(efg) abe cde acf bdf adg bcg
Thus e joins with 2 segments of the square to make two triangles, f joins with two segments, and g joins with two segments, covering all 6 segments. This graph is drawn for purposes of illustration. The coliniation is actually on triangles, not edges. If we were mapping edges to edges, the group would be S7. After all, we just drew a complete graph on 7 points. we might, for instance, map efg onto abc. However, abc is not one of the specified triples in our coliniation. The group consists of permutations that carry triples to triples, as listed above.
You'll notice a symmetry among triangles. If abe were the main triangle, and cdfg the complete square, e connects to fg and cd, a connects to cf and dg, and b connects to df and cg.
Our choice of abe was symbolically arbitrary. Any of the 7 triangles can act as the primary triangle, with 6 triangles connected to the remaining 6 edges. Use this to find the order of the group.
Declare one of the 7 triangles as primary, and map efg onto that triangle in 6 different ways. Now, how many ways can we rearrange the square? We can swap a for b and c for d, or a for c and b for d. That's a group of order 4. If a remains in position, b must as well, since there is no ace or ade triangle. Similarly, c and d are fixed. Once we know where a is, the square is established. Thus there are four ways to map abcd onto the complete square, once the primary triangle is selected. That gives a group of order 7×6×4 = 168.
This group is a permutation group on 7 letters. Its order is divisible by 7, so it has an element of order 7, which is a 7 cycle. Two examples are: efdagcb and efbcgad. These 7 cycles map triples to triples, as listed above, and they are independent of each other. We can now prove the group is simple.
Since G is transitive and 7 is prime, we can apply an earlier theorem. Thus every normal subgroup H of G contains all the elements of order 7. By the third Sylow theorem, the number of subgroups of order 7 is 1 mod 7, and divides 24. We produced two independent 7 cycles, so there must be 8 subgroups of order 7. These 8 groups define 48 elements of order 7, so |H| ≥ 49. since |H| divides 168, it is either 84 or 56.
If |H| = 84, then H cannot contain 8 instances of Z7 (third Sylow theorem). Thus |H| = 56.
Since H has 8 elements not of order 7, these form the Sylow subgroup of order 8 in H. This 8-group does not contain the 7 cycles of G, and is not normal in G, hence there are other conjugates of this 8-group floating around, outside of H. Something in G/H has even order, yet |G/H| = 3. Therefore the subgroup H cannot exist, and G is a simple group.
Sometimes G is presented as a permutation group on 7 letters, with a circular shift as a 7 cycle. recall our earlier 7 cycle efdagcb. Remap abcdefg to 0356142, and G contains the circular shifts of 0123456. If you relable the 7 triples, you get 013, 124, 235, 346, 450, 561, and 602. Thus the shift maps triples to triples, as it should.
When we relable the permutations on the abcd square, we get the involutions 3210465 and 5126403. It is also possible to swap f and g, and c and d. This remaps to 0143265. Verify that these involutions map triples to triples, as they should. None of these three permutations is a circular shift of another. In fact these three involutions span a subgroup of order 8.
Let r be the right circular shift, and let x y and z be our involutions, given below. Then verify the next three equations.
x = 0143265
y = 3120465
z = 5126403
rrrrxrrrrrrxrrx = y
rrrrrryrryrr = z
rrrzrzrrrrrzr = x
Any of the involutions, combined with the circular shift, generates the other two, and the group of order 8. Furthermore, zrzrr is an element of order 3. The group generated by the 7 cycle and any of these involutions is divisible by 7, 8, and 3, and is contained in our simple coliniation, hence it is the simple group of order 168.