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Let be a simple group of order 315. Sylow’s Theorem forces and . Now the left action of on for some Sylow 3subgroup induces an injective permutation representation . Let be a Sylow 5subgroup; note that is Sylow. We have , and ; but this implies that 3 divides 8, which is absurd. 

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Let be a simple group of order 495. Sylow’s Theorem forces , , and . Since the Sylow 5 and 11subgroups of are cyclic, contains at least elements, a contradiction. 

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Let be a simple group of order 735. Note that does not divide , so that no proper subgroup of has index at most 13. Sylow’s Theorem forces mod 49, so that by Lemma 13 and this previous exercise there exist such that is divisible by and some other prime. Thus ; in any case we have a contradiction, either because is normal in or because some subgroup has sufficiently small index. 

Let be a simple group of order 945. Sylow’s Theorem forces and . The left action of on , where is some Sylow 3subgroup, yields an injective permutation representation . Now let be a Sylow 7subgroup; then is also Sylow in . Now , while . This implies that 3 divides 2, which is absurd. 

Example on page 207 

Suppose is simple of order 1365. Then Sylow’s Theorem forces , , and . Since all Sylow subgroups are cyclic, has at least elements, a contradiction. 

Let be a simple group of order 1485. Sylow’s Theorem forces and . The left action of on , where is some Sylow 5subgroup, induces an injective permutation representation . Let be a Sylow 11subgroup; note that is also Sylow in . Now , while ; this implies that 3 divides 10, which is absurd. 

Let be a simple group of order 1575. Note that does not divide since the highest power of 5 dividing is ; thus no proper subgroup of has index at most 9. Now Sylow’s Theorem forces mod 25, so that by Lemma 13 and this previous exercise, there exist such that and normalize and is divisible by and another prime. Thus . In all but the first case either is normal in or its normalizer has sufficiently small index. Thus , and Sylow’s Theorem forces . However we know of at least two distinct Sylow 5subgroups in , namely and . 

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Example on page 205 

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Let be simple of order 2457. Sylow’s Theorem forces and . Now let be a Sylow 13subgroup of ; , so that for some Sylow 7subgroup . Now is a subgroup of , and since 7 does not divide 12, is abelian. Thus . But we know that , a contradiction. 

Let be a simple group of order 2475. Note that does not divide , so that no proper subgroup of has index at most 10. Now Sylow’s Theorem forces mod 25, so that by Lemma 13 and this previous exercise there exist such that and normalize and is divisible by and some other prime. Thus . In the last two cases, either is normal in or its normalizer has sufficiently small index. Thus . In either case, Sylow’s Theorem forces ; this is a contradiction since we know of at least two Sylow 5subgroups in , namely and . 

Let be a simple group of order 2625. Note that does not divide , so that no proper subgroup of has index at most 14. Sylow’s Theorem forces mod 25, so that by Lemma 13 and this previous exercise there exist such that is divisible by and some other prime. This implies that . In any case we have a contradiction, either because is normal or some subgroup has sufficiently small index. 

Let be simple of order 2775. Note that does not divide , so that no proper subgroup of has index at most 36. Now Sylow’s Theorem forces , , and . Since the Sylow 13 and 3subgroups of are cyclic, has at least elements, a contradiction. 

Let be simple of order 2835. Note that does not divide since the highest power of 3 which divides is . On the other hand, Sylow’s Theorem forces , so that has a subgroup of index 7, a contradiction. 

Let be simple of order 2907. Sylow’s Theorem forces and . Since Sylow 17 and 19subgroups are cyclic, has at least elements, a contradiction. 

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Example on page 203 

Let be a simple group of order 3465. Sylow’s Theorem forces and . Let be a Sylow 11subgroup. Now , so that for some Sylow 7subgroup P_7. Thus is a subgroup, and moreover because 7 does not divide 10, is abelian. Thus , and we have . The left action of on induces an injective permutation representation . Note that is Sylow in . Now while ; this is a contradiction since . 

Example on page 206 

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Let be a simple group of order 4455. Note that does not divide ; thus no proper subgroup of has index at most 10. Now Sylow’s Theorem forces and , and . Suppose now that has a subgroup of index 11. Then via the premutation representation induced by left multiplication of the cosets of this subgroup, we have , and notably the Sylow 11subgroups of are Sylow in . Now let be a Sylow 11subgroup. We have and , but this implies that 9 divides 10 by Lagrange, which is absurd. In particular, . Suppose now that and let be a Sylow 5subgroup. Then . By Sylow, we have . If is a Sylow 11subgroup, it is Sylow in as well, and we have . However, we know that , a contradiction by Lagrange. Thus , and moreover, . In particular, note that contains elements of order 5 or 11; there are 441 elements remaining. Now choose distinct such that is maximal. Now . We consider each case in turn. If , then all Sylow 3subgroups of intersect trivially, and contains elements in Sylow 3subgroups, a contradiction. If , we can carefully fill up some of the 55 Sylow 3subgroups of with the remaining elements. Let be a Sylow 3subgroup; note that . Now let be a second Sylow 3subgroup. Since the largest possible intersection of Sylow 3subgroups contains 3 elements, there are (at least) elements in which are not also in . Now let be a third Sylow 3subgroup; in the worst case, the intersections and do not overlap. Thus there are at least elements in which are not in or . Carrying on in this manner, after filling up seven Sylow 3subgroups, there are at least 504 elements in Sylow 3subgroups of , a contradiction. If , then using this previous exercise, is divisible by and another prime. Thus . We can see that in the last four cases, either is normal in , or its normalizer has sufficiently small index – either or 11. If , then by Sylow, . Since the Sylow 5subgroups of are Sylow in , this yields P_3 \cap Q_3 \leq N_G(P_5)$ for some Sylow 5subgroup . But Sylow’s Theorem forces , a contradiction. Thus we have . In this case, Sylow forces . Now let be the unique Sylow 3subgroup and the unique Sylow 11subgroup. In particular, normalizes , and . Note that, since , does not normalize ; thus if , . But does normalize , so that we have . This is a contradiction since is maximal among the intersections of distinct Sylow 3subgroups of . If , we have . In each case either is normal in , the index of is too small, or has a subgroup of index 11, each of which is a contradiction. 

Let be a simple group of order 4563. Note that does not divide , so that no proper subgroup of has index at most 25. Sylow’s Theorem then forces mod 9, so that by Lemma 13 and this previous exercise there exist Sylow 3subgroups and in such that is divisible by and another prime. But then , which yields either normal in or a subgroup of sufficiently small index. 

Let be a simple group of order 4725. Note that does not divide since the largest power of 5 dividing is ; thus no proper subgroup of has index at most 9. Now Sylow’s Theorem forces mod 25, so that by Lemma 13 and this previous exercise, there exist such that and both normalize and is divisible by and some other prime. Thus . We can see that in all but the first three cases, either is normal in or its normalizer has sufficiently small index. Thus . In each case, Sylow’s Theorem forces , but this is a contradiction since are distinct Sylow 5subgroups. 

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Let be a simple group of order 5355. Note that does not divide , so that no proper subgroup of has index at most 16. Sylow’s Theorem forces . Let be a Sylow 17subgroup; we have . Let be a Sylow 3subgroup; is also Sylow in , and is a subgroup. Moreover, because 3 does not divide 16 and 17 does not divide 8, is abelian. Thus . However, Sylow’s Theorem and the minimum index condition force , a contradiction. 

Let be a simple group of order 5445. Note that does not divide since the largest power of 11 which divides is ; thus no proper subgroup of has index at most 21. Sylow’s Theorem forces . Since mod 121, by Lemma 13 and this previous exercise there exist Sylow 11subgroups and in such that is divisible by and some other prime. Thus the index of in is either 1, 3, 5, or 15; in each case we have a contradiction, either because is normal in or a subgroup has sufficiently small index. 

Let be a simple group of order 6075. Note that does not divide , since the highest power of 3 which divides is . Thus no proper subgroup of has index at most 11. Sylow’s Theorem forces , and since mod 9, by Lemma 13 and this previous exercise there exist Sylow 3subgroups and in such that is divisible by and 5. But then either is normal in or has a subgroup of index 5, a contradiction. 

Let be a simple group of order 6375. Sylow’s Theorem forces and , with . If , then since the Sylow 17 and 3subgroups of intersect trivially, contains at least elements, a contradiction. If , let be a Sylow 3subgroup. Then , so that by Cauchy there is a Sylow 17subgroup of which normalizes . Now is a subgroup and 3 does not divide 16, so that is abelian. Thus . But we already know that , a contradiction. Thus . Now contains elements of order 17 and elements of order 3, so that there are at most elements of order a power of 5. Let be Sylow 5subgroups. The largest possible intersection of and contains 25 elements, so that , a contradiction. 

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Let be a simple group of order 6615. Note that does not divide since the highest power of 3 dividing is . Thus no proper subgroup of has index at most 8; this and Sylow’s Theorem force . Now let be a Sylow 3subgroup. We have . By Sylow again, ; let be the unique Sylow 5subgroup; note that is Sylow in as well. Now are both normal and intersect trivially, and by Lagrange. Thus by the recognition theorem for direct products. In particular, . However, Sylow also forces , a contradiction by Lagrange. 

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Let be a simple group of order 6825. Sylow’s Theorem forces and . Let be a Sylow 7subgroup; we have , so that by Cauchy, for some Sylow 13subgroup . Now is a subgroup, and since 7 does not divide 12, is abelian. Thus ; but we already know that , a contradiction. 

Let be a simple group of order 7371. Sylow’s Theorem forces and . Let be a Sylow 13subgroup. Now . By Cauchy, some Sylow 7subgroup normalizes . Thus is a subgroup, and moreover because 7 does not divide 12, is abelian. Thus . But we already know that , a contradiction. 

Let be a simple group of order 7425. Note that does not divide , so that no proper subgroup of has index at most 10. Sylow’s Theorem forces mod 25, so that by Lemma 13 and this previous exercise there exist such that and normalize and is divisible by and another prime. Thus . We can see that in the last three cases, either is normal in or its normalizer has sufficiently small index. Moreover, in the first three cases, by Sylow, a contradiction since are distinct Sylow 5subgroups. Thus . If , then we have a contradiction since are Sylow, thus Sylow’s Theorem forces . Since Sylow 5subgroups of are Sylow in , and has 11 Sylow 5subgroups, every Sylow 5subgroup of normalizes . We now consider the subgroup . Note that is divisible by and by another prime since contains more than one Sylow 5subgroup. We saw previously that there are seven possible orders for , taking into account the necessary value and the minimal index condition. Thus . Note that if , then by Sylow. Moreover, the unique Sylow 11subgroup is also Sylow in . Thus (for instance) the Sylow 5subgroup normalizes . However, since , this is absurd. Thus , and in fact . In particular, is generated by its Sylow 5subgroups. But then , since every element of can be written as a product of elements in Sylow 5subgroups and every elements of a Sylow 5subgroup normalizes . This is a contradiction becasue is a proper nontrivial subgroup. 

Let be a simple group of order 7875. Note that does not divide , since the largest power of 5 dividing is . Thus no proper subgroup of has index at most 14. Now Sylow’s Theorem forces mod 25, so that by Lemma 13 and this previous exercise there exist such that and normalize and is divisible by and another prime. Thus we have . We can see that in every case except , either is normal in or its normalizer has sufficiently small index. Now by the congruence and divisibility criteria of Sylow’s Theorem, but we know that has at least two Sylow 5subgroups namely and . 

Let be a simple group of order 8325. Note that does not divide , so that no proper subgroup of has index at most 36. Now Sylow’s Theorem forces mod 25. Then by Lemma 13 and this previous exercise, there exist such that and normalize and is divisible by and some other prime. Thus , . We can see that in all except the first two cases, either is normal in or its normalizer has sufficiently small index. In either of the remaining cases, , we have ; this is a contradiction since we know of at least two Sylow 5subgroups in , namely and . 

Let be a simple group of order 8505. Note that does not divide since the largest power of 3 which divides is ; thus has no proper subgroups of index at most 11. Now Sylow’s Theorem forces ; since mod 9, by Lemma 13 and this previous exercise there exist Sylow 3subgroups such that is divisible by and some other prime; there are three cases to consider, and we either have normal, has a subgroup of index 5, or has a subgroup of index 7, all of which are contradictions. 

Let be a simple group of order 8721. Sylow’s Theorem forces ; if is a Sylow 3subgroup we have an injective permutation representation induced by the left action of on . Let be a Sylow 17subgroup; note that is also Sylow in . Now , while . This yields a contradiction. 

Let be a simple group of order 8775. Sylow’s Theorem forces and . Let be a Sylow 13subgroup. Now . By Sylow’s Theorem, there exists a Sylow 5subgroup , and by cardinality considerations is Sylow in as well. Thus is a subgroup, and moreover because 5 does not divide 12 and 13 does not divide 24, is abelian. Thus . However, we already know that , a contradiction. 

Let be simple of order 8883. Sylow’s Theorem forces and . Since Sylow 7 and 47subgroups of intersect trivially, contains at least elements, a contradiction. 

Let be a simple group of order 8925. Note that Sylow’s Theorem forces , , and . Let be a Sylow 5subgroup. If , then . By Cauchy, we have for some Sylow 17subgroup . Thus is a subgroup. Now 5 does not divide 16 and 17 does not divide 24, so that is abelian. Thus . But we know that , a contradiction. If , then . By Cauchy, we have for some Sylow 7subgroup . Thus is a subgroup. Again because 5 does not divide 6 and 7 does not divide 24, is abelian, hence . However, we know that , a contradiction. 

Let be simple of order 9045. Sylow’s Theorem forces and . Since the Sylow 5 and 67subgroups of intersect trivially, contains at least elements, a contradiction. 

Let be a simple group of order 9405. Sylow’s Theorem forces and . Let be a Sylow 11subgroup. Now . By Cauchy, there exists a Sylow 19subgroup which normalizes ; now is a subgroup, and because 11 does not divide 18, is abelian. Thus . But we already know that , a contradiction. 

Let be a simple group of order 9477. Sylow’s Theorem forces . Since mod 9, Lemma 13 and this previous exercise imply that there exist such that is maximal in (hence nontrivial) and is divisible by and 13. But then is normal in , a contradiction. 

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Let be a simple group of order 9765. Note that does not divide , so that no proper subgroup of has index at most 30. This and Sylow’s Theorem force . In either case, 5 divides , where is some Sylow 3subgroup. By Cauchy, there exists a Sylow 5subgroup . Now is a subgroup, and because 3 does not divide 4 and 5 does not divide 8, is abelian. Thus . This, Sylow’s Theorem, and our observation on minimal subgroup indices imply that . Thus , and by Cauchy we have for some Sylow 7subgroup . Now is a subgroup and since 5 does not divide 6, is abelian. Thus ; but we already know that , so this is a contradiction. 