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Orchid XP v8 wrote:
> I was under the impression the result applies to all groups - which is
> why the groups of prime order are always cyclic.
It is true that every subgroup of a finite group has an order which
divides that of the group (this result has actually been around longer
than group theory itself). This implies that all prime order groups are
cyclic.
The converse, however, does not hold. Some finite groups do not have
any subgroups corresponding to some of their order's divisors.
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Kevin Wampler wrote:
> Orchid XP v8 wrote:
>> I was under the impression the result applies to all groups - which is
>> why the groups of prime order are always cyclic.
>
> It is true that every subgroup of a finite group has an order which
> divides that of the group (this result has actually been around longer
> than group theory itself). This implies that all prime order groups are
> cyclic.
>
> The converse, however, does not hold. Some finite groups do not have
> any subgroups corresponding to some of their order's divisors.
Ah, right. I guess I missed that technicallity.
Of course, intuitively, it totally makes sense that the order of a
subgroup would have to be a factor of the group's order. I just thought
that every group could be split into subgroups. So I guess there are
large non-prime groups that don't have any [nontrivial] subgroups then?
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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Orchid XP v8 wrote:
> Of course, intuitively, it totally makes sense that the order of a
> subgroup would have to be a factor of the group's order. I just thought
> that every group could be split into subgroups. So I guess there are
> large non-prime groups that don't have any [nontrivial] subgroups then?
This is not actually true either. If a group G is of order n, then G
does have a subgroup of every order which is a *prime* factor of n.
It's just that there's no guarantee for composite factors.
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Kevin Wampler wrote:
> Orchid XP v8 wrote:
>> Of course, intuitively, it totally makes sense that the order of a
>> subgroup would have to be a factor of the group's order. I just
>> thought that every group could be split into subgroups. So I guess
>> there are large non-prime groups that don't have any [nontrivial]
>> subgroups then?
>
> This is not actually true either. If a group G is of order n, then G
> does have a subgroup of every order which is a *prime* factor of n. It's
> just that there's no guarantee for composite factors.
So, in summary, G *definitely* has subgroups for every prime factor, and
*might* have subgroups for some or all of the composite factors as well?
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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From: Kevin Wampler
Subject: Re: Infinite sequences and probability
Date: 28 Apr 2009 16:25:59
Message: <49f76657@news.povray.org>
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Orchid XP v8 wrote:
> So, in summary, G *definitely* has subgroups for every prime factor, and
> *might* have subgroups for some or all of the composite factors as well?
I believe that is correct. In addition I think that G must have a
subgroup for every prime power which still divides the order of G.
Weather or not the composite factors have associated subgroups will
indeed depend on the particular group G represents.
As an example consider the group of rotations of a tetrahedron. It has
12 elements but no subgroup of order 6.
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Kevin Wampler wrote:
> I did like the Cantor set analogy by the way. I didn't forsee it when I
> was reading your post and was delighted when you pointed it out.
It popped into my head while reading Darren and Tim argue in the other
thread. Delighted me too to see the relationship.
>>> lim_{n->inf} a^n = 0 if 0 <= a < 1
>>
>> (Incidentally, I'm not sure I understand your limit. What is 'a' and
>> why is it fixed?
>
> For base m a would be (m-1)/m -- the probability that a single given
> digit of the random sequence doesn't contain a 1 in base m.
Ah, I get it.
> I'm enveious you had an integration theory class that got into measure
> theory. I don't recall such a course being available back when I was
> taking math classes, but it would have been pretty fun had there been.
Well, I think it's standard in the first semester of analysis at the
grad level. Some good schools probably do it at the undergrad. I learned
this stuff only recently - learning as I go along.
--
Lisa: Oedipus killed his father and married his mother.
Homer: Who payed for THAT wedding?
/\ /\ /\ /
/ \/ \ u e e n / \/ a w a z
>>>>>>mue### [at] nawaz org<<<<<<
anl
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>> So, in summary, G *definitely* has subgroups for every prime factor,
>> and *might* have subgroups for some or all of the composite factors as
>> well?
>
> I believe that is correct. In addition I think that G must have a
> subgroup for every prime power which still divides the order of G.
> Weather or not the composite factors have associated subgroups will
> indeed depend on the particular group G represents.
>
> As an example consider the group of rotations of a tetrahedron. It has
> 12 elements but no subgroup of order 6.
http://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)
It seems that for a *commutative* group, my original statements are
correct. I found many other interesting properties, but failed to find
any confirmation of your assertion about prime subgroup orders...
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Invisible wrote:
> http://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)
>
> It seems that for a *commutative* group, my original statements are
> correct. I found many other interesting properties, but failed to find
> any confirmation of your assertion about prime subgroup orders...
Wait a minute... this isn't finite sequences and probability! o_O
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>>> So, in summary, G *definitely* has subgroups for every prime factor,
>>> and *might* have subgroups for some or all of the composite factors
>>> as well?
>>
>> I believe that is correct. In addition I think that G must have a
>> subgroup for every prime power which still divides the order of G.
>> Weather or not the composite factors have associated subgroups will
>> indeed depend on the particular group G represents.
>>
>> As an example consider the group of rotations of a tetrahedron. It
>> has 12 elements but no subgroup of order 6.
>
> http://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)
>
> It seems that for a *commutative* group, my original statements are
> correct. I found many other interesting properties, but failed to find
> any confirmation of your assertion about prime subgroup orders...
Wait, here we go:
http://en.wikipedia.org/wiki/Sylow%27s_theorem
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