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 theorem and corollary are related to the general case; Theorem 1
proves existence while Corollary 1 provides/proves a methodology in which to fully
exploit the first theorem. 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, specifically: pn
≡ 1·(mod2)∀ primes p(= 2) and n ∈ N>0. Together
this paper should provide intuition on the existence of Sylow subgroups throughout
all primes and their combinations and moreover how to construct these 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)
= ——————————————————————————————————
—

4. 4 SHANE NICKLAS
Corollary 1
∀e ∈ E, (e · m) ∈ E as well, ∀m ∈ N.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Proof of Corollary 1. For any e ∈ E, we know qe
≡ 1(modp). Now for any m ∈ N:
(qe
)m
≡ [1(modp)]m
≡ 1m
(modp) ≡ 1(modp) ⇒ q(e·m)
≡ 1(modp) ⇒ (e · m) ∈ E
.
[Proof of Corollary 1]
———————————————————————————————————
Note on when e = 1
The best case is when e = 1. Meaning for some primes p and q, q ≡ 1(modp).
By Corollary 1, there exists a group with qm
Sylow p-subgroups ∀m ∈ N>0 Some
examples when e = 1 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 examining the equa-
tion: (p · n) + 1 for each case of n, starting with n = 1, 2, .. etc. If (p · n) + 1 is a
prime, then let q = (p · n) + 1 allowing Corollary 1 to apply and providing a much
wider range of Sylow p − subgroups available.
———————————————————————————————————
Theorem 2(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 2)