If is a nonabelian simple group of order , prove that . [Hint: Eliminate all orders but 60.]

In the following table, we list the integers , with a note justifying the assertion that no group of order is simple.

Factorization | Note | |

(prime) | ||

(prime) | ||

(prime) | ||

(prime) | ||

(prime) | ||

(prime) | ||

(prime) | ||

(prime) | ||

(prime) | ||

Lemma 3 | ||

(prime) | ||

(prime) | ||

Lemma 8 | ||

(prime) | ||

Lemma 4 | ||

(prime) | ||

(prime) | ||

(prime) | ||

Lemma 3 | ||

(prime) | ||

Lemma 1 | ||

Lemma 6 | ||

(prime) | ||

NA | ||

(prime) | ||

(prime) | ||

(prime) | ||

Lemma 9 | ||

(prime) | ||

(prime) | ||

Lemma 5 | ||

(prime) | ||

Lemma 7 | ||

Lemma 2 | ||

(prime) | ||

Lemma 10 | ||

Lemma 3 | ||

(prime) | ||

- : All such groups are cyclic, hence abelian.
- : No group of order , , distinct primes, is simple.
- : No group of order , and distinct primes, is simple.
- : No group of order , , , and distinct primes, is simple.
- : Every -group has a nontrivial center, which is normal.

Lemma 1: If is a finite group of order , where , then is not simple. Proof: By Sylow's Theorem, the number of Sylow -subgroups divides and is congruent to 1 mod . Thus , and the (unique) Sylow -subgroup is normal. Hence is not simple.

Lemma 2: If is a finite group of order , where , then is not simple. Proof: By Sylow's Theorem, divides and is congruent to 1 mod . Thus , and the (unique) Sylow -subgroup is normal. Hence is not simple.

Lemma 3: If is a finite group of order , with , then is not simple. Proof: Suppose otherwise. The number of Sylow 2-subgroups divides 3 and is not 1 since is simple. Thus . Note that acts on its Sylow 2-subgroups by conjugation, yielding a permutation representation . Let be the kernel of this action; note that for every Sylow 2-subgroup , we have . Since is normal in and is simple, there are two possibilities. If , we have for each Sylow 2-subgroup, thus a Sylow 2-subgroup is normal in , a contradiction. If , then the permutation representation is injective. But then , a contradiction. Thus is not simple.

Lemma 4: No group of order is simple. Proof: By Sylow's Theorem, divides 8 and is congruent to 1 mod 5. Thus , and the (unique) Sylow 5-subgroup of is normal. Hence is not simple.

Lemma 5: No group of order is simple. Proof: Suppose otherwise that is a simple group of order 80. By Sylow's Theorem, and . Since is simple, and . Now the Sylow 5-subgroups have prime order, and thus intersect trivially. Thus has elements of order 5. Let be a Sylow 2-subgroup. There exists a second Sylow 2-subgroup , and there exists an element not in ; then has at least 16 + 1 = 17$ elements in Sylow 2-subgroups. So has at least 81 elements, a contradiction. So no simple group of order 80 exists.

Lemma 6: No group of order is simple. Proof: Suppose to the contrary that is a simple group of order 56. Sylow's Theorem and the simplicity of force and . Since all Sylow 7-subgroups of intersect trivially, has at least elements of order 7. Let be a Sylow 2-subgroup, and let be a Sylow 2-subgroup distinct from . contains an element not in ; so contains at least elements in Sylow 2-subgroups. Hence , a contradiction. Thus no such group exists.

Lemma 7: No group of order is simple. Proof: By Sylow’s Theorem, divides 12 and is congruent to 1 mod 7; thus . The (unique) Sylow 7-subgroup is then normal.

Lemma 8: No group of order is simple for . Proof: Suppose to the contrary that is a simple group of this order. By Sylow’s Theorem, . If , then acts by conjugation on its (four) Sylow 3-subgroups, yielding a permutation representation with kernel . Note that for each Sylow 3-subgroup . If , then , so that is normal and , a contradiction. If , then divides . But 9 divides and does not divide , a contradiction. Thus no simple group of this order exists.

Lemma 9: No group of order is simple for . Proof: Suppose to the contrary that is a simple group of this order. By Sylow’s Theorem, . If , then acts by conjugation on its (four) Sylow 3-subgroups, yielding a permutation representation with kernel . Note that for each Sylow 3-subgroup . If , then , so that is normal in and , a contradiction. If , then ; but , a contradiction. So no simple group of this order exists.

Lemma 10: No group of order is simple. Proof: Suppose to the contrary that is a simple group of order 90. Sylow’s Theorem forces ; we may assume since is simple. Now acts on its (six) Sylow 5-subgroups by conjugation, yielding a permutation representation with kernel . Note that for each Sylow 5-subgroup . If , then , so that is normal in , a contradiction. If , we have . Note that every element of order 5 in is represented by a 5-cycle in ; thus . Clearly by order considerations. Now is a nontrivial normal subgroup of , a contradiction. Thus no simple group of order 90 exists.

Finally, note that if is a finite group of order with , then is not simple. If is a simple group of order 60, we saw in the text that .

## Comments

Typo in the last paragraph: n not 60.

Why do you have 2r 2x 2y 2W 2X . . . in the leftmost column of your table?

Thanks!

As for the letters in the left column- I’m not sure what’s going on. It’s correct in the page source. Weird…

Okay, the numbers show up correctly now.

To fix it, I put the messed up numbers in a pair of braces. Compare: (no braces) and (braces) .

Strangely, 37, 40, and 76 required two sets of braces. Compare: (no braces) , (one pair) , (two pairs) .

Strange.

> Strangely, 37, 40, and 76 required two sets of braces.

First Dummit and Foote, now wordpress. When will it end?!

I don’t understand “Note that every element in is represented by a 5-cycle in ” in Lemma 10.

That should read ‘every element of order 5 in ‘.

For some reason, WordPress isn’t letting me fix the typo at the moment.

Okay- it’s fixed now.

Just a thought, isn’t Z/pZ simple for prime p because there do not exist any nontrivial proper subgroups? If there are no nontrivial proper subgroups, there cannot possibly be any nontrivial proper normal subgroups.

True.

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