You are not logged in.

- Topics: Active | Unanswered

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Just got this as a homework assignment:

Prove that all groups of order < 120 are either not simple or of prime order, except 60.

5 pages of Latex later, I got it. Probably about the most fun I've had doing group theory in a while. Anyone wanna try it? No heavy machinery allowed (groups of order (p^a)(q^b) or of odd order).

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Use Sylow?

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Certainly a powerful tool, but not the only thing you use.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

For a start, all groups of order

where is an odd prime and are not simple. By Sylow, there would be a subgroup of order and such a subgroup would be normal because it has index 2. That is what I proved here: http://www.mathisfunforum.com/viewtopic.php?id=10264.This takes care of all groups with the following orders: 6, 10, 14, 18, 22, 26, 34, 38, 46, 50, 54, 58, 62, 74, 82, 86, 94, 98, 106, 118.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Indeed, you can use Sylow's Theorem directly to prove an even more general result:

If G = p^a * q and p > q, then G is not simple.

It might first be a good idea to prove that:

If G = pq, p > q, then G is not simple.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Yes, thats Corollary 9.25 in Humphreyss book (see this post). Now that I finally understand this, I can shoot down some more non-simple finite groups:

15, 21, 33, 35, 39, 51, 55, 57, 65, 69, 75, 77, 85, 87, 91, 93, 95, 111, 115, 119

Which leaves us with the following still to check:

8, 9, 12, 16, 18, 20, 24, 25, 27, 28, 30, 32, 36, 40, 42, 44, 45, 48, 49, 52, 56, 63, 64, 66, 68, 70, 72, 76, 78, 80, 81, 84, 88, 90, 92, 96, 99, 100, 102, 104, 108, 112, 114, 116, 117

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

I know. Ill list all the group orders and check them one by one. When I make another breakthrough, Ill update the list.

2 simple (prime order)

3 simple (prime order)

4 not simple (has normal subgroup of order 2)

5 simple (prime order)

6 not simple (Sylow 3-subgroup has index 2)

7 simple (prime order)

8 not simple: **see post #8 below**

9 not simple: **see posts #10 and #12 below**

10 not simple (Sylow 5-subgroup has index 2)

11 simple (prime order)

12 not simple: **see post #13 below**

13 simple (prime order)

14 not simple (Sylow 7-subgroup has index 2)

15 not simple (Sylow 5-subgroup has index 3 (smallest prime divisor))

16 not simple: **see post #12 below**

17 simple (prime order)

18 not simple (Sylow 3-subgroup has index 2)

19 simple (prime order)

20 not simple: **see post #14 below**

21 not simple (Sylow 7-subgroup has index 3 (smallest prime divisor))

22 not simple (Sylow 11-subgroup has index 2)

23 simple (prime order)

24 not simple: **see post #13 below**

25 not simple: **see post #12 below**

26 not simple (Sylow 13-subgroup has index 2)

27 not simple: **see post #12 below**

28 not simple: **see post #14 below**

29 simple (prime order)

30 not simple: **see post #22 below**

31 simple (prime order)

32 not simple: **see post #12 below**

33 not simple (Sylow 11-subgroup has index 3 (smallest prime divisor))

34 not simple (Sylow 17-subgroup has index 2)

35 not simple (Sylow 7-subgroup has index 5 (smallest prime divisor))

36 not simple: **see post #14 below**

37 simple (prime order)

38 not simple (Sylow 19-subgroup has index 2)

39 not simple (Sylow 13-subgroup has index 3 (smallest prime divisor))

40 not simple: **see post #20 below**

41 simple (prime order)

42 not simple: **see post #19 below**

43 simple (prime order)

44 not simple: **see post #14 below**

45 not simple: **see post #20 below**

46 not simple (Sylow 23-subgroup has index 2)

47 simple (prime order)

48 not simple: **see post #13 below**

49 not simple: **see post #12 below**

50 not simple (Sylow 5-subgroup has index 2)

51 not simple (Sylow 17-subgroup has index 3 (smallest prime divisor))

52 not simple: **see post #14 below**

53 simple (prime order)

54 not simple (Sylow 3-subgroup has index 2)

55 not simple (Sylow 11-subgroup has index 5 (smallest prime divisor))

56 not simple: **see post #22 below**

57 not simple (Sylow 19-subgroup has index 3 (smallest prime divisor))

58 not simple (Sylow 29-subgroup has index 2)

59 simple (prime order)

60 can be simple

61 simple (prime order)

62 not simple (Sylow 31-subgroup has index 2)

63 not simple: **see post #20 below**

64 not simple: **see post #12 below**

65 not simple (Sylow 11-subgroup has index 5 (smallest prime divisor))

66 not simple: **see post #19 below**

67 simple (prime order)

68 not simple: **see post #14 below**

69 not simple (Sylow 23-subgroup has index 3 (smallest prime divisor))

70 not simple: **see post #20 below**

71 simple (prime order)

72 not simple: **see post #32 below**

73 simple (prime order)

74 not simple (Sylow 37-subgroup has index 2)

75 not simple (Sylow 5-subgroup has index 3 (smallest prime divisor))

76 not simple: **see post #14 below**

77 not simple (Sylow 11-subgroup has index 7 (smallest prime divisor))

78 not simple: **see post #19 below**

79 simple (prime order)

80 not simple: **see post #25 below**

81 not simple: **see post #12 below**

82 not simple (Sylow 41-subgroup has index 2)

83 simple (prime order)

84 not simple: **see post #20 below**

85 not simple (Sylow 17-subgroup has index 5 (smallest prime divisor))

86 not simple (Sylow 43-subgroup has index 2)

87 not simple (Sylow 29-subgroup has index 3 (smallest prime divisor))

88 not simple: **see post #19 below**

89 simple (prime order)

90 not simple: **see post #31 below**

91 not simple (Sylow 13-subgroup has index 7 (smallest prime divisor))

92 not simple: **see post #14 below**

93 not simple (Sylow 31-subgroup has index 3 (smallest prime divisor))

94 not simple (Sylow 47-subgroup has index 2)

95 not simple (Sylow 19-subgroup has index 5 (smallest prime divisor))

96 not simple: **see post #13 below**

97 simple (prime order)

98 not simple (Sylow 7-subgroup has index 2)

99 not simple: **see post #19 below**

100 not simple: **see post #14 below**

101 simple (prime order)

102 not simple: **see post #19 below**

103 simple (prime order)

104 not simple: **see post #19 below**

105 not simple: **see post #22 below**

106 not simple (Sylow 53-subgroup has index 2)

107 simple (prime order)

108 not simple: **see post #14 below**

109 simple (prime order)

110 not simple: **see post #19 below**

111 not simple (Sylow 37-subgroup has index 3 (smallest prime divisor))

112 not simple: **see post #33 below**

113 simple (prime order)

114 not simple (Sylow 57-subgroup has index 2)

115 not simple (Sylow 23-subgroup has index 5 (smallest prime divisor))

116 not simple: **see post #14 below**

117 not simple: **see post #19 below**

118 not simple (Sylow 59-subgroup has index 2)

119 not simple (Sylow 17-subgroup has index 7 (smallest prime divisor))

*Last edited by JaneFairfax (2009-05-16 00:02:48)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

According to Humphreys (pp 4647) there are five nonisomorphic groups of order 8. Three of these are Abelian, being products of cyclic groups. Since they are Abelian, all their subgroups are normal (and they will definitely have subgroups of order 2).

The other two are the dihedral group and the quaternion group. In the former, the group of rotations are a subgroup of index 2. Im not familiar with properties of the quaternion group, but I do gather that there is an element of order 4. Hence the quaternion group also has a subgroup of index 2. So these two groups are not simple.

So I have convinced myself that all subgroups of order 8 are not simple. Ill go and update my post above.

*Last edited by JaneFairfax (2009-04-12 08:07:28)*

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Jane, all p-groups have a nontrivial center. That means not only are all groups of order 8 not simple, but any power of 2 (or any prime for that matter) as well.

Would you like to see a proof of this?

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Any Abelian group of order 9 will definitely have a subgroup of order 3; hence its not simple.

Suppose there is a nonabelian group of order 9. Every nonidentity element must have order 3, so we can list 7 of its elements straightaway:

Now, what can the element

be? It cant be any of the first six elements. Suppose (so ). ThenContradiction! Hence the elements of this group must be

So now the element

must be one of these. It obviously cant be any of the first six or . If then cancelling gives (contradiction); (again!). HenceWow, I have just made an amazing discovery on my own at four in the morning!!

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Sorry, Ricky. I was posting when you were.

Well, Ill be coming to the Sylow theorems in Humphreys very soon. Right now Im at the chapter on the orbitstabilizer theorem.

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Ricky wrote:

Jane, all p-groups have a nontrivial center. That means not only are all groups of order 8 not simple, but any power of 2 (or any prime for that matter) as well.

Would you like to see a proof of this?

Got it! Its Proposition 10.20 on page 94 of Humphreys!

And Corollary 10.22 on the following page says that every group of order the square of a prime is Abelian. Well, well. All my own work on groups of order 9 for nothing.

*Last edited by JaneFairfax (2009-04-12 21:19:34)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

The Sylow 2-subgroup

has index 3 in , so using a result in Humphreys, there is a subgroup normal in and contained in such that 3 divides and divides 3! = 6. Hence is either 3 or 6. This ensures that is never 1 or therefore is not simple!Thank goodness for 3! being a conveniently small number.

*Last edited by JaneFairfax (2009-04-13 07:00:28)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

The Sylow non-2-subgroup

has index 4 in , so there is a subgroup normal in and contained in such that 4 divides and divides 4! = 24. (in the case of the 108, ) is not simple.Yo, man! I hope Ricky is appreciating my hard work!

*Last edited by JaneFairfax (2009-05-11 20:58:48)*

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

From post #13:

...so using a result in Humphreys...

Can you post this result? I know of a similar result, but one must assume G is simple to use it (and hence you'd be going by contradiction).

Yo, man! I hope Ricky is appreciating my hard work!

Certainly. But perhaps it is wiser to read more on Sylow first before trying to prove more of these.

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Ricky wrote:

From post #13:

...so using a result in Humphreys...

Can you post this result? I know of a similar result, but one must assume G is simple to use it (and hence you'd be going by contradiction).

The result is Corollary 9.23 in the book by Humphreys:

Let

Hbe a subgroup of a groupGwith finite indexn. Then there exists a normal subgroupNofGcontained inHwithndividing |G:N| and |G:N| dividingn!.

Ricky wrote:

Yo, man! I hope Ricky is appreciating my hard work!

Certainly. But perhaps it is wiser to read more on Sylow first before trying to prove more of these.

Well, yes. Ive just started reading the chapter on the Sylow theorems in Humphreys. So now I know a few more tricks:

The number of Sylow -subgroups of is congruent to 1 modulo and divides . If the Sylow -subgroup is unique, then it is normal in .

Now Ill go away and work on these; in a day or two, Ill come back and give the results.

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Okay, lets start at the beginning.

Let be a group, a subgroup and the set of all left cosets of in .

Any element

induces a permutation by for each left coset . Yes, its well defined, and its bijective. Moreover, forms a group isomorphic to a subgroup of the symmetric group .Now define a homomorphism

by . Yup, its a homomorphism all right. The kernel of is this peculiar subgroup of :[align=center]

[/align]This is the *N* that is referred to in Humphreyss Corollary 9.23. By imposing special properties on the subgroup

As an aside, you could also impose the condition that

be the trivial group . The result is Cayleys theorem: Every group is isomorphic to a subgroup of a symmetric group. This is Corollary 9.24 in Humphreys.*Last edited by JaneFairfax (2009-04-15 01:50:16)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

All right, lets go one step further and prove Corollary 9.23.

Since is normal in and is a subgroup of , is a (normal) subgroup of and . divides .

The second assertion is proved by the fact that

is isomorphic to the image of in the previous post, which is a subgroup of , which is isomorphic to a subgroup of . divides .Well, Ricky, I hope you are checking everything I say and ascertaining that it makes sense. You have to, because Im going through a bottle of rosé wine while posting all this stuff so I could easily have messed up something here and there.

*Last edited by JaneFairfax (2009-04-14 04:16:57)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

These are groups of orders

where is a prime and . I will show that the Sylow -subgroup of any such group is unique.The number of Sylow

-subgroups is of the form . Each of these is cyclic of prime order; therefore distinct Sylow -subgroups have only the identity in common. So the total number of elements in the union of all the Sylow -subgroups is . This cannot exceed the total number of elements in .Hence

; in other words, there is only 1 Sylow -subgroup! This must be normal in since it is equal to all its conjugates in ; therefore is not simple.Well! This is the first time I am diligently making use of the Sylow tools; previously Ive only used Humphreyss Corollay 9.23 with only the knowledge that Sylow

-subgroups exist. Im fairly happy with myself for this.*Last edited by JaneFairfax (2009-04-15 07:20:52)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

For 90, I made a mistake so forget it for the time being.

In the other cases, the Sylow

-subgroup for the max prime divisor is unique, therefore normal, as can be ascertained by patiently checking each individual case.*Last edited by JaneFairfax (2009-05-12 06:07:51)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Incidentally its been pointed out to me that my result in #19 could also be proved by considering the fact that divides . If then and so would not divide . Hence divides only if .

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

If

then has 1 or 6 Sylow 5-subgroups and 1 or 10 Sylow 3-subgroups, each Sylow subgroup being cyclic of prime order. If there are 6 Sylow 5-subgroups, the union of these subgroups has elements; if there are 10 Sylow 3-subgroups, their union will have elements, all distinct from the elements of the other union except the identity. Since is only 30, therefore cannot have 6 Sylow 5-subgroups and 10 Sylow 3-subgroups at the same time; hence must have either a unique (therefore normal) Sylow 5-subgroup or a unique (∴ normal) Sylow 3-subgroup.If

then has 1 or 15 Sylow 7-subgroups and 1 or 21 Sylow 5-subgroups (each Sylow subgroup being cyclic of prime order). By a similar argument to the have, we find that must have either a normal Sylow 7-subgroup or a normal Sylow 5-subgroup.If

then has 1 or 8 Sylow 7-subgroups (cylic of order 7). If 8, then the union of the the Sylow 7-subgroups has 48 non-identity elements. This leaves 56 − 48 = 8 other elements in the group (including the identity), which must then make up the unique Sylow 2-subgroup. Hence must have either a normal Sylow 7-subgroup or a normal Sylow 2-subgroup.*Last edited by JaneFairfax (2009-05-12 06:03:28)*

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Post #20 is incorrect for groups of order 90. It is possible to have six Sylow 5-subgroups. I'll probably be checking the rest over once the semester is...

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Oops, youre right. I was a bit careless there.

PS: I now realize that I meant to take the largest-order Sylow subgroup instead.

*Last edited by JaneFairfax (2009-05-12 00:25:18)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

The Sylow 2-subgroup has index 5 and so by Humphreyss Corollary 9.23 there is a normal subgroup

with divisible by 5 and dividing 5! = 120. must also divide 80. is not simple.Offline