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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Simple groups

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..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

Use Sylow?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Simple groups

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..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Simple groups

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..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

Yes, that’s Corollary 9.25 in Humphreys’s 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

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

I know. I’ll list all the group orders and check them one by one. When I make another breakthrough, I’ll 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)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

According to Humphreys (pp 46–47) 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. I’m 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. I’ll go and update my post above.

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

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Simple groups

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?

---

"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..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

Any Abelian group of order 9 will definitely have a subgroup of order 3; hence it’s 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 can’t be any of the first six elements. Suppose

(so

). Then

Contradiction! Hence the elements of this group must be

So now the element

must be one of these. It obviously can’t be any of the first six or

. If

then cancelling gives

(contradiction);

(again!). Hence

Wow, I have just made an amazing discovery on my own – at four in the morning!!

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

Sorry, Ricky. I was posting when you were.

Well, I’ll be coming to the Sylow theorems in Humphreys very soon. Right now I’m at the chapter on the orbit–stabilizer theorem.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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! It’s 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)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Simple groups

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.

---

"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..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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 H be a subgroup of a group G with finite index n. Then there exists a normal subgroup N of G contained in H with n dividing |G : N| and |G : N| dividing n!.

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. I’ve 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 I’ll go away and work on these; in a day or two, I’ll come back and give the results.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

Okay, let’s 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, it’s well defined, and it’s bijective. Moreover,

forms a group isomorphic to a subgroup of the symmetric group

.

Now define a homomorphism

by

. Yup, it’s a homomorphism all right. The kernel of

is this peculiar subgroup of

:

[align=center]

[/align]

This is the N that is referred to in Humphreys’s Corollary 9.23. By imposing special properties on the subgroup

, you can have a normal subgroup

with special properties. Corollary 9.23 is obtained by imposing the condition

on

.

As an aside, you could also impose the condition that

be the trivial group

. The result is Cayley’s 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)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

All right, let’s 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 I’m 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)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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 I’ve only used Humphreys’s Corollay 9.23 with only the knowledge that Sylow

-subgroups exist. I’m fairly happy with myself for this.

Last edited by JaneFairfax (2009-04-15 07:20:52)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

Incidentally it’s 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

.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Simple groups

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...

---

"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..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

Oops, you’re 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)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Simple groups

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

with

divisible by 5 and dividing 5! = 120.

must also divide 80.

is not simple.