Hamilton–Jacobi equation: Difference between revisions
en>SporkBot m TFD outcome; please use Category:Equations; Category:Theoretical physics |
|||
Line 1: | Line 1: | ||
In mathematics, '''Schreier's lemma''' is a [[theorem]] in [[group theory]] used in the [[Schreier–Sims algorithm]] and also for finding a [[presentation of a group|presentation]] of a [[subgroup]]. | |||
==Definition== | |||
Suppose <math>H</math> is a [[subgroup]] of <math>G</math>, which is finitely generated with generating set <math>S</math>, that is, ''G'' = <math>\scriptstyle \langle S\rangle</math>. | |||
Let <math>R</math> be a right [[Transversal (combinatorics)|transversal]] of <math>H</math> in <math>G</math>. In other words, <math>R</math> is (the image of) a [[Section (fiber bundle)|section]] of the quotient map <math>G \to H\backslash G</math>, where <math>H\backslash G</math> denotes the set of '''right cosets''' of <math>H</math> in <math>G</math>. | |||
We make the definition that given <math>g</math>∈<math>G</math>, <math>\overline{g}</math> is the chosen representative in the transversal <math>R</math> of the coset <math>Hg</math>, that is, | |||
:<math>g\in H\overline{g}.</math> | |||
Then <math>H</math> is generated by the set | |||
:<math>\{rs(\overline{rs})^{-1}|r\in R, s\in S\}</math> | |||
== Example == | |||
Let us establish the evident fact that the group '''Z'''<sub>3</sub> = '''Z'''/3'''Z''' is indeed cyclic. Via [[Cayley's theorem]], '''Z'''<sub>3</sub> is a subgroup of the [[symmetric group]] ''S''<sub>3</sub>. Now, | |||
: <math>\Bbb{Z}_3=\{ e, (1\ 2\ 3), (1\ 3\ 2) \}</math> | |||
: <math>S_3= \{ e, (1\ 2), (1\ 3), (2\ 3), (1\ 2\ 3), (1\ 3\ 2) \}</math> | |||
where <math>e</math> is the identity permutation. Note ''S''<sub>3</sub> = <math>\scriptstyle\langle</math>{ ''s''<sub>1</sub>=(1 2), ''s''<sub>2</sub> = (1 2 3) }<math>\scriptstyle\rangle</math>. | |||
'''Z'''<sub>3</sub> has just two cosets, '''Z'''<sub>3</sub> and ''S''<sub>3</sub> \ '''Z'''<sub>3</sub>, so we select the transversal { ''t''<sub>1</sub> = ''e'', ''t''<sub>2</sub>=(1 2) }, and we have | |||
: <math>\begin{matrix} | |||
t_1s_1 = (1\ 2),&\quad\text{so}\quad&\overline{t_1s_1} = (1\ 2)\\ | |||
t_1s_2 = (1\ 2\ 3) ,&\quad\text{so}\quad& \overline{t_1s_2} = e\\ | |||
t_2s_1 = e ,&\quad\text{so}\quad& \overline{t_2s_1} = e\\ | |||
t_2s_2 = (2\ 3) ,&\quad\text{so}\quad& \overline{t_2s_2} = (1\ 2). \\ | |||
\end{matrix}</math> | |||
Finally, | |||
: <math>t_1s_1\overline{t_1s_1}^{-1} = e</math> | |||
: <math>t_1s_2\overline{t_1s_2}^{-1} = (1\ 2\ 3)</math> | |||
: <math>t_2s_1\overline{t_2s_1}^{-1} = e </math> | |||
: <math>t_2s_2\overline{t_2s_2}^{-1} = (1\ 2\ 3).</math> | |||
Thus, by Schreier's subgroup lemma, { e, (1 2 3) } generates '''Z'''<sub>3</sub>, but having the identity in the generating set is redundant, so we can remove it to obtain another generating set for '''Z'''<sub>3</sub>, { (1 2 3) } (as expected). | |||
==References== | |||
* Seress, A. Permutation Group Algorithms. Cambridge University Press, 2002. | |||
[[Category:Theorems in group theory]] | |||
[[Category:Lemmas]] |
Revision as of 13:59, 27 January 2014
In mathematics, Schreier's lemma is a theorem in group theory used in the Schreier–Sims algorithm and also for finding a presentation of a subgroup.
Definition
Suppose is a subgroup of , which is finitely generated with generating set , that is, G = .
Let be a right transversal of in . In other words, is (the image of) a section of the quotient map , where denotes the set of right cosets of in .
We make the definition that given ∈, is the chosen representative in the transversal of the coset , that is,
Example
Let us establish the evident fact that the group Z3 = Z/3Z is indeed cyclic. Via Cayley's theorem, Z3 is a subgroup of the symmetric group S3. Now,
where is the identity permutation. Note S3 = { s1=(1 2), s2 = (1 2 3) }.
Z3 has just two cosets, Z3 and S3 \ Z3, so we select the transversal { t1 = e, t2=(1 2) }, and we have
Finally,
Thus, by Schreier's subgroup lemma, { e, (1 2 3) } generates Z3, but having the identity in the generating set is redundant, so we can remove it to obtain another generating set for Z3, { (1 2 3) } (as expected).
References
- Seress, A. Permutation Group Algorithms. Cambridge University Press, 2002.