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freedom641@yahoo.com
science forum beginner
Joined: 18 May 2005
Posts: 13
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 Posted: Sat Jul 15, 2006 2:12 pm Post subject:
A problem in elementary group theory
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Dear All,
I need some help with the origin (and also the solution) of the
following problem:
Let G be a finite group of order n and let S be a non-empty subset
of G.
Prove that S^n is a subgroup of G. (where by S^n we mean the set
whose
elements are any product of n elements of S.)
Thanks. |
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nacir
science forum beginner
Joined: 17 Jul 2006
Posts: 1
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| Posted: Mon Jul 17, 2006 7:45 pm Post subject:
Re: A problem in elementary group theory
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freedom641@yahoo.com wrote:
| Quote: | Dear All,
I need some help with the origin (and also the solution) of the
following problem:
Let G be a finite group of order n and let S be a non-empty subset
of G.
Prove that S^n is a subgroup of G. (where by S^n we mean the set
whose
elements are any product of n elements of S.)
Thanks.
nacirhmidouch: |
hi i found your probleme, but i do not unedrstand ,what did you mean S^n |
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Peter L. Montgomery
science forum Guru Wannabe
Joined: 01 May 2005
Posts: 181
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| Posted: Tue Jul 18, 2006 1:15 am Post subject:
Re: A problem in elementary group theory
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In article <e9gpbt$b2n$1@dizzy.math.ohio-state.edu> "nacir" <nacirhmidouch@yahoo.com> writes:
| Quote: |
freedom641@yahoo.com wrote:
Dear All,
I need some help with the origin (and also the solution) of the
following problem:
Let G be a finite group of order n and let S be a non-empty subset
of G.
Prove that S^n is a subgroup of G. (where by S^n we mean the set
whose
elements are any product of n elements of S.)
Thanks.
nacirhmidouch:
hi i found your probleme, but i do not unedrstand ,what did you mean S^n
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S^n means all products s_1 s_2 ... s_n where each s_i in S.
S^n includes the identity. If s in S (recall S is nonempty),
then s^n = 1 by Lagrange's theorem. Let s_1 = s_2 = ... = s_n = s.
We still need to show S^n is closed under multiplication.
That is, a product of 2n elements of S is a product
of n elements of S.
If we achieve this, then S^n is closed under inversion since
(s_1 s_2 ... s_n)^{-1} = s_n^(n-1) s_{n-1}^(n-1) ... s_1^(n-1)
is a product of n*(n-1) elements of S, which can be reduced
to a product of n elements of S.
--
Why do storms (e.g., Katrina) have human names,
while earthquakes (e.g., Loma Prieta) have place names?
pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA |
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William Elliot
science forum Guru
Joined: 24 Mar 2005
Posts: 1906
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| Posted: Tue Jul 18, 2006 11:42 am Post subject:
Re: A problem in elementary group theory
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On Mon, 17 Jul 2006, Peter L. Montgomery wrote:
| Quote: | nacirhmidouch@yahoo.com> writes:
freedom641@yahoo.com wrote:
Let G be a finite group of order n and let S be a non-empty subset
of G.
Prove that S^n is a subgroup of G.
S^n means all products s_1 s_2 ... s_n where each s_i in S.
S^n includes the identity. If s in S (recall S is nonempty),
then s^n = 1 by Lagrange's theorem. Let s_1 = s_2 = ... = s_n = s.
We still need to show S^n is closed under multiplication.
That is, a product of 2n elements of S is a product
of n elements of S.
If S has one element, then this is obvious as S^n = {e} |
If S has two elements a,b that commute, then r in S^n is of the form
r = a^j b^k; j + k = n, j,k in Z+
The product of two elements r,s then has the form
rs = a^(j+j') b^(k+k'), j + k = n = j' + k', j,k,j',k' in Z+
j + j' + k + k' = 2n
n <= j + j' or n <= k + k'
rs = a^(j+j' - n) b^(k+k') in S^n
or
rs = a^(j+j') b^(k+k' - n) in S^n
| Quote: | If we achieve this, then S^n is closed under inversion since
(s_1 s_2 ... s_n)^{-1} = s_n^(n-1) s_{n-1}^(n-1) ... s_1^(n-1)
is a product of n*(n-1) elements of S, which can be reduced
to a product of n elements of S.
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Yves De Cornulier
science forum beginner
Joined: 07 Sep 2005
Posts: 8
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| Posted: Tue Jul 18, 2006 5:45 pm Post subject:
Re: A problem in elementary group theory
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freedom641@yahoo.com wrote:
| Quote: | Let G be a finite group of order n and let S be a non-empty subset
of G.
Prove that S^n is a subgroup of G. (where by S^n we mean the set
whose elements are any product of n elements of S.)
|
The sequence (S^k) strictly increases until it stabilizes:
S < S^2 < ... S^{k-1} < S^k =S^{k+1} =...
Thus S^k is a sub-semigroup of the group G; as elements have
finite order, S^k is a subgroup.
On the other hand, clearly #(S^i)>= i for all i<= k.
Thus #(S^k)>=k. In particular, n>=k, so that S^n=S^k is a subgroup.
--
Yves |
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Robert B. Israel
science forum Guru
Joined: 24 Mar 2005
Posts: 2151
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| Posted: Wed Jul 19, 2006 2:15 am Post subject:
Re: A problem in elementary group theory
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In article <e9j6mt$207$1@dizzy.math.ohio-state.edu>,
Yves de Cornulier <decornul@clipper.ens.fr> wrote:
| Quote: |
freedom641@yahoo.com wrote:
Let G be a finite group of order n and let S be a non-empty subset
of G.
Prove that S^n is a subgroup of G. (where by S^n we mean the set
whose elements are any product of n elements of S.)
The sequence (S^k) strictly increases until it stabilizes:
S < S^2 < ... S^{k-1} < S^k =S^{k+1} =...
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If by < you mean "is a subset", this is not necessarily true.
Thus there's no reason for S^2 to contain S. On the
other hand, if you're talking about cardinality, it might happen
that card(S) = card(S^2) but S is not a subgroup. Example for both:
let S be the set of all odd permutations on n letters.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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Yves De Cornulier
science forum beginner
Joined: 07 Sep 2005
Posts: 8
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| Posted: Wed Jul 19, 2006 1:45 pm Post subject:
Re: A problem in elementary group theory
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Thanks to Robert Israel to point out that my proof was incomplete
(except if S contains 1).
Here are the required modifications.
| Quote: | The sequence (S^k) strictly increases until it stabilizes:
S < S^2 < ... < S^{k-1} < S^k =S^{k+1} =...
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This is true for cardinalities.
| Quote: | Thus S^k is a sub-semigroup of the group G; as elements have
finite order, S^k is a subgroup.
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Not necessarily, as Robert also noticed. However, as S^n
does contain 1 and #(S^n.S^n)=#S^n, it holds that S^n is a
subgroup.
[While S^k is a coset of S^n, contained in the subgroup H generated
by S, which may be bigger than S. Clearly H normalizes S^n.]
| Quote: | On the other hand, clearly #(S^i)>= i for all i<= k.
Thus #(S^k)>=k. In particular, n>=k, so that S^n=S^k is a subgroup.
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--
Yves |
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