# Uniqueness of the Simple Group of Order 168

Our final task awaits us: proving there is a unique simple group of order 168 and a unique simple group of order 360, which are isomorphic to Failed to parse (Can't write to or create math temp directory): \mathrm{PSL}(2,7)

and Failed to parse (Can't write to or create math temp directory): A_6
respectively. To do so will require some very complicated arguments with Sylow subgroups, and while they require a strong stomach, it is quite amazing how much can be deduced about the structure of Failed to parse (Can't write to or create math temp directory): G
simply by supposing that Failed to parse (Can't write to or create math temp directory): G
is a simple group of order 168 or 360.


First, we look at the case of simple groups of order 168. Our aim is to show that any simple group of order 168 is isomorphic to Failed to parse (Can't write to or create math temp directory): \mathrm{PSL}(2,7) . We do so by finding a subgroup of order 24, and using its seven conjugates as seven "points" in a geometry, and other subgroups as "lines", with the "incidence relation" being a type of intersection. This geometry will turn out to be the so-called Fano plane, of which Failed to parse (Can't write to or create math temp directory): \mathrm{PSL}(2,7)

is the automorphism group.


## Theorem 8.10

If Failed to parse (Can't write to or create math temp directory): G
and Failed to parse (Can't write to or create math temp directory): H
are both simple groups of order Failed to parse (Can't write to or create math temp directory): 168


, then Failed to parse (Can't write to or create math temp directory): G \cong H .

### Proof

(This proof is adapted from Smith and Tabachnikova, pp.141-144.)

#### Cases

Let Failed to parse (Can't write to or create math temp directory): G

be a simple group of order Failed to parse (Can't write to or create math temp directory): 168=2^3 \cdot 3 \cdot 7
(which we know exists, by consideration of its order). As 168 is not prime, Failed to parse (Can't write to or create math temp directory): G
must not be abelian. Sylow's Theorems allow us to conclude that

• Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_2(G)| = 3, 7 \text{ or } 21
• Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(G)| = 4, 7 \text{ or } 28

, and;

• Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_7(G)| = 8

.

(We have eliminated the possibility of a unique Sylow Failed to parse (Can't write to or create math temp directory): p -subgroup immediately since Failed to parse (Can't write to or create math temp directory): G

is simple.)


#### Sylow 3-subgroups

First, we will show that

Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(G)| = 28

. If Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(G)| = 4

then the normalizer is a subgroup of index 4, but

Failed to parse (Can't write to or create math temp directory): 168 \nmid 4!

, and Poincaré's argument gives us a contradiction. We will demonstrate that Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(G)| = 7

leads to a contradiction, and thus conclude that Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(G)| = 28


.

Assume, for a contradiction, that Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(G)| = 7 . As Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_7(G)| = 8 , any Failed to parse (Can't write to or create math temp directory): P \in \mathrm{Syl}_7(G)

satisfies

Failed to parse (Can't write to or create math temp directory): |G : \mathrm{N}_G(P)| = 8

, and thus Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(P)|=21 . Applying Sylow's Theorems to the Sylow 3-subgroups of Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(P)

leads us to conlude that Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(P)
contains either 1 or 7 subgroups of order 3.


We will show that each of these possibilities leads to a contradiction.

If Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(P)

contains only one Sylow 3-subgroup, Failed to parse (Can't write to or create math temp directory): Q
say, then

Failed to parse (Can't write to or create math temp directory): Q \trianglelefteq \mathrm{N}_G(P)

, and so Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(P) \leq \mathrm{N}_G(Q) . Since Failed to parse (Can't write to or create math temp directory): Q

is a Sylow 3-subgroup not just of Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(P)
but of Failed to parse (Can't write to or create math temp directory): G


, we know that

Failed to parse (Can't write to or create math temp directory): |G:\mathrm{N}_G(Q)|=7 \text{ or } 28

(since there are 7 or 28 Sylow 3-subgroups in Failed to parse (Can't write to or create math temp directory): G , and thus

Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(Q)| = 6 \text{ or } 24

. But Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(P)| = 21

does not divide 6 or 24, and we have a contradiction.


Now consider the case in which Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(P)

contains seven Sylow 3-subgroups. All these Sylow subgroups must be contained in Failed to parse (Can't write to or create math temp directory): P


. The subgroup generated by all seven Sylow 3-subgroups must have at least 15 elements, and hence must be the whole of Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(P)

that is, Failed to parse (Can't write to or create math temp directory)
\mathrm{N}_G(P)
is generated by the set of all Sylow 3-subgroups of Failed to parse (Can't write to or create math temp directory): G


. Since all the Sylow 3-subgroups are conjugate, this implies that Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(P)

is a normal subgroup of Failed to parse (Can't write to or create math temp directory): G


, but Failed to parse (Can't write to or create math temp directory): G

is simple, and so we have a contradiction. Hence, Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(G)| = 28


.

#### Sylow 2-subgroups

If Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_2(G)| = 3

then the normalizer is a subgroup of index 3, but

Failed to parse (Can't write to or create math temp directory): 168 \nmid 3!

, and Poincaré's argument gives us a contradiction. We will demonstrate that Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_2(G)| = 7

leads to a contradiction, and thus conclude that Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_2(G)| = 21


.

Assume that Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_2(G)| = 7 . Pick Failed to parse (Can't write to or create math temp directory): Q \in \mathrm{Syl}_2(G) , which by Sylow's theorems satisfies Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(Q)|=24 . Therefore, the number of distinct subgroups of order 3 in Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q)

must be 1 or 4. If there were a unique Sylow 3-subgroup, then the normalizer in Failed to parse (Can't write to or create math temp directory): G
of that Sylow 3-subgroup would have an order no less than 24, but we know that such a normalizer must have order 6. Thus we can assume that there must be four Sylow 3-subgroups in Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q)


.

Note that by the conjugacy of the Sylow 3-subgroups, every subgroup in Failed to parse (Can't write to or create math temp directory): G

with order 3 must lie in one of the seven normalizers of a Sylow 2-subgroup. We know that there are 7 distinct conjugates of Failed to parse (Can't write to or create math temp directory): Q
in Failed to parse (Can't write to or create math temp directory): G


, and therefore at most 7 distinct conjugates of Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q) , each of which contains 4 subgroups of order 3. However, Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(G)|=28 , so there must be 7 distinct conjugates of Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q) , whose intersection does not contain any elements of order 3 (since otherwise there would not be 28 Sylow 3-subgroups).

This implies that if Failed to parse (Can't write to or create math temp directory): Q_1, Q_2

are distinct, then Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q_1)
and Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q_2)
must also be distinct. We also know that their intersection Failed to parse (Can't write to or create math temp directory): R=\mathrm{N}_G(Q_1) \cap \mathrm{N}_G(Q_2)
cannot contain an element of order 3, and so Failed to parse (Can't write to or create math temp directory): |R|
is some power of 2. Now,

Failed to parse (Can't write to or create math temp directory): 168 = |G| \geq |\mathrm{N}_G(Q_1)\mathrm{N}_G(Q_2)| = \frac{|\mathrm{N}_G(Q_1)||\mathrm{N}_G(Q_2)|}{|\mathrm{N}_G(Q_1) \cap \mathrm{N}_G(Q_2)|} = \frac{24^2}{|R|}

.

Therefore Failed to parse (Can't write to or create math temp directory): |R| \geq 24^2/168 = \tfrac{24}{7} , which tells us that Failed to parse (Can't write to or create math temp directory): |R| = 8

or Failed to parse (Can't write to or create math temp directory): 4


. However, if Failed to parse (Can't write to or create math temp directory): |R|=8

then Failed to parse (Can't write to or create math temp directory): Q_1 = R = Q_2


, since Failed to parse (Can't write to or create math temp directory): Q_1

and Failed to parse (Can't write to or create math temp directory): Q_2
contain a unique Sylow 2-subgroup.


Thus Failed to parse (Can't write to or create math temp directory): |R| = 4

and is a normal subgroup of both Failed to parse (Can't write to or create math temp directory): Q_1
and Failed to parse (Can't write to or create math temp directory): Q_2


, as it has index 2 in them. Thus every element of Failed to parse (Can't write to or create math temp directory): Q_1

and Failed to parse (Can't write to or create math temp directory): Q_2
normalize Failed to parse (Can't write to or create math temp directory): R


, that is Failed to parse (Can't write to or create math temp directory): Q_1,Q_2 \leq \mathrm{N}_G(R) , and so

Failed to parse (Can't write to or create math temp directory): 8 \Big| |\mathrm{N}_G(R)| \Big| 168

, and since Failed to parse (Can't write to or create math temp directory): Q_1 \neq Q_2

we also have Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(R)| > 8


. Poincaré's argument implies that Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(R) \leq 24 = 168/7 , as the order of Failed to parse (Can't write to or create math temp directory): G

must divide Failed to parse (Can't write to or create math temp directory): |G:\mathrm{N}_G(R)|!


, and

Failed to parse (Can't write to or create math temp directory): 168 \nmid n!
for Failed to parse (Can't write to or create math temp directory): n < 7


. Hence the only possibility remaining is that Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(R)|=24 , and Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(R)

contains at least 2 (and hence, by Sylow, exactly 3) subgroups of order 8.


Next, recall that Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(Q)| = 24 , but contains only one subgroup of order 8, unlike Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(R) , which contains 3. Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(R)

must contain a Sylow 3-subgroup, and since each of the Sylow 3-subgroups occurs in one of the normalizers of a Sylow 2-subgroup, by replacing Failed to parse (Can't write to or create math temp directory): Q
by a suitable conjugate, we can assume that the groups Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q)
and Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(R)
(which are distinct) share a common subgroup of order 3.


As the intersection of Failed to parse (Can't write to or create math temp directory): Q

and any Sylow 2-subgroup of Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(R)
has order 4, we have that Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(Q) \cap \mathrm{N}_G(R)|
divides 24, and is divisible by 12. Since Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q) \neq \mathrm{N}_G(R)


, it cannot have order 24, and so must have order 12.

Set Failed to parse (Can't write to or create math temp directory): H = \mathrm{N}_G(Q) \cap \mathrm{N}_G(R) . Failed to parse (Can't write to or create math temp directory): H

has index 2 in Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(R)
and Failed to parse (Can't write to or create math temp directory): \mathrm{N}_G(Q)


, and thus

Failed to parse (Can't write to or create math temp directory): H \vartriangleleft \mathrm{N}_G(Q)

and

Failed to parse (Can't write to or create math temp directory): H \vartriangleleft \mathrm{N}_G(R)

.

We now conclude that Failed to parse (Can't write to or create math temp directory): H

is a normal subgroup of Failed to parse (Can't write to or create math temp directory): \langle \mathrm{N}_G(R), \mathrm{N}_G(Q) \rangle


. We must have

Failed to parse (Can't write to or create math temp directory): \langle \mathrm{N}_G(R), \mathrm{N}_G(Q) \rangle > 24

, thus the index of Failed to parse (Can't write to or create math temp directory): \langle \mathrm{N}_G(R), \mathrm{N}_G(Q) \rangle

in the simple group must be less than 7. Poincaré's argument implies that Failed to parse (Can't write to or create math temp directory): \langle \mathrm{N}_G(R), \mathrm{N}_G(Q) \rangle = G


, since again

Failed to parse (Can't write to or create math temp directory): 168 \nmid n!
for Failed to parse (Can't write to or create math temp directory): n < 7


. Thus Failed to parse (Can't write to or create math temp directory): H \vartriangleleft G

is a proper non-trivial normal subgroup, which is a contradiction. So we can conclude (at last!) that

Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_2(G)|=21

.

#### The Story So Far

We have so far shown that the number of Sylow Failed to parse (Can't write to or create math temp directory): p -subgroups is as follows:

prime Failed to parse (Can't write to or create math temp directory): p Failed to parse (Can't write to or create math temp directory): |P| Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_p(G)| Failed to parse (Can't write to or create math temp directory): |\mathrm{N}_G(P)|
2 8 21 8
3 3 28 6
7 7 8 28

#### A subgroup of order 24

Our next aim is to show that Failed to parse (Can't write to or create math temp directory): G

has a subgroup of order 24, isomorphic to Failed to parse (Can't write to or create math temp directory): S_4


. We do this by examining how the Sylow 2-subgroups intersect.

The possibility that every pair of distinct elements has a trivial intersection is quickly dismissed, as if this was the case then there would be Failed to parse (Can't write to or create math temp directory): 21 \times 7 = 147

non-identity elements contained in Failed to parse (Can't write to or create math temp directory): G
with 2-power order, and also Failed to parse (Can't write to or create math temp directory):  2 \times 28 = 56
elements of order 3 in Failed to parse (Can't write to or create math temp directory): G


. Failed to parse (Can't write to or create math temp directory): G

has only

Failed to parse (Can't write to or create math temp directory): 168 < 203 = 147 + 56

elements, so we have a contradiction. Therefore, there exist Failed to parse (Can't write to or create math temp directory): T_1,T_2 \in \mathrm{Syl}_2(G)

with Failed to parse (Can't write to or create math temp directory): |T_1 \cap T_2| = 2 \text{ or } 4


. Choose Failed to parse (Can't write to or create math temp directory): T_1

and Failed to parse (Can't write to or create math temp directory): T_2
so that Failed to parse (Can't write to or create math temp directory): T_1 \cap T_2
has maximal order. Set Failed to parse (Can't write to or create math temp directory): U = T_1 \cap T_2


, and set Failed to parse (Can't write to or create math temp directory): N = \mathrm{N}_G(U) .

#### The intersection of the Sylow 2-subgroups

Next, we want to examine how elements of Failed to parse (Can't write to or create math temp directory): \mathrm{Syl}_2(G)

intersect. Our aim is to prove that there are distinct Sylow 2-subgroups Failed to parse (Can't write to or create math temp directory): S_1,S_2 \in \mathrm{Syl}_2(G)
with Failed to parse (Can't write to or create math temp directory): |S_1 \cap S_2| = 4


. The possibility that every pair of distinct elements has a trivial intersection is quickly dismissed, as if this was the case then there would be Failed to parse (Can't write to or create math temp directory): 21 \times 7 = 147

non-identity elements contained in Failed to parse (Can't write to or create math temp directory): G
with 2-power order, and also Failed to parse (Can't write to or create math temp directory):  2 \times 28 = 56
elements of order 3 in Failed to parse (Can't write to or create math temp directory): G


. Failed to parse (Can't write to or create math temp directory): G

has only

Failed to parse (Can't write to or create math temp directory): 168 < 203 = 147 + 56

elements, so we have a contradiction.

Therefore, there exist Failed to parse (Can't write to or create math temp directory): T_1,T_2 \in \mathrm{Syl}_2(G)

with

Failed to parse (Can't write to or create math temp directory): |T_1 \cap T_2| = 2 \text{ or } 4

. Let Failed to parse (Can't write to or create math temp directory): U = T_1 \cap T_2 . If Failed to parse (Can't write to or create math temp directory): |U|=4

then we are finished, so assume Failed to parse (Can't write to or create math temp directory): |U|=2


.

For Failed to parse (Can't write to or create math temp directory): i = 1,2 , set

Failed to parse (Can't write to or create math temp directory): M_i = N_{T_i}(U)

. By the normalizer condition, Failed to parse (Can't write to or create math temp directory): U

is a proper subgroup of each Failed to parse (Can't write to or create math temp directory): M_i


, so we know that each Failed to parse (Can't write to or create math temp directory): M_i

has order either 4 or 8. Let Failed to parse (Can't write to or create math temp directory): V = \langle M_1 , M_2 \rangle

as Failed to parse (Can't write to or create math temp directory)
U \vartriangleleft M_i
 for each Failed to parse (Can't write to or create math temp directory): i


, Failed to parse (Can't write to or create math temp directory): U \trianglelefteq V . As Failed to parse (Can't write to or create math temp directory): G

is simple, Failed to parse (Can't write to or create math temp directory): V \neq G


. By Poincare's theorem, we must have Failed to parse (Can't write to or create math temp directory): |V| \leq 168/7 = 24 , and as 4 divides Failed to parse (Can't write to or create math temp directory): |V|

divides 168, and Failed to parse (Can't write to or create math temp directory): |V| > 4


, we know that

Failed to parse (Can't write to or create math temp directory): |V| = 8, 12 \text{ or } 24

.

• If Failed to parse (Can't write to or create math temp directory): |V|=8

, then Failed to parse (Can't write to or create math temp directory): V

is a Sylow 2-subgroup of Failed to parse (Can't write to or create math temp directory): G
and Failed to parse (Can't write to or create math temp directory): |V \cap T_1| = 4


, so we are done.

• Suppose that Failed to parse (Can't write to or create math temp directory): |V|=12

. Then Failed to parse (Can't write to or create math temp directory): V

must have either 1 or 4 subgroups of order 3. A single subgroup of order 3 is not possible as the size of its normalizer must be 6. Thus Failed to parse (Can't write to or create math temp directory): V
must contain four subgroups of order Failed to parse (Can't write to or create math temp directory): 3


, and hence eight elements of order 3. The four remaining elements must comprise the unique Sylow 2-subgroup of Failed to parse (Can't write to or create math temp directory): V . This implies that Failed to parse (Can't write to or create math temp directory): T_1 \cap V = T_2 \cap V , which is a contradiction, and so Failed to parse (Can't write to or create math temp directory): |V| \neq 12 .

• Suppose that Failed to parse (Can't write to or create math temp directory): |V|=24

. If both Failed to parse (Can't write to or create math temp directory): T_1,T_2 \leq V , then:

• Failed to parse (Can't write to or create math temp directory): 24 = |V| \geq |T_1T_2| = |T_1||T_2|/|T_1 \cap T_2| = 64/2 = 32

,

As we know Failed to parse (Can't write to or create math temp directory): M_1,M_2 \leq V

, we can assume without loss of generality that Failed to parse (Can't write to or create math temp directory): |T_1 \cap V| = 4 . By Sylow's Theorems there exists Failed to parse (Can't write to or create math temp directory): S \in \mathrm{Syl}_2(V)

with Failed to parse (Can't write to or create math temp directory): T_1 \cap V \leq S


. So as Failed to parse (Can't write to or create math temp directory): |S|=8 , we know that Failed to parse (Can't write to or create math temp directory): S \in \mathrm{Syl}_2(G)

and Failed to parse (Can't write to or create math temp directory): M_1 \leq T_1 \cap S \leq T_1 \cap V


.

Now,

Failed to parse (Can't write to or create math temp directory): 4 \le |M_1| \leq |T_1 \cap S| \leq |T_1 \cap V| = 4

, and so

Failed to parse (Can't write to or create math temp directory): |T_1 \cap S| = 4

, and we are done.

#### Home Stretch

Depiction of the Fano plane

We now know that there is a pair of Sylow 2>-subgroups of Failed to parse (Can't write to or create math temp directory): G

whose intersection is a subgroup of order 4. As this intersection cannot be a normal subgroup of Failed to parse (Can't write to or create math temp directory): G


, we know that Failed to parse (Can't write to or create math temp directory): Y = \langle S_1, S_2 \rangle

has order 24, where Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_2(Y)| = 3
and Failed to parse (Can't write to or create math temp directory): |\mathrm{Syl}_3(Y)| = 4


.

There are exactly 7 conjugates of Failed to parse (Can't write to or create math temp directory): Y

in Failed to parse (Can't write to or create math temp directory): G
and 28 Sylow 3-subgroups of Failed to parse (Can't write to or create math temp directory): G


, so the intersection of distinct conjugates of Failed to parse (Can't write to or create math temp directory): Y

must be a 2-group. Also, as there are 21 Sylow 2-subgroups of Failed to parse (Can't write to or create math temp directory): G


, so each will occur in exactly one conjugate of Failed to parse (Can't write to or create math temp directory): Y .

Any two distinct conjugates Failed to parse (Can't write to or create math temp directory): Y_1, Y_2

of Failed to parse (Can't write to or create math temp directory): Y
intersect in a subgroup Failed to parse (Can't write to or create math temp directory): W_{12}


, which has order 4, and will be contained in a unique Sylow 2-subgroup of both Failed to parse (Can't write to or create math temp directory): Y_1

and Failed to parse (Can't write to or create math temp directory): Y_2


. These Failed to parse (Can't write to or create math temp directory): 2 -groups generate a group Failed to parse (Can't write to or create math temp directory): Z

of order 24, which has 3 Sylow Failed to parse (Can't write to or create math temp directory): P


-subgroups, two of which we already know about. We know that Failed to parse (Can't write to or create math temp directory): Y_1, Y_2, Y_3

are the only conjugates of Failed to parse (Can't write to or create math temp directory): Y
which share a Sylow 2-subgroup with Failed to parse (Can't write to or create math temp directory): Z


. Note that Failed to parse (Can't write to or create math temp directory): Z

has 7 conjugates.


To finish off the proof, we construct a "geometry" - our geometry consists of a set of "points", Failed to parse (Can't write to or create math temp directory): P , and a collection of "lines", Failed to parse (Can't write to or create math temp directory): L , and some rules which describe which points and lines are incident.

The "points" in our geometry are the conjugates of Failed to parse (Can't write to or create math temp directory): Y , and the "lines" are the conjugates of Failed to parse (Can't write to or create math temp directory): Z . A point and a line meet if and only if they share a Sylow Failed to parse (Can't write to or create math temp directory): 2 -subgroup. So, every point is incident to 3 lines, and every line is incident to 3 points.

One possible pictorial realization of this is an equilateral triangle, the perpendicular bisectors of the sides, and an inscribed circle - 7 lines, which are the sides of the triangle, the bisectors and the circle; and 7 points where lines intersect, which are the centre of the triangle, the corners, and the middle points of the sides of the triangle. (See the picture.)

Failed to parse (Can't write to or create math temp directory): G

acts (transitively) by conjugation on the vertices in this geometry, preserving the way in which points and lines co-incide. We can observe that the automorphism group of this geometry will have an order of at most 168, and is therefore a copy of Failed to parse (Can't write to or create math temp directory): G


.

Thus any two simple groups of order 168 are isomorphic.

Failed to parse (Can't write to or create math temp directory): \Box

And so the following:

## Result

 This theorem is used directly in our classification at the end of the course. Click here to see the complete classification table. The orders that are proved directly from this theorem are highlighted by the colour of this box. If Failed to parse (Can't write to or create math temp directory): G is a simple group of order 168, then Failed to parse (Can't write to or create math temp directory): G \cong PSL(2,7)  .

 Previous Page: Special Cases IV: The Last Six Cases Chapter 8: Classification of Finite Simple Groups of Order at most 500 Uniqueness of the Simple Group of Order 168 Chapters 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 - Classification of Finite Simple Groups of Order at most 500 · Appendices Sections