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freedom641@yahoo.com
science forum beginner

Joined: 18 May 2005
Posts: 13

Posted: Sat Jul 15, 2006 2:12 pm    Post subject: A problem in elementary group theory

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.
nacir
science forum beginner

Joined: 17 Jul 2006
Posts: 1

Posted: Mon Jul 17, 2006 7:45 pm    Post subject: Re: A problem in elementary group theory

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
Peter L. Montgomery
science forum Guru Wannabe

Joined: 01 May 2005
Posts: 181

Posted: Tue Jul 18, 2006 1:15 am    Post subject: Re: A problem in elementary group theory

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

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
William Elliot
science forum Guru

Joined: 24 Mar 2005
Posts: 1906

Posted: Tue Jul 18, 2006 11:42 am    Post subject: Re: A problem in elementary group theory

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.
Yves De Cornulier
science forum beginner

Joined: 07 Sep 2005
Posts: 8

Posted: Tue Jul 18, 2006 5:45 pm    Post subject: Re: A problem in elementary group theory

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
Robert B. Israel
science forum Guru

Joined: 24 Mar 2005
Posts: 2151

Posted: Wed Jul 19, 2006 2:15 am    Post subject: Re: A problem in elementary group theory

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} =...

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
Yves De Cornulier
science forum beginner

Joined: 07 Sep 2005
Posts: 8

Posted: Wed Jul 19, 2006 1:45 pm    Post subject: Re: A problem in elementary group theory

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} =...

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.

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.

--
Yves

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