THE EXISTENCE AND UNIQUENESS OF A SIMPLE GROUP
OF ORDER 168
TIMOTHY L. VIS
Abstract. The study of simple groups has been, and continues to be, the
most significant area of work in group theory. In this paper, we give a proof
of the existence and uniqueness of a simple group of order 168. We begin by
presenting (without proof) the major results from group theory upon which
this proof draws. We then deduce many of the properties a simple group of
order 168 must satisfy. We use these properties to show that such a group, if
it exists, must be unique. Finally, we give a concrete realization of the group
as the collineation group of the Fano plane from finite geometry.
1. Preliminaries
The following results will play an important role in our discussion of the properties a simple group of order 168 would necessarily satisfy. Since we are concerned
primarily with the proof of the uniqueness and existence of the simple group of
order 168, we present these results without proof. However, the proofs of these
results are, in general, instructive in the study of simple groups.
Theorem 1.1 (Index Theorem). Let G be a group with subgroup H of index
|G : H| = k. If |G| 6 k!, then G contains a normal subgroup K. In particular,
G is not simple.
Definition 1.2. If G is a group of order pa m, where p is a prime not dividing m,
then a subgroup H of G of order |H| = pa is called a Sylow p-subgroup of G.
Notation 1.3. The number of Sylow p-subgroups of G will be denoted np .
Theorem 1.4 (Sylow’s Theorem). Let G be a group of order pa m, where p is a
prime not dividing m. Then the following hold.
(1) G has a Sylow p-subgroup.
(2) If P and Q are distinct Sylow p-subgroups of G, then there exists some
g ∈ G such that Q = gP g −1 . That is, any two Sylow p-subgroups of G are
conjugate in G.
(3) The number of Sylow p-subgroups of G, np satisfies np ≡ 1 mod p. Further,
np is the index in
G of the normalizer NG (P ) of any Sylow p-subgroup P ,
and as such, np m.
Corollary 1.5. np 6= 1 for any simple group G.
Theorem 1.6. If np ≡
6 1 mod pk there exist distinct Sylow p-subgroups P and R
of G such that P ∩ R is of index ≤ pk−1 in both P and R.
Date: February 26, 2007.
1
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TIMOTHY L. VIS
Theorem 1.7. If G has more than one Sylow p-subgroup, and P and Q are Sylow
p-subgroups with |P ∩ Q| is maximal, then NG (P ∩ Q) has more than one Sylow
p-subgroup and any two distinct Sylow p-subgroups of NG (P ∩ Q) intersect in P ∩Q.
Theorem 1.8. If G is a group with subgroup H, then NG (H) /CG (H) is isomorphic to some subgroup of Aut (H).
Theorem 1.9. If G is a group of order 12, then either G has a unique Sylow
3-subgroup or G ∼
= A4 .
Theorem 1.10. The following properties hold for the group S4 .
(1) S4 has a unique normal 2-subgroup isomorphic to the Klein 4-group and
three other conjugate Klein 4-groups.
(2) Any 2 of the non-normal Klein 4-groups in S4 generate S4 .
(3) The normal Klein 4-group and any other Klein 4-group in S4 generate a
copy of D8 .
(4) Sylow 2-subgroups of S4 are isomorphic to the dihedral group D8 , and there
are 3 such subgroups.
(5) The normalizer of a Sylow 3-subgroup of S4 is isomorphic to S3 .
Theorem 1.11. The number of conjugates of a subset S in a group G is the index
of the normalizer of S, |G : NG (S)|. In particular, the number of conjugates of an
element s of G is the index of the centralizer of s, |G : CG (s)|.
Theorem 1.12. If U and W are normal subgroups of a Sylow p-subgroup P of G
then U is conjugate to W in G if and only if U is conjugate to W in NG (P ).
Theorem 1.13 (First Isomorphism Theorem). If φ is a homomorphism from a
group G into a group H, then kerφ G, and G/kerφ ∼
= φ (G).
Theorem 1.14 (Second Isomorphism Theorem). If G is a group with subgroups
A and B such that A ≤ NG (B). Then AB ≤ G, B AB, A ∩ B A, and
AB/B ∼
= A/A ∩ B.
Theorem 1.15. If V is an elementary Abelian group of order pe , then Aut(V ) ∼
=
GLe (GF (p)). This group has order (pe − 1) (pe − p) · · · pe − pe−1 .
Theorem 1.16. The following are true of characteristic subgroups.
(1) Characteristic subgroups are normal.
(2) If H is the unique subgroup of G of a given order, then H is characteristic
in G.
(3) If K is a characteristic subgroup of H and H G, then K G.
2. Properties of a Simple Group of Order 168
Suppose that G is a simple group, |G| = 168 = 23 · 3 · 7. In this section we
determine conditions which G must necessarily satisfy in order to be simple. What
follows requires no knowledge of G other than the assumptions that G is simple and
that G has order 168. The arguments that follow are adapted, in large part, from
[1]. For easy reference, we state each individual fact as a separate proposition.
Proposition 2.1. G has no proper subgroup of index less than 7.
Proof. The index theorem (1.1) implies that if H is a proper
subgroup of G of index
k, 168k!. Now if k < 7, k!|6!, which implies that |G| 6!. But 168 does not divide
720 = 6!. Thus, if H is a proper subgroup of G, H has index at least 7.
SIMPLE GROUP OF ORDER 168
3
Proposition 2.2. G has exactly 8 Sylow 7-subgroups; i.e. n7 = 8
Proof. By Sylow’s Theorem (1.4), the number of Sylow
7-subgroups of G, n7 sat
isfies n7 ≡ 1 mod 7 and n7 168
=
24.
To
satisfy
n
24,
n
7
7 ∈ {1, 2, 3, 4, 6, 8, 12, 24}.
7
Given these options the requirement that n7 ≡ 1 mod 7 restricts n7 to {1, 8}. The
simplicity of G then requires that n7 = 8.
Corollary 2.3. G has 48 elements of order 7.
Proof. Since, by Proposition 2.2 G has 8 Sylow 7-subgroups, and since each of these
contains 6 elements of order 7 different from those contained in any other Sylow
7-subgroup, there are at least 48 elements of order 7. Since every element of order
7 in G generates some Sylow 7-subgroup of G, there are at most 48 elements of
order 7 in G. That is, G has 48 elements of order 7.
Corollary 2.4. If P is a Sylow 7-subgroup of G, then |NG (P )| = 21.
Proof. By Sylow’s Theorem (1.4), n7 = 8 is the index in G of NG (P ). But since
|G : NG (P )| · |NG (P )| = 168, this implies that |NG (P )| = 21.
Corollary 2.5. If P is a Sylow 7-subgroup of G, no element of order 2 normalizes
P.
Proof.
If x normalizes P , then, by definition, x ∈ NG (P ). But then |x| 21. Since
2 6 21, |x| =
6 2.
Proposition 2.6. G has no elements of order 14.
2
Proof. Suppose
7 x ∈ G, |x| = 14. Then x ∈ P2, where P is some Sylow 7-subgroup
of G and x = 2. More specifically, P = hx i. But then if p ∈ P , p = x2m and
vice versa. So x7 px−7 = x7 x2m x−7 = x2m = p, and x7 ∈ NG (P ), in violation of
Corollary 2.5. Thus, G has no element of order 14.
Proposition 2.7. G has no elements of order 21.
Proof. Suppose x ∈ G, |x| = 21. Then x7 = 3 and x7 ∈ P , where P is some Sylow
3-subgroup of G. But then P = hx7 i. Thus, if p ∈ P , p = x7m
and vice versa. So
x3 px−3 = x3 x7m x−3 = x7m = p, and x3 ∈ NG (P ). Since x3 = 7, 7 |NG (P )|. By
Sylow’s Theorem (1.4), then 7 6 n3 . Since n3 |56, it follows that n3 |8. Since Sylow’s
Theorem restricts n3 to 3k + 1 for integer values of k, it follows that n3 = 4. But
then, |G : NG (P )| = 4, in violation of Proposition 2.1. Thus, G has no element of
order 21.
Corollary 2.8. If P is a Sylow 7-subgroup of G, then |CG (H)| = 7.
Proof. Let p be a generator of P . Then CG (p) = CG (P ). Since |P | = 7 and
|NG (P )| = 21, we have either |CG (P )| = 7 or |CG (P )| = 21. In the latter case,
P CG (P ) with index 3, so that the centralizer would be Abelian. But then
CG (P ) ∼
= Z21 , which is impossible since G has no element of order 21.
Corollary 2.9. An element of G of order 7 has exactly 24 conjugates.
Proof. By Theorem 1.11, the number of conjugates is the index of the centralizer.
Since the centralizer of an element of order 7 has order 7, the element has 168
7 = 24
conjugates.
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TIMOTHY L. VIS
Corollary 2.10. G contains 2 conjugacy classes of elements of order 7.
Proof. By Corollary 2.3, G contains 48 element of order 7. By Corollary 2.9, an
element of order 7 has 24 conjugates. It follows then that G contains 48
24 = 2
conjugacy classes of elements of order 7.
Proposition 2.11. n3 = 28.
Proof. By Slogs Theorem (1.4), n3 satisfies n3 ≡ 1 mod 3 and n3 168
3 = 56. To
satisfy n3 56, n3 ∈ {1, 2, 4, 7, 8, 14, 28, 56}. Given these options, the requirement
that n3 ≡ 1 mod 3 restricts n3 to {1, 7, 28}. The simplicity of G eliminates n3 = 1.
Let R be a Sylow 7-subgroup of G. By Proposition 2.4, |NG (R)| = 21. But then
there is some x ∈ NG (R) with |x| = 3. This x must be in some Sylow 3-subgroup
Q of G. Then Q normalizes R in G.
By Sylow’s Theorem (1.4) all Sylow 3-subgroups of G are conjugate. Since
conjugation is an isomorphism, it follows that any Sylow 3-subgroup normalizes
some Sylow 7-subgroup in G. Specifically, if y maps Q to Q′ by conjugation, then
Q′ normalizes yRy −1 .
Suppose now that n3 = 7. Then |NG (P )| = 24 = 23 · 3. Let P be some Sylow
3-subgroup of G. So P normalizes a Sylow 7-subgroup S of G. Let T be a Sylow
2-subgroup of NG (P ). For every t ∈ T , tP t−1 normalizes tSt−1 by the preceding
argument, so that P normalizes tSt−1 for all t ∈ T .
Now consider the subgroup T acting by conjugation on the set of Sylow 7subgroups of G. By Corollary 2.5, no element of order 2 normalizes a Sylow 7subgroup of G. As such, since all non-identity elements of T have a power of order
2, no non-identity element of T normalizes any Sylow 7-subgroup of G. Now let R
be, as before, a Sylow 7-subgroup of G and let t1 , t2 ∈ T be distinct non-identity
−1
elements of G. Then if t1 ·R = t2 ·R, we have t−1
1 t2 ·R = R, so that t1 t2 normalizes
R. But only the identity normalizes R, so this is a contradiction. Thus, each t ∈ T
maps R to a distinct Sylow 7-subgroup of G. Since n7 = 8 = |T |, it follows that
T acts transitively on the set of Sylow 7-subgroups of G. Thus, every Sylow 7subgroup of G is of the form tSt−1 , so that P normalizes all Sylow 7-subgroups of
G.
Thus, P is a subgroup of the intersection of the normalizers of all Sylow 7subgroups, and thus a proper subgroup of G. Further, suppose that x is in this
intersection. If R is a Sylow 7-subgroup of G, then xRx−1 = R. Consider then
−1
gxg −1 for some g ∈ G. gxg −1 R gxg −1
= gxg −1 Rgx−1 g −1 . But this is just
′ −1 −1
gxR x g
for some other Sylow 7-subgroup R′ . Since x normalizes all Sylow
7-subgroups, this is gR′ g −1 . But note that g −1 Rg = R′ , so that this is simply
R. Thus, this intersection is a normal subgroup of G. Thus, the simplicity of G
requires that this intersection be trivial, a contradiction, and n3 = 28.
Corollary 2.12. G has 56 elements of order 3.
Proof. Since n3 = 28, and each Sylow 3-subgroup of G contains 2 elements of order
3 distinct from those contained in any other Sylow 3-subgroup of G, G contains
at least 56 elements of order 3. Since any element of order 3 generates a Sylow
3-subgroup of G, there are at most 56 element of order 3. Thus, G contains exactly
56 elements of order 3.
Corollary 2.13. If P is a Sylow 3-subgroup of G, |NG (P )| = 6.
SIMPLE GROUP OF ORDER 168
5
Proof. By Sylow’s Theorem (1.4), n3 = 28 is the index in G of NG (P ). But since
|G : NG (P )| · |NG (P )| = 168, this implies that |NG (P )| = 6.
Lemma 2.14. There is a pair of distinct Sylow 2-subgroups of G with a non-trivial
intersection.
168
Proof. By Sylow’s
Theorem
(1.4)
n
satisfies
n
≡
1
mod
2
and
n
2
2
2
2 = 84.
To satisfy n2 84, n2 ∈ {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}. Given these options,
the requirement that n2 ≡ 1 mod 2 restricts n2 to {1, 3, 7, 21}. Simplicity of G
eliminates n2 = 1. If n2 = 3, then the normalizer of a Sylow 2-subgroup has
index 3, in contradiction to Proposition 2.1. Thus, n2 is either 7 or 21. But
then n2 6≡ 1 mod 23 , so by Theorem 1.6 there exists a pair of Sylow 2-subgroups
whose intersection has index at most 22 in each. Since the order of each is 23 , the
intersection is non-trivial.
Proposition 2.15. The normalizer of the largest intersection of a pair of Sylow
2-subgroups is isomorphic to S4 .
Proof. Over all pairs of distinct Sylow 2-subgroups with nontrivial intersection, let
T1 and T2 be chosen so that U = T1 ∩T2 has maximum order. Now let N = NG (U ).
Since the order of a Sylow 2-subgroup is 8, the order of U is either 2 or 4. Then U
is either Z2 or Z4 .
U is clearly a normal subgroup of N , so by Theorem 1.8, the group N/CN (U ) is
isomorphic to a subgroup of Aut (U ). The restriction on the order of U guarantees
that Aut (U ) ∈ {{1} , Z2 , S3 }. In particular, the factor group N/CN cannot have
order divisible by 7. If then N has a subgroup of order 7, this subgroup must
lie within CN (U ). Thus, this subgroup commutes with all elements of U . In
particular, this subgroup is a Sylow 7-subgroup of G which is normalized by an
element of order 2 in U , in contradiction
of Corollary 2.5.
Thus, the order of N is
not divisible by 7, and |N | ∈ 2, 22 , 23 , 2 · 3, 22 · 3, 23 · 3 .
By Theorem 1.7 N has more than one Sylow 2-subgroup. Thus, the order of N
cannot be a power of 2, otherwise N would be the unique Sylow 2-subgroup of N .
Further, by Theorem 1.7 any two distinct Sylow 2-subgroups of N intersect in the
subgroup U , so that the order of N is divisible by 22 . Thus |N | ∈ 22 · 3, 23 · 3 .
Now suppose that P is a Sylow 3-subgroup of N . Then P must also be a Sylow
3-subgroup of G, and its normalizer in G, NG (P ) has order 6. Since NN (P ) ≤
NG (P ), it follows that |NN (P )| ∈ {3, 6}. Since |N | ∈ {12, 24}, |N : NN (P )| ∈
{2, 4, 8}. However, Sylow’s Theorem (1.4) guarantees that the index is the number
of Sylow 3-subgroups in N and restricts that number to {1, 4}. Thus, |N : NN (P )| =
4, and N has 4 Sylow 3-subgroups, which must, by Sylow’s Theorem be permuted
transitively by elements of N under conjugation.
Now |N | ∈ {12, 24}. If |N | = 12, then the existence of 4 Sylow 3-subgroups
implies that N ∼
= A4 by Theorem 1.9. Since A4 has a unique Sylow 2-subgroup and
N does not, it follows that |N | =
6 12, and thus, that |N | = 24 (and |NN (P )| = 6).
Consider N acting by conjugation on its Sylow 3-subgroups, and let K be the
associated kernel. Then K is the intersection of the normalizers of each of the
Sylow 3-subgroups of N . Suppose that K 6= {1}. Then, since K ∈ NN (P ), |K| 6.
Additionally, since P does not normalize another Sylow 3-subgroup of N , P 6≤ K.
Since |P | = 3, |K| = 2. Since K is the kernel of a homomorphism (associated to
the group action), the group N/K exists and has order 12. But N/K has multiple
6
TIMOTHY L. VIS
Sylow 2-subgroups, corresponding to the Sylow 2-subgroups of N factored out by
K and four Sylow 3-subgroups, corresponding to the four Sylow 3-subgroups of
N . This is a contradiction to Theorem 1.9, so that |K| = 1, and thus K = {1}.
But then N permutes the four Sylow 3-subgroups in all possible ways, so that
N∼
= S4 .
Corollary 2.16. The largest intersection of a pair of Sylow 2-subgroups is a Klein
4-group.
Proof. Using all notation as in the proof of Proposition 2.15, note that S4 has the
Klein 4-group as its unique normal 2-subgroup (Theorem 1.10. Since U is a normal
2-subgroup of N and N ∼
= S4 , U must be a Klein 4-group.
Corollary 2.17. Sylow 2-subgroups of G are isomorphic to D8 .
Proof. Again, using all notation as in the proof of Proposition 2.15, note that a
Sylow 2-subgroup of N has order 8, and is thus a Sylow 2-subgroup of G. Since
Sylow 2-subgroups of S4 are isomorphic to the dihedral group D8 (by Theorem 1.10
and all Sylow 2-subgroups of G are isomorphic (by conjugation), then the Sylow
2-subgroups of G are isomorphic to D8 as desired.
Corollary 2.18. G has 42 elements of order 4.
Proof. Each Sylow 2-subgroup is isomorphic to D8 and thus contains 2 elements
of order 4, for a total of at most 42 elements of order 4. On the other hand, no
intersection of two Sylow 2-subgroups of G contains an element of order 4, so that
there are exactly 42 elements of order 4.
Corollary 2.19. G has a unique conjugacy class of elements of order 4.
Proof. Any two elements of order 4 in the same Sylow 2-subgroup of G are conjugate within D8 , the structure of that Sylow 2-subgroup of G. Further, by Sylow’s
Theorem (1.4), any 2 Sylow 2-subgroups are conjugate. That is, an element of
order 4 in one Sylow 2-subgroup is conjugate to some element of order 4 in any
other Sylow 2-subgroup, and thus, to every other element of order 4 in G.
Corollary 2.20. The normalizer in G of a Sylow 3-subgroup is isomorphic to S3 .
Proof. Again, using all notation as in the proof of Proposition 2.15, note (by Theorem 1.10 that the normalizer of a Sylow 3-subgroup of S4 is isomorphic to S3 .
Since the normalizer of a Sylow 3-subgroup in N has order 6 and the normalizer
of a Sylow 3-subgroup in G has order 6, and since a Sylow 3-subgroup of N is a
Sylow 3-subgroup of G, it follows that the normalizer of any Sylow 3-subgroup of
G is isomorphic to S3 .
Corollary 2.21. All elements of order 3 in G are conjugate.
Proof. Any element of order 3 is conjugate to its inverse in S3 . Since the normalizer
of a Sylow 3-subgroup in G is isomorphic to S3 , any element of order 3 is conjugate
to the other element of order 3 in that Sylow 3-subgroup of G. Further, Sylow’s
Theorem (1.4) states that all Sylow 3-subgroups of G are conjugate. Thus, an
element of order 3 is conjugate to some element of order 3 in any other Sylow
3-subgroup of G. But then all elements of order 3 in G are conjugate.
Corollary 2.22. G has no elements of order 6.
SIMPLE GROUP OF ORDER 168
7
Proof. Suppose g is anelement of order 6 in G. Then hg 2 i is a Sylow 3-subgroup
of G andg ∈ NG hg 2 i . But then NG hg 2 i contains an element of order 6. But
NG hg 2 i ∼
= S3 , and S3 has no element of order 6. Thus, G has no element of order
6.
Proposition 2.23. The centralizer of the generator of the center of a Sylow 2subgroup of G is that Sylow 2-subgroup of G.
Proof. By Corollary 2.5, no element of order 2 in G commutes with an element
of order 7. Suppose an element of order 2 commutes with an element of order
3. It would follow that their product has order 6, which violates Corollary 2.22.
Thus, no element of order 2 commutes with any element of odd order (since the
only admissible odd orders are 3, 7, and 21, and no element of order 21 exists by
2.7). Let T be a Sylow 2-subgroup of G. Then, by Corollary 2.17, T ∼
= D8 , and
Z (T ) = hzi, where z is a particular element of order 2. Consider then CG (z). Since
z has order 2, no element of odd order is in CG (z), and |CG (z)| ∈ {2, 4, 8}. But
T ≤ CG (z), and |T | = 8, so that CG (z) = T .
Corollary 2.24. The normalizer of a Sylow 2-subgroup of G is that Sylow 2subgroup of G
Proof. Clearly a Sylow 2-subgroup T of G normalizes itself. Further, any element
normalizing T must normalize its center. But this implies that such an element
would commute with the generator of the center, so that the normalizer is a subgroup of the centralizer of the generator of the center. By Proposition 2.23, this
centralizer is simply T . That is, NG (T ) ≤ T ≤ NG (T ), so that NG (T ) = T .
Corollary 2.25. n2 = 21.
Proof. With all notation as in the proof of Proposition 2.23 Suppose g normalizes
T . Then g normalizes Z (T ); that is, g commutes with z. But then g ∈ T and
NG (T ) = T , so that |NG (T )| = 8. By Sylow’s Theorem (1.4) n2 = |G : NG (T )|.
But this is simply 168
8 = 21.
Proposition 2.26. G has a unique conjugacy class of elements of order 2.
Proof. Since, by Proposition 2.23, there is an element z of order 2 in G with CG (z)
a Sylow 2-subgroup of G, so that |CG (z)| = 8. But then z has 168
8 = 21 distinct
conjugates in G by Theorem 1.11.
By Corollary 2.18 there are 42 elements of order 4 in G. By Corollary 2.3 there
are 48 elements of order 7 in G. Finally, by Corollary 2.12, there are 56 elements
of order 3 in G. Thus, 146 of the nonidentity elements of G have order other than
2, so that at most 21 elements of G have order 2, and the 21 conjugate elements
discussed are the only elements of order 2. Thus, the elements of G of order 2 form
a unique conjugacy class.
Proposition 2.27. The two Klein 4-groups contained in a Sylow 2-subgroup of G
are not conjugate in G.
Proof. By Theorem 1.12, it suffices to show that these two Klein 4-groups, which we
denote by U and W , are not conjugate in NG (T ), where T is a Sylow 2-subgroup
of G. But recall from Corollary 2.24 that NG (T ) = T ∼
= D8 . Since each Klein
4-group has index 2 in D8 , it follows that each is normal and conjugate only to
8
TIMOTHY L. VIS
itself, so that U and W are not conjugate in T . Thus, U and W are not conjugate
in G.
Proposition 2.28. If T is a Sylow 2-subgroup of G containing a Klein 4-group U
as the intersection with another Sylow p-subgroup, then the normalizer of the other
Klein 4-group in T is isomorphic to S4 .
Proof. Let W be the other Klein 4-group in T , and let W = hz, wi, where A (T ) =
hzi. By Proposition 2.26 w is conjugate in G to z. Then CG (w) ∼
= CG (z) and is a
Sylow 2-subgroup of G containing W , distinct from U . But then W is the maximal
intersection of these two Sylow 2-subgroups and, by Corollary 2.15 NG (W ) ∼
= S4
as desired.
Corollary 2.29. If U is a Klein 4-group contained in a Sylow 2-subgroup of G,
then U has 7 conjugates.
Proof. By Corollary 2.15 and Proposition 2.15, NG (U ) ∼
= S4 . Then, by Theorem
1.11, U has 168
=
7
conjugates.
24
Corollary 2.30. G is isomorphic to a subgroup of A7 .
Proof. Since G has a subgroup isomorphic to S4 , and thus of index 7, there exists
a homomorphism from G into Sk with kernel contained in S4 . Then, by the First
Isomorphism Theorem (1.13) G is isomorphic to a subgroup of S7 . However, if G
is not also isomorphic to a subgroup of A7 , the intersection of G and A7 within S7
would be a normal subgroup of G. Hence, G is isomorphic to a subgroup of A7 . Proposition 2.31. Sylow 3-subgroups of G and Sylow 7-subgroups of G are cyclic.
Proof. Sylow 3-subgroups and Sylow 7-subgroups of G have order 3 and 7 respectively. The only groups of those orders are cyclic.
3. Relations of the Klein 4-Groups in G
With these characteristics of a simple group of order 168 known, we now proceed
to prove that if such a group exists, it is unique. Recall from Proposition 2.27, that
the two Klein 4-groups in a given Sylow 2-subgroup are not conjugate. Fix a Sylow
2-subgroup P and let U and W be its Klein 4-groups.
Notation 3.1. Denote the 7 conjugates of U by Ui and the 7 conjugates of W by
Wi .
Proposition 3.2. Each Wi is normalized by exactly 3 of the Uj .
Proof. Consider NG (Wa ). By Corollary 2.15, NG (Wa ) ∼
= S4 . Further, Wa must
be the unique normal Klein 4-group in S4 . But this S4 contains three further
Klein 4-groups by Theorem 1.10. Since the product of Wa with any of these 3 is a
subgroup of order 8, and thus a Sylow 2-subgroup, it follows that these three are
among the Uj and Wi . But none of these are conjugate to Wa in S4 , and thus
cannot be conjugate to Wa in G (by Theorem 1.12). That is, these are among the
Uj . Further there are no other Klein 4-groups in S4 , so that these are the only
subgroups Uj normalizing Wa .
Corollary 3.3. Each Ui normalizes exactly 3 of the Wj .
SIMPLE GROUP OF ORDER 168
9
Proof. Suppose that Ua normalizes Wb . Then Ua Wb ∼
= D8
= D8 , so that Wb Ua ∼
and Wb normalizes Ua . By Proposition 3.2 and the symmetry of the Ui and Wj
established in the proof of Proposition 2.28, each Ui is normalized by exactly 3 of
the Wj . But then those 3 Wj are each normalized by Ui . Further, if any other Wk
were normalized by Ui , Ui would be normalized by 4 of the Wj , a contradiction of
Proposition 3.2.
Proposition 3.4. For any Ui 6= Uj , there exists at most one Wk that is normalized
by both Ui and Uj .
Proof. Suppose that both Ua and Ub normalize Wc . So Ua and Ub are non-normal
Klein 4-groups in the S4 that is NG (Wc ). But then Ua and Ub generate NG (Wc )
by Theorem 1.10. Suppose that Ua and Ub also both normalize Wd . Then Ua and
Ub generate NG (Wd ) so that NG (Wc ) = NG (Wd ). Since Wc is the unique normal
Klein 4-group in NG (Wc ) and Wd is the unique normal Klein 4-group in NG (Wd ),
it follows that Wc = Wd . That is, at most one Wk is normalized by both Ui and
Uj for distinct Ui , Uj .
Corollary 3.5. For any Wi 6= Wj , there exists at most one Uk that normalizes
both Wi and Wj .
Proof. Since Ua normalizes Wb if and only if Wb normalizes Ua (as in the proof of
Corollary 3.3), and since the symmetry of the Ui and Wj , along with Proposition
3.4, guarantee that at most one Uk is normalized by both Wi and Wj , at most one
Uk normalizes both Wi and Wj .
Proposition 3.6. For any Ui 6= Uj , there exists exactly one Wk that is normalized
by both Ui and Uj .
Proof. Suppose that no Wj is normalized by both Ua and Ub . There are 7 Wj , and
are normalized by each of Ua and Ub . Thus, a unique Wc is normalized by neither.
But then the 3 Ui that normalize Wc must each normalize exactly one of the Wj
normalized by each of Ua and Ub . 5 of the Ui are thus accounted for. The remaining
Ui must each normalize 3 of the Wj which are among those normalized by Ua and
Ub . But in order for this to happen, some pair (Wd , We ) must be normalized by
two of the Ui , a contradiction of Corollary 3.5.
Corollary 3.7. For any Wi 6= Wj , there exists exactly one Uk that normalizes both
Wi and Wj .
Proof. By Proposition 3.6 and symmetry of the Ui with the Wj , there exists exactly
one Uk that is normalized by both Wi and Wj . Since Ui normalizes Wj if and only if
Wj normalizes Ui , there exists exactly one Uk that normalizes both Wi and Wj . Each element g ∈ G acts by conjugation on the Ui and on the Wj . Since
conjugation is an inner automorphism of G, this action preserves the relation of
normalizing. That is, if Ua normalizes Wb , gUa g −1 normalizes gWb g −1 . Further,
by the definition of the Ui and Wj , the action of conjugation maps each of the Ui
to some Ui and each of the Wi to some Wj .
10
TIMOTHY L. VIS
4. The Projective Plane of Order 2
In the preceding section, we determined a great deal about the relationship
between the Klein 4-groups of G. In this section we describe a geometry realized
by the Klein 4-groups of G.
Definition 4.1. A projective plane is a point-line incidence geometry satisfying
the following axioms.
(1) Any 2 points are incident with a unique line.
(2) Any 2 lines are incident with a unique point.
(3) There exist at least 4 points such that no 3 are incident with a common
line.
The following classic result in finite geometry gives other conditions on the existence of a projective plane.
Theorem 4.2. For any projective plane π having a finite number of points, there
exists some integer n such that each line in π is incident with exactly n points, each
point is incident with exactly n lines, and π has exactly n2 + n + 1 points.
Definition 4.3. A finite projective plane having n points incident with each line
is said to have order n.
From the definition of a projective plane (Definition 4.1) and Theorem 4.2 it is
clear that the smallest possible order of a projective plane is 2. In fact, such a plane
exists and is unique.
Definition 4.4. The unique projective plane of order 2, is called the Fano plane.
The Fano plane has an elegant visual presentation which follows.
We note the following properties of the Fano plane.
(1) Every line is incident with exactly 3 points and every point is incident with
exactly 3 lines.
(2) Every pair of points is incident with a unique line, and every pair of lines
is incident with a unique point.
We also note the following definition and theorem regarding automorphisms of
projective planes.
SIMPLE GROUP OF ORDER 168
11
Definition 4.5. An automorphism of a projective plane is a permutation of the
points and lines that preserves the incidence relation. That is, if φ is an automorphism of the plane, then the point P is incident with the line l if and only if the
point φ (P ) is incident with the line φ (l).
Theorem 4.6. Any automorphism of the Fano plane is determined by its action
on 3 non-collinear points.
Finally, we note the following useful characterization of the Fano plane.
Theorem 4.7. The geometry induced by taking the 1-dimensional vector subspaces
3
of GF (2) as points and the 2-dimensional vector subspaces as lines and incidence
as subspace inclusion and containment is the Fano plane.
5. Uniqueness of a Simple Group of Order 168
In the preceding two sections, we presented properties of the relation between
Klein 4-subgroups of G and of the Fano plane. Notice, however, that if we replace
the Ui by points, the Wi with lines, and the relation “is normalized by” with the
relation “is incident with” in the discussion of the Klein 4-groups of G that we obtain
the properties of the Fano plane. We can, thus, dispense with the cumbersome
terminology from G and instead use their geometrical analogues from the Fano
plane in the remaining discussion.
Note that this analogue extends to the discussion of G acting by conjugation on
the Ui and the Wj . The analogue of this is that of G acting on the points and
lines of the Fano plane. Further, this action preserves the incidence relation in the
Fano plane, since conjugation preserves the normalizing relation between the Klein
4-groups of G.
Proposition 5.1. The kernel of this action by G on the Fano plane is the identity.
Proof. Any element g of G normalizing each of the Ui would necessarily be in the
intersection of their normalizers. But this intersection cannot have elements of
order 4 (a consequence of Corollary 2.16). Thus, such |g| = 2 and, as g ∈ NG (Ui ),
g ∈ V , where V is a Klein 4-group of NG (Ui ). Since g normalizes each of the Ui ,
the structure of S4 ensures that g ∈ Ui . But then g cannot normalize all of the Wj .
By way of the correspondance established, g is not in the kernel of this action. Corollary 5.2. G is isomorphic to some subgroup of the automorphism group of
the Fano plane.
Proof. The group action described is a homomorphism from G into the automorphism group of the Fano plane. Since the kernel is trivial, the First Isomorphism
Theorem (1.13) guarantees that G is isomorphic to the range of this action.
Proposition 5.3. The order of the automorphism group of the Fano plane is at
most 168.
Proof. Any automorphism of the Fano plane is determined by its action on a triple
of non-collinear points by Theorem 4.6. Given a triple of non-collinear points, there
are 7 choices for the image of the first point, 6 for the second, and 4 for the third,
for a total of 168 possible choices. That is, there are at most 168 automorphisms
of the Fano plane.
12
TIMOTHY L. VIS
Theorem 5.4. If G exists, it is isomorphic to the automorphism group of the Fano
plane and is thus unique.
Proof. By Corollary 5.2, G must be isomorphic to a subgroup of the automorphism
group of the Fano plane. By Proposition 5.3, this automorphism group has order
at most 168. Thus, if G exists, it cannot be isomorphic to a proper subgroup of the
automorphism group and must be isomorphic to the automorphism group itself.
Since the automorphism group is unique, G is unique if it exists.
6. Existence of a Simple Group of Order 168
In order to establish the existence of a simple group of order 168, 2 things remain
to be proved: that the automorphism group of the Fano plane indeed has order 168,
and that this group is indeed simple.
Proposition 6.1. The automorphism group of the Fano plane has order 168.
Proof. Consider the construction of the Fano plane given by Theorem 4.7. By
3
Theorem
1.15, the
automorphism
group of GF (2) is GL3 (GF (2)) and has order
23 − 1 23 − 2 23 − 22 = 7 · 6 · 4 = 168.
Consider the action of GL3 (GF (2)) on the points and lines of the Fano plane.
Suppose that some element of GL3 (GF (2)) fixes all points and lines of the Fano
3
plane. Such an element fixes all 1-dimensional subspaces of GF (2) . But these
1-dimensional subspaces each consist of a single vector and its scalar multiples.
The only scalar multiple of a vector over GF (2) other than that vector is the
zero vector. Since elements of GL3 (GF (2)) are invertible, the generator of a 1dimensional vector subspace must be mapped to itself by an element fixing that
vector subspace. But then an element fixing all 1-dimensional vector subspaces
fixes all vectors. Thus, the only element of GL3 (GF (2)) acting as the identity on
the Fano plane is the identity element.
But then GL3 (GF (2)) is isomorphic to the automorphism group of the Fano
plane. Since this group has order 168, the automorphism group of the Fano plane
must also have order 168.
Proposition 6.2. The group GL3 (GF (2)) is simple.
Proof. For sake of contradiction, suppose H is a proper nontrivial normal subgroup
of GL3 (GF (2)). Let P be a point in the Fano plane and let N be its stabilizer
in GL3 (GF (2)). Since 168 is the maximum number of triples of non-collinear
points, a triple of non-collinear points (P, P ′ , P ′′ ) is mapped to every other triple of
non-collinear points in the plane by some element of GL3 (GF (2)). In particular,
the action of GL3 (GF (2)) on the points of the Fano plane is transitive. As such,
|GL3 (GF (2)) : N | = 7.
Consider the intersection of all conjugates of N . This intersection must fix all
points of the Fano plane, and, as such, is trivial. Then H 6≤ N . It follows then
that N < HN . But since N has index 7, this implies that HN = GL3 (GF (2)).
By
the Second Isomorphism Theorem (1.14), |H : H ∩ N | = |HN : N |, so that
7 |H|. Since GL3 (GF (2)) has a subgroup of index 7, it is isomorphic to a subgroup of S7 . Now the normalizers of a Sylow 7-subgroup of S7 have order 42, so
that GL3 (GF (2)) has no normal Sylow 7-subgroup, as such would require order
168. Then, by the same counting arguments as in the proof of Proposition 2.2, it
follows that for GL3 (GF (2)), n7 = 8. Now if H had a normal Sylow 7-subgroup,
SIMPLE GROUP OF ORDER 168
13
this subgroup would be characteristic in H and thus normal in GL3 (GF (2)) by
Theorem 1.16, a contradiction. Thus,
for H, n7 6= 1. Then n7 must be 8 in H, so
that |H| is divisible by 8. Then 56 |H|. Since H is proper, |H| = 56.
Consider then the number of elements in H. Since n7 = 8 in H, there are 48
elements in H of order 7. There remain only 8 other elements of H, so that a
Sylow 2-subgroup of H, which must exist, must be unique, and thus characteristic.
Then this is normal in GL3 (GF (2)), so that GL3 (GF (2)) has a unique Sylow
2-subgroup. Consider the subset of GL3 (GF (2)) consisting of lower triangular
matrices. In order to be invertible, all that is required is that the diagonal entries
be nonzero, and thus be 1. The 3 entries below the main diagonal can each be
1 or 0, so that there are 8 such matrices. But multiplication of lower triangular
matrices produces lower triangular matrices, so that this subset is in fact a subgroup
of order 8, that is, a Sylow 2-subgroup of GL3 (GF (2)). The same argument holds
for the upper triangular matrices. But the upper and lower triangular matrices
intersect only trivially, so that we have produced distinct Sylow 2-subgroups of
GL3 (GF (2)). This is a contradiction, and no proper nontrivial normal subgroup
H can exist. That is, GL3 (GF (2)) is simple.
Theorem 6.3. A simple group of order 168 exists.
Proof. Proposition 6.1 establishes that the automorphism group of the Fano plane
has order 168 and is isomorphic to GL3 (GF (2)). Proposition 6.2 establishes that
GL3 (GF (2)) is simple. Thus, GL3 (GF (2)) is a simple group of order 168.
Theorem 6.4. There exists a unique (up to isomorphism) simple group of order
168.
Proof. Theorem 6.3 gives the existence of such a simple group and Theorem 5.4
establishes its uniqueness.
References
[1] D. Dummit and R. Foote. Abstract Algebra. John Wiley and Sons, 2004.