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#1 2008-09-25 07:37:29

Ricky
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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..."
 

#2 2008-09-25 07:46:46

JaneFairfax
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Re: Simple groups

Use Sylow?


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#3 2008-09-25 07:50:11

Ricky
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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..."
 

#4 2008-10-09 07:48:06

JaneFairfax
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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. smile

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. http://www.makephpbb.com/oceanhunter/images/smiles/applause.gif


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#5 2008-10-09 09:12:11

Ricky
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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..."
 

#6 2009-04-13 04:31:08

JaneFairfax
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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: big_smile

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


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A: Click here for answer.
 

#7 2009-04-13 05:47:25

JaneFairfax
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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. smile

  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 22:02:48)


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#8 2009-04-13 06:02:18

JaneFairfax
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Re: Simple groups

smile

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. big_smile

Last edited by JaneFairfax (2009-04-13 06:07:28)


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#9 2009-04-13 13:00:20

Ricky
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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..."
 

#10 2009-04-13 13:22:43

JaneFairfax
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Re: Simple groups

smile

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

eek

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


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#11 2009-04-13 13:25:24

JaneFairfax
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Re: Simple groups

Sorry, Ricky. I was posting when you were. dizzy

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. smile


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#12 2009-04-13 19:09:59

JaneFairfax
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Re: Simple groups

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! It’s Proposition 10.20 on page 94 of Humphreys! big_smile

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. roll

Last edited by JaneFairfax (2009-04-13 19:19:34)


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#13 2009-04-13 20:00:33

JaneFairfax
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Re: Simple groups

smile

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! big_smile

Thank goodness for 3! being a conveniently small number. wink

Last edited by JaneFairfax (2009-04-14 05:00:28)


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#14 2009-04-14 04:40:58

JaneFairfax
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Re: Simple groups

smile

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 (2009-05-12 18:58:48)


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#15 2009-04-14 10:11:37

Ricky
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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..."
 

#16 2009-04-15 00:59:15

JaneFairfax
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Re: Simple groups

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 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!.

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. 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. cool


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#17 2009-04-15 01:30:38

JaneFairfax
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Re: Simple groups

Okay, let’s start at the beginning. smile


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
:



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
. cool

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. up

Last edited by JaneFairfax (2009-04-15 23:50:16)


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A: Click here for answer.
 

#18 2009-04-15 02:11:26

JaneFairfax
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Re: Simple groups

All right, let’s go one step further and prove Corollary 9.23. smile


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 rosι wine while posting all this stuff – so I could easily have messed up something here and there. roflol

Last edited by JaneFairfax (2009-04-15 02:16:57)


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#19 2009-04-15 23:45:14

JaneFairfax
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Re: Simple groups

smile

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. http://www.makephpbb.com/oceanhunter/images/smiles/applause.gif

Last edited by JaneFairfax (2009-04-16 05:20:52)


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A: Click here for answer.
 

#20 2009-04-16 19:28:57

JaneFairfax
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Re: Simple groups

smile

For 90, I made a mistake so forget it for the time being. http://i188.photobucket.com/albums/z198/fairjane/Blind.gif

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. wink

Last edited by JaneFairfax (2009-05-13 04:07:51)


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A: Click here for answer.
 

#21 2009-04-23 05:24:59

JaneFairfax
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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
.


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A: Click here for answer.
 

#22 2009-04-23 17:50:58

JaneFairfax
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Re: Simple groups

smile

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 (2009-05-13 04:03:28)


Q: Who wrote the novels Mrs Dalloway and To the Lighthouse?

A: Click here for answer.
 

#23 2009-04-24 14:51:19

Ricky
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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..."
 

#24 2009-04-24 23:03:47

JaneFairfax
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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. wink

Last edited by JaneFairfax (2009-05-12 22:25:18)


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A: Click here for answer.
 

#25 2009-05-12 19:22:42

JaneFairfax
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Re: Simple groups

smile

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.


Q: Who wrote the novels Mrs Dalloway and To the Lighthouse?

A: Click here for answer.
 

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