Finite Groups, Simple Groups of Low Order

Simple Groups of Low Order

Let's see if there are any simple groups of order < 240, besides the ones we know and love: the cyclic groups of prime order, A5, and the coliniation of order 168.

We don't have to worry about p groups; they all have nontrivial centers, which are normal subgroups. This rules out orders like 25, 27, 121, etc.

If the order of the group is npk, and there is no r dividing n with r = 1 mod p, other than 1, then the sylow subgroup of order pk is unique by the third Sylow theorem, and is a proper normal subgroup. This rules out orders like 20, 35, 84, etc.

If |G| = npk, and |G| does not divide n!, then the group cannot be simple, as shown by the strong Cayley theorem. This rules out orders like 12, 36, 80, etc. Actually |G| must divide n!/2, since a simple group does not have odd permutations.

This leaves, for our consideration, orders 30, 56, 60, 72, 90, 105, 112, 120, 132, 144, 168, 180, and 210.

Order 56

Let G be a simple group of order 56. There are 8 subgroups of order 7, consuming 8×6 = 48 distinct elements of order 7. This leaves just enough for the subgroup of order 8, which is required. This sylow subgroup is unique, hence normal in G.

Order 132

The group of order 132 is similar. Twelve sylow subgroups consume 120 elements of order 11, and 4 sylow subgroups consume 8 elements of order 3. This leaves just enough for the sylow subgroup of order 4, which is unique and normal.

Order 30

If Z3 and Z5 are not normal in the simple group of order 30, there are 24 elements of order 5 and 20 elements of order 3, exceeding 30.

Order 105

The group of order 105 has 15 groups of order 7, consuming 90 elements of order 7. At the same time we need 21 groups of order 5, and we can't accommodate that with the 15 elements remaining.

Order 72

If |G| = 72, there are four instances of the sylow subgroup of order 9, and they are all conjugate. Thus the normalizer of such a subgroup has index 4 in G. However, 72 does not divide into 4!.

Order 90

The sylow subgroup Z5 has 6 conjugates in a simple group of order 90. The normalizer of this subgroup has index 6, hence G embeds in S6. Call this set S, containing the elements 0 through 5. Let K be the subgroup of G that stabilizes 0. If all of G fixes 0 then G embeds in S5, which is impossible, since 90 does not divide into 120. If the orbit of 0 is less than 6 then K is a subgroup with index < 6. Yet G does not embed in S5 or less, so 0 has an orbit of size 6. In other words, G is transitive on S. Since K has size 15, in contains elements of order 3 and 5. These correspond to cycles of lengths 3 and 5 within the elements 1 through 5 (0 being fixed). By an earlier theorem, this generates A5, of order 60. This is a contradiction, hence G is not simple.

In general, a simple group that embeds in Sn, where n is minimal, is transitive. If not then the stabilizing subgroup has an index < n, G acts on the cosets of this subgroup, and G becomes a permutation group on fewer than n elements.

Order 112

If G has order 112, it embeds in S7. It has an element of order 7, which is a 7 cycle in S. It also has an involution, whence we can apply the previous theorem. There is no new simple group here.

Order 120

When |G| = 120 the normalizer of Z5 has index 6. Thus G embeds in S6. If G fixes 0 then G embeds in A5, which is impossible, hence G is transitive on S. The stabilizer has order 20, and includes a 5 cycle and an involution. The involution must be two transpositions running in parallel, an even permutation. Thanks to an earlier theorem, K = D5. This has order 10, and is normal in K. Let x be any element in K, but not in D5. If x has order 4 it is a 4 cycle, an odd permutation. If x has order 10 we cannot squeeze it into a permutation on 5 elements. Finally x has order 2, another involution. This also flips the 5 cycle over, using a different pivot point; else we get A5 back again. However, such a permutation is already part of D5.

Order 60

If G has order 60, let H be the largest proper subgroup of G. If the index of H is less than 5, cayley's strong theorem prevents G from being simple. When the index is 5, G embeds in A5, and G = A5. Let's look for a subgroup H with index > 5. There are 6 copies of Z5, all conjugate, hence the normalizer of Z5 is a subgroup with index 6. Now G embeds in S6, and is transitive.

Let K be the stabilizing subgroup fixing element 0. Now K has order 10, and includes an element of order 5, which becomes a 5 cycle in S. This is true no matter which element is fixed by K.

If G has 4 Sylow subgroups of order 3 then G embeds in S4, hence G has 10 subgroups of order 3. If G has a 3 cycle that cycle embeds in a 5 cycle, and this produces A5. Thus all elements of order 3 are two 3 cycles running in parallel. If two distinct group elements share a cycle, the "other" cycle must run in opposite directions. Compose the two permutations to get an isolated 3 cycle, which is impossible. If one 3 cycle is established, the other 3 cycle can only run in one direction.

Let K fix 0 and let a 5 cycle shift 1 through 5. Select a subgroup of order 3. It carries 0 to x and then y. This can be done in 5 choose 2 = 10 ways. The other 3 cycle, disjoint from the orbit of 0, can only run in one direction. We need all 10 of these combinations, to satisfy the 10 Sylow subgroups. This includes the permutation that moves 0 to 1 to 2, and (perhaps) 3 to 4 to 5. Apply this permutation, then left shift 1 through 5, giving an isolated 3 cycle and generating A5. If we run 5 to 4 to 3, combine this with the involution that swaps 1 and 5, and 2 and 4. This must exist, since K has order 10. The two permutations span a subgroup of order 12. (This is easier to see with the symbolically equivalent permutations 120534 and 542310.) The index is 5, and that puts us back into the world of 5 elements, hence A5.

Order 180

If G has order 180 it might have 6 copies of Z5. The normalizer of Z5 has index 6, and G embeds in S6. Actually G embeds in A6, and has index 2 in A6. Since A6 is simple it cannot have a normal subgroup of index 2, thus there are more than 6 subgroups of order 5. There must be 36 subgroups, consuming 144 elements.

If there are four subgroups of order 9 then G embeds in S4, which is impossible; hence there are 10 subgroups of order 9. If these are cyclic then 6 of the 9 elements can be considered generators. These are distinct among the 10 groups, consuming 60 elements. That gives 144+60, which exceeds 180, hence the sylow subgroup is Z3*Z3. (Remember that groups of order 9 are abelian.)

Conjugation moves any of the ten subgroups of order 9 onto any of the other subgroups of order 9, preserving the structure of G. If one subgroup intersects no others (other than the identity element), then they are all distinct, consuming 80 elements. Clearly every subgroup intersects another.

Suppose they form distinct pairs, each pair intersecting in its own instance of Z3. That gives 5×14 = 70 elements; still too many. If 3 subgroups intersect in an instance of Z3 there are 10/3 such clusterings, which is not an integer. If we want n Sylow groups to intersect in one subgroup Z3, n must be 5 or 10. This consumes 64 and 62 elements respectively. This is still to many, hence each sylow subgroup intersects others in multiple instances of Z3.

If 3 subgroups intersect pairwise, giving 3 distinct intersections of Z3, then the three generators x y and z commute, and generate a group of order 27, which is impossible. In other words, there are no triangles.

If each group joins exactly two others then they form one ring of 10 or two rings of 5. Either way we need 60 elements of order 3, which is impossible. Thus each subgroup joins at least 3 others.

Start with one subgroup, consuming 8 non-identity elements. Join this one with three others, each consuming 6 new elements. That's 26 elements of order 3. The 3 adjoined groups don't intersect each other, as that would produce a group of order 27, yet every group joins at least 3 others, via at least two intersections. Add a group to one of these three, giving 6 new elements. That's 32 elements, and even if there are no more elements of order 3 (unlikely), we only have 4 elements left for the group of order 4, whence it is a normal subgroup.

This completes the proof for 180.

Order 144

Since 144 does not divide into 6!/2, or 7!/2, G must embed in at least 8 letters.

If g lives in 8 letters There's no room for a 9 cycle, hence each subgroup of order 9 is Z/3*Z/3. When constrained to 8 digits, such a group is generated by two independent 3 cycles. This moves 6 digits and fixes 2. Each subgroup of order 9 is in two stabilizers. At the same time each stabilizer has size 18, with one subgroup of order 9. That's 8/2 = 4 subgroups of order 9; nowhere near 16. Therefore G embeds in A9. This is as high as we need go, since there are 9 sylow groups of order 16.

If an element of order 3 or 9 can be implemented as a permutation on 8 digits then it is part of the stabilizing subgroup of the ninth digit. Yet that stabilizer has order 16. All elements of order 3 or 9 move all 9 digits. That rules out a 9 cycle, since its cube is a 3 cycle.

The sylow subgroup is Z/3*Z*3. Each element of order 3 is three 3 cycles running in parallel. Let the first generator shift 012 345 678, all to the right. The second generator consists of three parallel 3 cycles, and commutes with the first. If 0 moves to 1, then out of the first cycle, the permutations do not commute. In general, you can't take one step within a group of 3, then leave. Suppose the second permutation also has the same 3 cycles, running the same or opposite directions. They can't all run the same, or opposite, so shift 012 to the right, like the first generator, and the rest to the left. Multiply these generators together and get an isolated 3 cycle, which lives within a stabilizer, which is a contradiction. Therefore the second generator moves 0 to 3 to 6. Permutations commute, so 0 moves to 3 to 4, or 0 to 1 to 4, hence 1 moves to 4, then 7. Finally 2 to 5 to 8, and you have your sylow group.

Ok, where do we go from here?

Order 168

If |G| = 168 we have 8 instances of the sylow 7 group. Thus G embeds in S8. Since G has an element of order 7, it has a 7 cycle in S. If G embeds in S7, bring in an involution, which must be an even permutation, and apply the previous theorem, whence G has to be the simple group already described. Therefore G is transitive on all 8 elements.

Now the stabilizer, call it K, has order 21. There's no room for a permutation with period 21, so everything in K has order 3 or 7. The Sylow subgroup of order 7 is unique. The other 14 elements have order 3. That's 7 subgroups of order 3, all conjugate. Clearly K contains a 7 cycle. If K also contains a 3 cycle we have A7. Thus every element of order 3 consists of two disjoint 3 cycles running in parallel. Again, there are 14 such elements in K, and 8 different stabilizing subgroups, and each element of order 3 fixes two digits and is in two stabilizing subgroups. That's 14*8/2 = 56 such elements. By the Sylow theorems, there can be no more; we've found them all.

The group contains an involution, which swaps 2 or 4 pairs. If it only swaps 2 pairs we can embed it in a 7 cycle, and that generates the simple coliniation of order 168. So every involution swaps 4 pairs of digits. Number the digits so that 0 and 1 are not swapped with each other, and find the element of order 3 that fixes 0 and 1. Suppose 0 and 1 are swapped with two elements in the same 3 cycle. Let the cycles be 234 and 567, with 0 and 1 swapping for 2 and 3. Apply the involution then the right shift of the 3 cycles.

Write one of these period-3 permutations down and the others are conjugates of this one. If two permutations combine to make anything other than these, we are done. Let x be a permutation of order 3, with 2 3 cycles, and label the elements so that 0 is fixed by x. Let r perform the right shift. If x moves 1 to 6, then x, followed by r, which is written rx, just to confuse you, moves 0 to 1, and back to 0. That's a transposition, not a circular shift, nor part of a 3 cycle. Suppose x moves 1 to 5. Now rx moves 0 to 1 to 6, 6 being in the other orbit. This cannot be one of our 7 shifts, so it must be a conjugate of x. Thus rx moves 6 back to 0, meaning x moves 6 to 6, which is impossible. Let x move 1 to 4, so rx moves 0 to 1 to 5. Again, rx has to take 5 back to 0, so x moves 5 to 6. Since rx fixes something, it has to be 3, i.e. x moves 4 to 3, and 3 to 1. That means x moves 5 to 6 to 2. That's fine, but x and rx aren't conjugates of each other; they don't look like each other. In x, the digit before 5->6 backs up 1; in rx, the digit before 0->1 backs up 3. So let x take 1 to 3, and 4 to 6, so that rx takes 0 to 1 to 4 to 0. Either 4 or 6 must be fixed by rx, but if one is, they both are. Finally let x map 1 to 2, and 3 to 6. Let x map 6 back to 5, so that rx fixes 6. To our surprise, rx and xr are both conjugates of x.

Ok, where do we go from here?