈ P such that k = (0, b), g =
(x, a), g−1
= (−a−1
· x, a−1
). Note that since g /∈ P ⇒ x = 0.
(g ∗ k ∗ g−1
) = ((x, a) ∗ (0, b)) ∗ g−1
= (x + (a · 0), a · b) = (x + 0, a · b) ∗ g−1
=
= (x, a · b) ∗ (−a−1
· x, a−1
) = (x − (a · b) · a−1
· x, a · b · a−1
) = (x − b · x, b) = (∗)
Let b = 1 ⇒ (∗) = (x · (1 − b), b) ∈ P, since b = 1 , x(1 − b) = 0 iff x = 0 ⇒⇐
by our assumption that x = 0. Thus g ∈ P ⇒ NG(P) ⊂ P and since its given that
P ⊂ NG(P) ⇒ P = NG(P). Thus by Sylow’s Theorem,
|Sylp(G)| = [G : NG(P)] = [G : P] = (p · qei
)/p = qei
[Proof of Lemma 3]
Q.E.D (Theorem 1)
———————————————————————————————————
Primes with Mod-1 Relations
Quick Notation: If primes p and q satisfy q ≡ 1(modp), then p and q are going
to be called mod1(M1)-related.
———————————————————————————————————

4. 4 SHANE NICKLAS
Theorem 2
If (p, q) are M1-related then there exists a group with qn
Sylow p-subgroups
∀n ∈ N>0
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Proof of Theorem 2. We know p and q satisfy: q ≡ 1(modp) → qn
≡ 1n
(modp) ≡
1(modp)∀n ∈ N Therefore by Theorem 1, E = N and there exists a group with qn
Sylow p-subgroups ∀n ∈ N.
[Proof of Theorem 2]
———————————————————————————————————
Note on M1-Relations
Some examples of M1-related primes are (in the form (p,q)): (3,7), (3,13), (3,
19), (5,11), (5,31), (5,41), (7,43), (7,71), (11, 23), (11, 67), (11, 89), (13, 53), (13,
79),(13,157), etc. These relations can be easily found for any prime p simply by
eamining the equation: (p · n) + 1 and each case of n, starting with n = 1, 2, .. etc.
If (p·n)+1 is a prime, then you’ve found a pair of M1-related primes thus Theorem
2 applies and provides a much wider range of Sylow p − subgroups available.
———————————————————————————————————
Theorem 3(2-Case)
For every positive odd integer, there exists a Group with that amount of Sylow
2-subgroups.
———————————————————————————————————
Notations(2)
• Let ”n” represent any positive odd integer, such that it has the following
prime factorization: n = pe1
1 · pe2
2 · ... · pem
m
• Denote the set:
(3) G = {(x, A)|x = (x1, x2, ..., xm), A = (−1)a
· Im}
Where each xi ∈ Fp
ei
i
and Im represents the m-dimensional identity matrix.
Moreover since each xi has pei
i possibillities and A has 2 possibillities ⇒
|G| = 2 · (pe1
1 · pe2
2 · ... · pem
m ) = 2 · n
• Denote the subset P ⊂ G, such that
(4) P = (0, L) L = (−1)l
· Im
Moreover since 0 is fixed and L has 2 possibillities ⇒ |P| = 2
———————————————————————————————————

5. SPECIFIC FINITE GROUPS 5
Lemma 4
(G, * ) is a group via the operation: (x, A) ∗ (y, B) = (x + Ay, AB)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Proof of Lemma 4. Want to show G is a group by satisfaction of the 4 axioms:
i.) Closure
ii.) Associativity
iii.) Existence of Identity
iv.) Existence of Inverses
[Closure] For g1, g2 ∈ G where
g1 = (x, A) = ((x1, x2, ..., xm), (−1)a
· Im)
g2 = (y, B) = ((y1, y2, ..., ym), (−1)b
· Im)
g1 ∗ g2 = (x + Ay, AB) = (x + (−1)a
y, (−1)a+b
· Im)
Clearly each (xi +(−1)a
·yi) ∈ Fp
ei
i
and ((−1)a+b
·Im) ∈ {Im, −Im} ⇒ (g1 ∗g2) ∈
G
[Associativity] For the same g1, g2 ∈ G and g3 ∈ G such that g3 = (z, C) we
have:
(g1 ∗ g2) ∗ g3 =
= (x + Ay, AB) ∗ g3 = (x + Ay + ABz, ABC)
= (x + A(y + Bz), A(BC)) = (g1 ∗ (y + Bz, BC)
= g1 ∗ (g2 ∗ g3)
[Existence of Identity] Consider e ∈ G such that e = (0, Im),
e ∗ g1 = (0 + (Im · x), Im · A) = (0 + x, A) = (x, A) = g1
g1 ∗ e = (x + (A · 0), A · Im) = (x + 0, A) = (x, A) = g1
[Existence of Inverses] First note that since A = ±Im ⇒ A2
= Im ⇒ A = A−1
.
For g1, h ∈ G, where h = (−Ax, A), we have:
g1 ∗ h = (x, A) ∗ (−Ax, A) = (x − A2
· x, A2
) = (x − x, Im) = (0, Im) = e
h ∗ g1 = (−Ax, A) ∗ (x, A) = (−Ax + Ax, A2
) = (0, Im) = e
⇒ h = g−1
1 ∈ G
The above axioms are satisfied thus (G, *) is a group.
[Proof of Lemma 4]
———————————————————————————————————

6. 6 SHANE NICKLAS
Lemma 5
P is a Sylow 2-subgroup of G
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Proof of Lemma 5. First to prove P < G by showing ∀t1, t2 ∈ P, (t1 ∗ t−1
2 ) ∈ P.
Given t1 = (0, (−1)a
·Im), t2 = (0, (−1)b
·Im) ⇒ t−1
2 = t2. Thus (t1∗t2) = (t1∗t−1
2 ) =
= (0 + (−1)a
· Im · 0, (−1)a+b
· Im) = (0 + 0, (−1)a+b
· Im) = (0, (−1)a+b
· Im)
Clearly then (t1 ∗ t2) = (t1 ∗ t−1
2 ) ∈ P ⇒ P < G.
Moreover since |G| = 2 · n and since 2 n (”n” is odd), 21
is the highest power of 2
for |G| and given |P| = 2 ⇒ P is a Sylow 2-subgroup of G.
[Proof of Lemma 5]
———————————————————————————————————
Lemma 6
|Syl2(G)| = n
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Proof of Lemma 6. Want to show ∀g ∈ NG(P), g ∈ P. Proceed by contridiction,
assume g /∈ P. Consider (g ∗ k ∗ g−1
) ∈ P, ∀k ∈ P such that k = (0, (−1)a
· Im), g =
(x, (−1)b
· Im), g−1
= (−(−1)b
· Im · x, (−1)b
· Im). Note that since g /∈ P ⇒ x =
0 ⇒ ∃xi = 0. Without loss of generality, let x1 = 0.
(g ∗ k ∗ g−1
) = (x + (−1)a
· Im · 0, (−1)a+b
· Im) ∗ g−1
= (x + 0, (−1)a+b
· Im) ∗ g−1
=
Let ”b” be even ⇒ (∗) = (x + x, Im) = (2 · x, Im) ∈ P, clearlyIm = ±Im but
we have 2 · x = 0 ⇒ 2 · x1 = 2 · x2 = ... = 2 · xm = 0 ⇒ ∀i, xi = 0 ⇒⇐ by
our assumption that x1 = 0. Thus g ∈ P ⇒ NG(P) ⊂ P and since its given that
P ⊂ NG(P) ⇒ P = NG(P). Thus by Sylow’s Theorem,
|Syl2(G)| = [G : NG(P)] = [G : P] = (2 · n)/2 = n
[Proof of Lemma 6]
Q.E.D (Theorem 3)