Suppose is a simple group of order 720. Find as many properties of as you can (Sylow numbers, isomorphism type of Sylow subgroups, conjugacy classes, etc.) Is there such a group?

[Disclaimer: I had lots of help from a discussion about an old proof by Burnside, as well as a proof by Derek Holt.]

Note that . Sylow’s Theorem forces the following.

Note that does not divide , so that no proper subgroup has index at most 5.

Suppose has a subgroup of index 6. Then via the action of on we have ; however, , a contradiction. Thus no subgroup has index 6.

Suppose . Now if is a Sylow 5-subgroup, then . Since 3 does not divide 4 and 5 does not divide 8, is abelian. Moreover, if is a Sylow 3-subgroup, then is also Sylow in . Then . Thus . Now every Sylow 3-subgroup of normalizes some Sylow 5-subgroup, and no Sylow 5-subgroup is normalized by two Sylow 3-subgroups. Likewise for Sylow 5s and 3s. Thus if , we have ; as otherwise either or some Sylow 3-subgroup normalizes distinct Sylow 5s.

Now acts on by conjugation, and . Using the Orbit-Stabilizer Theorem, each orbit of this action has order 1, 15, or 45. Only one orbit has order 1; namely , since any other orbit of order 1 consists of a normal Sylow 5-subgroup in . There are not enough elements left for an orbit of order 45. Thus there is an orbit of order 15, which is precisely . Thus acts transitively on , and we have for all , .

Now let , where with . Moreover choose distinct from both and . Now for some . Let . Then . Thus we have , hence . So . Since and is normal, we have , so that .

Now , and clearly is normal. Since is nontrivial, is nontrivial, and is not simple, a contradiction. Thus . Now , and has elements of order 5.

Now suppose . Then for each ; every group of this order is abelian, so that for some Sylow 5-subgroup . But this is a contradiction since . Thus .

Now suppose . Then if , then . Now acts on by conjugation, an each orbit of this action has order 1, 3, or 9. Clearly is one orbit of order 1; if there is another, say normalizes , then is Sylow and we have . Thus there is a unique orbit of order 1 under this action. Now if the remaining orbits all have order 9, we have a contradiction, since has no integer solution. Thus there exists an orbit of order 3. Let , and consider . Now normalizes since and is (necessarily) abelian. We claim also that normalizes . To see this, note that , and that has a unique Sylow 3-subgroup, namely , which then contains , so that . Thus contains more than one Sylow 3-subgroup. Now . In all but the two cases and , either has a normal subgroup, has a subgroup of sufficiently small index, or does not have enough Sylow 3-subgroups. We handle these cases in turn. Note that in this discussion, it matters only that is a subgroup of order 3 and that contains more than one Sylow 3-subgroup.

Suppose . Then is a group of order 12, and as we saw in this previous exercise, acts on by conjugation (by coset representatives). Moreover, has 4 Sylow 3-subgroups by the Lattice Isomorphism Theorem. By the lemma, . Since has no subgroups of index 2, by the Orbit Stabilizer Theorem, the orbits of this action have order 1 or 3. However, the identity element of is in an orbit of order 1; thus all orbits have order 1. Thus this action is trivial; that is, for all and , ; thus . Note that the four Sylow 3-subgroups of intersect in , so that has elements of 3-power order, and 10 elements remain. Now . Let . Suppose . If some element of has order 4, then each Sylow 2-subgroup has two elements of order 4, all distinct. But then , a contradiction. Thus all elements in Sylow 2-subgroups of have order 2. Now must contain at least 4 elements of order 2; let these comprise the set . Now , , and are 12 mutually distinct elements of order 2, 6, and 6, respectively, so that , a contradiction. Suppose now that . Again if some element has order 4, then has elements of order 4 and 2 \cdot 6 = 12$ elements of order 12, so that , a contradiction. Thus the elements in Sylow 2-subgroups of have order 2. There are at least 4 of these, say in the set . Now , , and contain (at least) 12 distinct elements of order 2, 6, and 6, respectively, so that , a contradiction. Thus . Let be the unique Sylow 2-subgroup. Now for some Sylow 2-subgroup of , and is properly contained in . Thus 8 divides , and since is normal in , also divides . Thus . In all but one case, either has a normal subgroup or a subgroup of sufficiently small index. Thus we have . Now has index 2, and thus is normal. Moreover, (see this previous exercise) is characteristic in , and thus normal in . But then , a contradiction.

Suppose . Note that, by the N/C Theorem, . Then we have ; in either case, there exists an element of order 2 which centralizes . Now write , and note that has 10 conjugates since . Consider the action of on the ten cosets in by left multiplication. Via this action, we have . Now is a product of 3-cycles. Note that since , every conjugate of is contained in 4 Sylow 3-subgroups, and that every Sylow 3-subgroup contains a conjugate of (by Sylow’s Theorem, since ). Moreover, in this case there are 40 Sylow 3-subgroups. By the pigeonhole principle, every Sylow 3-subgroup of contains exactly one conjugate of . Suppose now that fixes . Then , so that ; thus in fact (since is contained in some Sylow 3-subgroup of which also contains ), and we have . Thus fixes only , and has cycle shape . As shown in the beginning of this paragraph, let have order 2. Now , so that is obtained from by applying entrywise. In particular, permutes the orbits of ; it must interchange two and fix the third. Thus has cycle shape or ; in either case, contains an odd permutation, so that is not simple. Thus .

We are left with . Now , and acts by conjugation on , and via this action we have . Let and suppose is cyclic; say . Then is necessarily a 9-cycle. Moreover, by the N/C Theorem we have , so that . In either case, by Cauchy there exists an element of order 2 such that . However, we saw in the discussion on page 127 of the text that ; a contradiction. Thus (indeed each of the Sylow 3-subgroups) is elementary abelian. Now let be an element of order 3; is a product of 3-cycles. Suppose fixes more than one element- that is, is a product of one or two 3-cycles. Note first that if fixes a Sylow 3-subgroup , then and in fact, since is the unique Sylow 3-subgroup in , we have . Thus, letting , we have . Since is abelian, . Now since fixes exactly 7 Sylow 3-subgroups of , contains 7 Sylow 3-subgroups. But mod 3, a contradiction. Suppose now that is a product of two 3-cycles. Now is divisible by and, since exactly four Sylow 3-subgroups contain , is divisible by as well. Thus . We can see that in all but the cases and , either is normal in or a subgroup of has sufficiently small index. In the two remaining cases, contains four Sylow 3-subgroups and has order 36 or 72; we eliminated these possibilities in our discussion about the case . Thus must be a product of three 3-cycles. Suppose now that there exists an element of order 2 such that . Now must permute the orbits of , so that it must interchange two orbits and fix the third. Thus has cycle shape or , and in either case contains an odd permutation. Thus if is an element of order 3, there does not exist an element of order 2 which centralizes it. Finally, note that if with , then . That is, . In particular, . Thus, , and hence intersects every other Sylow 3-subgroup trivially. By this previous exercise, all Sylow 3-subgroups intersect trivially. Thus contains elements of order 3.

Now since and using Lagrange, every element of has order 1, 2, 3, 4, 5, 6, 8, 9, or 10. We showed previously that no element has order 9. If has order 6, then is an element of order 3 and is an element of order 2 which centralizes it, a contradiction. Thus no element has order 6. Now let have order 5; then is a product of one or two 5-cycles. If is a single 5-cycle, then for some Sylow 3-subgroup ; this is a contradiction. Thus is a product of two 5-cycles. Suppose now that there exists an element which centralizes ; that is, . Now must permute the orbits of . Suppose fixes each orbit of ; the restriction of to an orbit of is then either the identity or a 5-cycle, contradicting the fact that has order 2. Thus transposes the orbits of and is thus a product of five 2-cycles; then is an odd permutation and is not simple, a contradiction. Thus no element of order 2 centralizes an element of order 5. In particular, no element of has order 10 (for the same reason that no element has order 6). Thus every element of is contained in a Sylow subgroup. Now suppose is an element of order 8 which normalizes a Sylow 3-subgroup; then must have the cycle shape , so that contains an odd permutation and is not simple. Thus no element of order 8 normalizes a Sylow 3-subgroup. Now , every element has order 1, 2, 3, 4, 5, or 8, and there are 80 and 144 elements of order 3 and 5, respectively. Thus the remaining 496 elements are contained in Sylow 2-subgroups. Recall also from our proof classifying the groups of order 20 that only one group of this order does not contain an element of order 10; namely , where , , and . This group is thus isomorphic to the normalizers of Sylow 5-subgroups in .

Let have order 3. Now is contained in, and thus normalizes, a unique Sylow 3-subgroup . Without loss of generality, we can say that (via the action of on ) is a product of three 3-cycles; say . Now let . We have and , so that must permute the orbits of . There are 6 choices we can make for , and any one of them generates (together with ). So without loss of generality, let . Since , by Cauchy there exists an element of order 2 which normalizes , say . There are 7 possibilities for ; note that since , .

- Suppose . If , then is an odd permutation. If , then latex u(3) = 6$. If , then . If , then . If , then and . If , then is an odd permutation. If , then . if , then is an odd permutation. If , then and .
- Suppose . If , then is an odd permutation. If , then . If , then . If , then . If , then and . If , then is an odd permutation. If , then . If , then is an odd permutation. If , then and .
- Suppose . If , then is an odd permutation. If , then we must have ; but we also have , a contradiction. Similarly, if , then . If , then and , a contradiction. If , then and , a contradiction. If , then is an odd permutation. If , then and , a contradiction. If , then is an odd permutation. If , then and , a contradiction.
- Suppose . If , then is an odd permutation. If , then , a contradiction. If , then , a contradiction. If , then . If , then and , a contradiction. If , then is an odd permutation. If , then , a contradiction. If , then is an odd permutation. If , then and , a contradiction.
- Suppose . If , then is an odd permutation. If , then , a contradiction. If , then . If , then . If , then and . If , then is an odd permutation. If , then . If , then is an odd permutation. If , then and .
- Suppose . If , then is an odd permutation. If , then , a contradiction. If , then , a contradiction. If , then , a contradiction. If , then and , a contradiction. If , then is an odd permutation. If , then , a contradiction. If , then is an odd permutation. If , then and , a contradiction.

Thus we must have ; by a similar argument, . Hence corresponds to the matrix . In particular, if is a Sylow 2-subgroup, then and contains a unique element of order 2. Moreover, we know that contains no element of order 8 since such an element would have cycle shape , and thus be odd. We know from our classification of order 8 groups that ; thus contains exactly 9 elements of order 2 and elements of order 4. Moreover, every element of order 2 in must fix some Sylow 3-subgroup, since otherwise it has cycle shape . We have shown that every element of order 2 which fixes a Sylow 3-subgroup has cycle shape , and thus *every* element of of order 2 has this cycle shape.

Consider now ; since and have the same orbits, must permute the orbits of . If fixes an orbit (of order 3) then is a 2-cycle. If fixes all of the orbits of , then has cycle shape and is thus an odd permutation. So must transpose two orbits of and fix the third. Let and be arbitrary disjoint 3-cycles with orbits and ; if fixes then is one of , , and . If transposes and , then is one of , , and . Choosing which orbit of to fix and which permutation of each orbit of orbits divide , there are possibilities for .

- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .
- Suppose . Then , so that .

The remaining nine possibilities,

satisfy , and in fact must correspond to the nine elements of order 2 in . Now for each , there are six solutions of the equation in . In , if , then the cycle shape of is either or , and the latter is not possible. If and , then an orbit of is a union of two orbits of . There are 3 different ways to choose pairs of orbits of , and for each choice, we get two solutions: and . Thus the only solutions of the equation in are , , , and their inverses. Fixing , let , , and .

At this point, note the following for all nonidentity elements .

- If , then has cycle shape .
- If , then has cycle shape .
- If , then has order 2 and thus has cycle shape . So has cycle shape or , the latter being odd and thus impossible.
- If , then does not normalize a Sylow 3-subgroup, and hence has cycle shape .
- If , then has cycle shape or , the latter being impossible.

In no case does fix more than 3 elements. Recall the elements , , and for later.

If , then there are at most elements of 2-power order, and does not contain enough elements. Similarly, if , then there are at most elements of 2-power order. Thus . Let be chosen such that is maximal.

- If , then the Sylow 2-subgroups of are pairwise disjoint and contains elements of 2-power order, a contradiction.
- If , then consider . Since is a normal subgroup of order 2, we know also that is central in . In particular, there cannot be an element of order 3 in , as otherwise would have an element of order 6. Moreover, since is contained in a 2-group, it is properly contained in its normalizer in , which is then contained in . So is divisible by and not divisible by 3; thus , .
- If , then is normal for some Sylow 2-subgroup . But then , violating the maximalness of .
- If , then by Sylow’s Theorem, . However, in our classification of groups of order 20, we saw that if a group of order 20 has a unique Sylow 2-subgroup, then it is abelian. Thus has an element of order 10, a contradiction.
- If , then by Sylow’s Theorem, we have . But then divides for some Sylow 5-subgroup , which is also Sylow in ; this is a contradiction.
- If , then by Sylow’s Theorem, . If , then as in the previous case, divides for some Sylow 5-subgroup . If , then since Sylow 5-subgroups intersect trivially, contains elements of order 5. Now suppose . Then for some Sylow 2- and 5-subgroups and , respectively, of , a contradiction since .
- Suppose . Now properly contains , and of course ; similarly for the normalizer of in . Moreover, for some Sylow 2, and must be distinct from at least one of and . But then either or has order 4, violating the maximalness of .
- If , then by order considerations we have , contradicting the maximalness of .

- If , then since is properly contained in , 8 divides . Thus , .
- If , then , contradicting the maximalness of .
- If , then we have normal for some Sylow 2-subgroup . But then , violating the maximalness of .
- If , then has a subgroup of sufficiently small index.
- If , then is normal.
- If , then by Sylow’s Theorem, , so that 8 divides for some Sylow 5-subgroup .
- Suppose . If , then we have dividing for some Sylow 5-subgroup , a contradiction. Thus . Since the Sylow 5-subgroups intersect trivially, contains elements of order 5. If , then there are at least 16 elements of 2-power order, so that contains at least elements, a contradiction. Thus . Now has 9 conjugates, and each conjugate of is normal in a (unique) Sylow 2-subgroup of . Similarly, every Sylow 2-subgroup of normalizes a conjugate of . However, there are 45 Sylow 2-subgroups in , so that some conjugate of must be normalized by more than one Sylow 2-subgroup, a contradiction.
- Suppose . If , then we have 3 dividing for some Sylow 2-subgroup ; this is a contradiction since . Thus . If and are Sylow 2-subgroups in (which are also Sylow in ), we clearly have ; if this inclusion is proper, we contradict the maximalness of . Thus for all Sylow 2-subgroups . Thus there are elements of 2-power order in . Now . Suppose , and let be the (unique) Sylow 3-subgroup in . Now is normal for some Sylow 3-subgroup , so that ; either case yields a contradiction. If , then contains elements of order 3. But then contains elements of order 1, 2, 3, 4, or 8; no other orders are possible, so that does not contain enough elements. If , then contains elements of order 3, so that contains too many elements.
- Suppose . Now is a group of order 24 which does not contain an element of order 6; thus, by this previous exercise, . In this previous exercise, we found that , and since Sylow subgroups of are Sylow in , . We claim that has a unique normal subgroup of order 4 which is isomorphic to . To see this, recall that every normal subgroup (of an arbitrary group) is a union of conjugacy classes. If is a normal subgroup of order 4, then its elements have order 2 or 4. The elements of these orders in have cycle shape , , or , which have orders 6, 6, and 3, respectively. Thus there is only one possible conjugacy class of elements of order 2 in , which (with the identity) comprise all of . Thus we have , and moreover is characteristic in . Now each normalizer of a conjugate of contains three subgroups of order 8 (which, incidentally, are isomorphic to ). Each of these is contained in a Sylow 2-subgroup of , and in fact is contained in a
*unique*Sylow 2-subgroup of , as otherwise we violate the maximalness of .Now every Sylow 2-subgroup contains as normal subgroups isomorphic copies of , which has 5 elements of order 2, and of , which has 1 element of order 2. Thus there is an element of order 2 which normalizes but is not in . (Recall the definitions of and above.) If fixes 1, then , so that , a violation of Lagrange. Thus must not fix 1. By our calculation above of the elements of order 2 which normalizes Sylow 3-subgroups, we may assume that appears in the cycle decomposition of .

Now permutes the three order 4 subgroups of , and thus must fix at least one. Note that ; however, this implies that some element of does not fix 1, a contradiction.

- Suppose . If , then we have Sylow 2-subgroups. Thus , and this subgroup has order 8, violating the maximalness of . Thus . Since the Sylow 2-subgroups here intersect pairwise in (otherwise would violate the maximalness of ), contains either or elements of 2-power order. Similarly, by Sylow’s Theorem and since the Sylow 3-subgroups of intersect trivially, contains either or elements of order 3. Since every element of is contained in a Sylow subgroup, we see that the only possibility is and . Let be a Sylow 3-subgroup. Then is a group of order 36. In a lemma to this previous exercise, we showed that a group of order 36 which does not have a unique Sylow 3-subgroup contains an element of order 6, a contradiction. Thus is normal, so that in fact , and using Cauchy, we have an element of order 6, for another contradiction.

- If , then since is normal in both and , is divisible by and another prime. Now . We can see that in all but the cases and , either is normal in or some subgroup has sufficiently small index. Suppose . By Sylow’s theorem, . If , then since the Sylow 5-subgroups in are Sylow in , we have for some Sylow 5-subgroup . This is a contradiction of Lagrange. If , then since the Sylow 5-subgroups of intersect trivially, contains elements of order 5. However it also contains at least 17 elements of 2-power order, since ; this is a contradiction, and thus . Suppose . Note that ; say that , , and are the Sylow 2-subgroups of . Now is the pairwise intersection of , , and , so that contains precisely nonidentity elements of 2-power order. By Sylow’s Theorem, . If , then for some subgroup of order 3. Since is also normal in some Sylow 3-subgroup, we have . Neither of these can occur, either because would then be normal of would have sufficiently small index. If , then since Sylow 3-subgroups intersect trivially, there are precisely elements of order 3 in . Now the elements of order 3 or 2-power number , but because , no other orders are possible. Thus does not contain enough elements. If , then since Sylow 3-subgroups intersect trivially, contains precisely elements of order 3. But then contains elements, a contradiction.

Thus , and no simple group of order 720 exists.