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Sagar Kolte science forum beginner
Joined: 14 Apr 2005
Posts: 31

Posted: Sun Jul 16, 2006 2:48 pm Post subject:
Re: Finite # of subgroups > Finite group



Quote:  Hello,
I'm working my way through some group theory on my
own, and was looking
for advice on the following proof. I think my
approach works, but it
seems like there must be an easier way. This problem
is in the chapter
of the text on cyclic subgroups, so that's what I'm
trying to use.
Given: Group G with a finite number of subgroups
Prove: Group G is finite
I will prove the contrapositive  if G is infinite,
then we can
generate an infinite number of distinct subgroups
from it.
Choose a in G where a != e. (G is infinite, so this
a must exist.)
From a we can construct the cyclic subgroup <a>.
Case 1: <a> is finite
(Is this case really valid? Can I get a finite cyclic
subgroup other
than {e} out of an infinite group?)
Choose b in G such that b is not in <a>. This b must
exist because G is
of infinite order, while <a> is finite. Then we have
a new subgroup <b
distinct from <a>. This process can be continued so
long as we keep
generating finite cyclic subgroups.
Case 2: <a> is infinite
Then we can construct a new subgroup <a^2>. a is not
in <a^2>, because
if it were then there would be an n in Z such that
(a^2)^n = a
a^2n = a
a^(2n1) = e
But that would mean that <a> is finite, which is a
contradiction. Thus
a is not in <a^2>, so <a^2> is distinct from <a>. By
the same
reasoning, <a^3> is distinct from either <a^2> or
a>, and so on...
Therefore, from an infinite group G, we can generate
an infinite number
of subgroups.
And the contrapositive also holds, that if a group G
has a finite
number of subgroups, then G itself must be finite.
Valid, I think, but kind of messy. There must be an
easier way...

Case 1: All elements of the groups have finite order
Suppose the number of subgroups is finite then the number of cyclic subgroups is also finite also each cyclic subgroup is finite. each element belongs to atleast one cyclic subgroup therefore union of all cyclic subgroups is the whole group. But the union of finite number of sets with finite elements is finite, a contradiction to the fact that G is infinite.
Case2: Atleast one element has infinite order
Then the cyclic subgroup generated by this element has infinitely many subgroups. 

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G.E. Ivey science forum Guru
Joined: 29 Apr 2005
Posts: 308

Posted: Sun Jul 16, 2006 11:29 am Post subject:
Re: Finite # of subgroups > Finite group



Quote:  Brian VanPelt wrote:
I think that this is a good place to anchor the
entire argument. Here,
you've shown that any subgroup generated by a
single element must be
finite.
I have? I didn't think I showed that. At the point
you clipped, I
(thought I) was showing that *if* you picked an a
from G such that <a
is infinite (which seems possible) that you could
take that one
infinite subgroup and generate infinitely more
infinite subgroups by
using powers of a as generators. My last line that
you quoted is
following the (ultimately contradictory) supposition
that a is in
a^2>.
I follow your reasoning from there, but at least one
of us (quite
possibly me) is not understanding what I originally
posted.
Todd
Ah! Now I understand William Eliot's point (which I had just asked about!). You have shown that if <a> is infinite, then <a^2>, <a^3>, etc. are all distinct subgroups and so there would exist an infinite number of subgroups. Therefore, GIVEN THAT THERE ARE A FINITE NUMBER OF SUBGROUPS, each <a> must be finite. Of course, if there are only a finite number of subgroups, then there must be only a finite number of subgroups generated by a single element. Since G is equal to the union of all such subgroups, and the union of a finite number of finite sets in finite, G is finite. 


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G.E. Ivey science forum Guru
Joined: 29 Apr 2005
Posts: 308

Posted: Sun Jul 16, 2006 11:24 am Post subject:
Re: Finite # of subgroups > Finite group



Quote:  From: tppytel@gmail.com
Given: Group G with a finite number of subgroups
Prove: Group G is finite
For all g in G, o(g), <g> finite; { <g> : g in G }
finite
G = \/{ <g> : g in G } finite union of finite sets

For all g in G such that order of g is finite? How does it follow that G= union of all such subgroups? 


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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Sun Jul 16, 2006 9:24 am Post subject:
Re: Finite # of subgroups > Finite group



On Sat, 15 Jul 2006 tppytel@gmail.com wrote:
Quote:  William Elliot wrote:
Given: Group G with a finite number of subgroups
Prove: Group G is finite
For all g in G, o(g), <g> finite; { <g> : g in G } finite
G = \/{ <g> : g in G } finite union of finite sets
I don't know what "o(g)" means 

o(g) = ord(g) = smallest positive integer with g^n = e.
Quote:  I don't understand how you know that <g> is finite.
Because o(g) is finite. 
Please do not over clip the context or you will lose the line of thought. 

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Jim Heckman science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 121

Posted: Sun Jul 16, 2006 7:13 am Post subject:
Re: Finite # of subgroups > Finite group



On 15Jul2006, tppytel@gmail.com
wrote in message <1153014425.273629.30430@s13g2000cwa.googlegroups.com>:
Quote:  Hello,
I'm working my way through some group theory on my own, and was looking
for advice on the following proof. I think my approach works, but it
seems like there must be an easier way. This problem is in the chapter
of the text on cyclic subgroups, so that's what I'm trying to use.
Given: Group G with a finite number of subgroups
Prove: Group G is finite
I will prove the contrapositive  if G is infinite, then we can
generate an infinite number of distinct subgroups from it.

Yes, that's the correct approach.
[...]
Quote:  Valid, I think, but kind of messy. There must be an easier way...

I didn't go through your proof, but the usual way to prove this is:
If G has an element x of infinite order, then it has an infinite
set of subgroups, namely {<x>,<x^2>,<x^3>,...}.
On the other hand, if G does not have an element of infinite order,
then <x> is finite for every x in G. Now, G is the union of these
<x>'s, so if G is infinite there must be an infinite number of
them, since the union of a finite number of finite sets is clearly
finite.

Jim Heckman 

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tppytel@gmail.com science forum beginner
Joined: 16 Jul 2006
Posts: 3

Posted: Sun Jul 16, 2006 6:47 am Post subject:
Re: Finite # of subgroups > Finite group



Brian VanPelt wrote:
Quote: 
I think that this is a good place to anchor the entire argument. Here,
you've shown that any subgroup generated by a single element must be
finite.

I have? I didn't think I showed that. At the point you clipped, I
(thought I) was showing that *if* you picked an a from G such that <a>
is infinite (which seems possible) that you could take that one
infinite subgroup and generate infinitely more infinite subgroups by
using powers of a as generators. My last line that you quoted is
following the (ultimately contradictory) supposition that a is in
<a^2>.
I follow your reasoning from there, but at least one of us (quite
possibly me) is not understanding what I originally posted.
Todd 

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tppytel@gmail.com science forum beginner
Joined: 16 Jul 2006
Posts: 3

Posted: Sun Jul 16, 2006 6:33 am Post subject:
Re: Finite # of subgroups > Finite group



William Elliot wrote:
Quote: 
Given: Group G with a finite number of subgroups
Prove: Group G is finite
For all g in G, o(g), <g> finite; { <g> : g in G } finite
G = \/{ <g> : g in G } finite union of finite sets

Ah... a little too terse for me to follow  I'm not a math major, I
don't look at this all day. First off, I don't know what "o(g)" means 
I'm sure I'm just missing some expected notation here. Second (maybe
related to the above), I don't understand how you know that <g> is
finite.
Could you spell it out just a tiny bit more for me?
Thanks,
Todd 

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Brian VanPelt science forum beginner
Joined: 08 May 2005
Posts: 42

Posted: Sun Jul 16, 2006 6:24 am Post subject:
Re: Finite # of subgroups > Finite group



On 15 Jul 2006 18:47:05 0700, tppytel@gmail.com wrote:
Quote:  Hello,
I'm working my way through some group theory on my own, and was looking
for advice on the following proof. I think my approach works, but it
seems like there must be an easier way. This problem is in the chapter
of the text on cyclic subgroups, so that's what I'm trying to use.
Given: Group G with a finite number of subgroups
Prove: Group G is finite
I will prove the contrapositive  if G is infinite, then we can
generate an infinite number of distinct subgroups from it.
Choose a in G where a != e. (G is infinite, so this a must exist.)
From a we can construct the cyclic subgroup <a>.
Case 1: <a> is finite
(Is this case really valid? Can I get a finite cyclic subgroup other
than {e} out of an infinite group?)
Choose b in G such that b is not in <a>. This b must exist because G is
of infinite order, while <a> is finite. Then we have a new subgroup <b
distinct from <a>. This process can be continued so long as we keep
generating finite cyclic subgroups.
Case 2: <a> is infinite
Then we can construct a new subgroup <a^2>. a is not in <a^2>, because
if it were then there would be an n in Z such that
(a^2)^n = a
a^2n = a
a^(2n1) = e

I think that this is a good place to anchor the entire argument. Here,
you've shown that any subgroup generated by a single element must be
finite. Now, look at the group G itself.
Isn't it true that G is the union of every < g >, where g ranges
through every element of G? Since G has only a finite number of
subgroups, there are only a finite number of < g > ' s, and each one
of those is finite.
I think this will do it for you.
Brian
Quote: 
But that would mean that <a> is finite, which is a contradiction. Thus
a is not in <a^2>, so <a^2> is distinct from <a>. By the same
reasoning, <a^3> is distinct from either <a^2> or <a>, and so on...
Therefore, from an infinite group G, we can generate an infinite number
of subgroups.
And the contrapositive also holds, that if a group G has a finite
number of subgroups, then G itself must be finite.
Valid, I think, but kind of messy. There must be an easier way... 


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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Sun Jul 16, 2006 5:38 am Post subject:
Re: Finite # of subgroups > Finite group



From: tppytel@gmail.com
Quote:  Given: Group G with a finite number of subgroups
Prove: Group G is finite

For all g in G, o(g), <g> finite; { <g> : g in G } finite
G = \/{ <g> : g in G } finite union of finite sets
 

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tppytel@gmail.com science forum beginner
Joined: 16 Jul 2006
Posts: 3

Posted: Sun Jul 16, 2006 1:47 am Post subject:
Finite # of subgroups > Finite group



Hello,
I'm working my way through some group theory on my own, and was looking
for advice on the following proof. I think my approach works, but it
seems like there must be an easier way. This problem is in the chapter
of the text on cyclic subgroups, so that's what I'm trying to use.
Given: Group G with a finite number of subgroups
Prove: Group G is finite
I will prove the contrapositive  if G is infinite, then we can
generate an infinite number of distinct subgroups from it.
Choose a in G where a != e. (G is infinite, so this a must exist.)
Quote:  From a we can construct the cyclic subgroup <a>.

Case 1: <a> is finite
(Is this case really valid? Can I get a finite cyclic subgroup other
than {e} out of an infinite group?)
Choose b in G such that b is not in <a>. This b must exist because G is
of infinite order, while <a> is finite. Then we have a new subgroup <b>
distinct from <a>. This process can be continued so long as we keep
generating finite cyclic subgroups.
Case 2: <a> is infinite
Then we can construct a new subgroup <a^2>. a is not in <a^2>, because
if it were then there would be an n in Z such that
(a^2)^n = a
a^2n = a
a^(2n1) = e
But that would mean that <a> is finite, which is a contradiction. Thus
a is not in <a^2>, so <a^2> is distinct from <a>. By the same
reasoning, <a^3> is distinct from either <a^2> or <a>, and so on...
Therefore, from an infinite group G, we can generate an infinite number
of subgroups.
And the contrapositive also holds, that if a group G has a finite
number of subgroups, then G itself must be finite.
Valid, I think, but kind of messy. There must be an easier way... 

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