Talk:Permutation group
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[edit] Examples of permutation groups
I added two examples from my own work. Others may argue that this is too egocentric. Cullinane 21:54, 10 August 2005 (UTC)
[edit] Permutations of infinite sets?
I would be interested to know about permutations (and permutation groups) of infinite sets (eg: Z or Q). The article says that "if M is any finite or infinite set, then the group of all permutations of M is often written as Sym(M)". I'm interested to know how a permutation would be defined for an uncountable set like R.
If you know anything about this topic please include some information here (or start another article about it).
- Sorry, forgot to sign that last commentDonkeyKong the mathematician (in training) 06:48, 28 May 2006 (UTC)
-
- A permutation of M may be defined as a bijection of M into itself (an automorphism in the category of sets), thus S(M) is the set of all such permutations for a given set M. Of course one cannot "explicitely" write the correspondence table
[ x1 x2 ... ]
f = [ ]
[ f(x1) f(x2) ... ]
-
- of such a permutation, if it has an infinite support; however, this is possible if it has a finite support (supp f = { x | f(x) <> x }), in which case one would only write the elements on which it acts nontrivially. In that case, one can also decompose it into cycles and define its order etc. in the usual manner. It's easy to see that the subset So(M) = { f ∈ S(M) | suppf is finite } is a (normal) subgroup of S(M). — MFH:Talk 03:02, 11 November 2006 (UTC)
[edit] Inversions and transpositions
{NOTE that the permutation 4,3,1,2 has five inversions but only three transpositions. There is an error in the text} —Preceding unsigned comment added by 01001 (talk • contribs)
[edit] Isomorphisms and the symmetric group
I disagree with the following paragraph:
"If (G,M) and (H,M) such that both G and H are isomorphic as groups to Sym(M), then (G,M) and (H,M) are isomorphic as permutation groups; thus it is appropriate to talk about the symmetric group Sym(M) (up to isomorphism)."
If M is finite and a group G of permutations of M is isomorphic to Sym(M) as a group, then it has the same cardinality and hence it contains all the permutations of M, hence it is also equal to Sym(M). Using this argument, we can conclude that the symmetric group Sym(M) is simply unique. However, the uniqueness of Sym(M) was already clear in the introduction when it was defined as (the) group of all the permutations of M with function composition as an operation.
If M is infinite, I think the statement is false in general, but anyway I think the paragraph wasn't meant to address this case.
In conclusion, I would delete the paragraph. Marcosaedro (talk) 07:00, 7 March 2009 (UTC)
- I agree that the statement is either trivial or false, so not very informative in either case. There is even a finite counterexample to the less trivial claim for permutation representations (Sym(6) has two non-isomorphic permutation representations on 6 element sets), so I think that even if the statement were corrected, it would still be confusing and still would not say anything enlightening. JackSchmidt (talk) 15:22, 7 March 2009 (UTC)
- I also don't understand your example. It's the first time I hear about permutation representations.
- Update: I found a definition in the article on group representation. It says that a group representation of a group G on a set M is an homomorphism from G to Sym(M). This definition seems counterintuitive to me, because it doesn't require the homomorphism to be inyective, and it therefore doesn't encode all the knowledge of G.
- Update: What I would have called representation is actually called faithful representation.
- Do you mean that there are two morphisms from G=Sym(6) to Sym(X) (being X a six-element set) whose images are isomorphic as groups but are not conjugates of each other? I would expect those morphisms to be non-injective, because with a cardinality argument we can see that injectivity implies surjectivity. Anyway, which is the claim that this example intends to disprove?
- MetaWiki: I'm quite new to Wikipedia. I understand that with this comment I'm drifting away from the purpose of improving the article, because the paragraph is unlikely to reappear. Is this talk page suitable for this kind of discussion? (And do you have the time?) Marcosaedro (talk) 22:44, 7 March 2009 (UTC)
- If X is a six element set, then there are two injective homomorphisms f, g from Sym(6) to Sym(X) such that there is at least one π in Sym(6), where f(π) is not conjugate in Sym(X) to g(π). Of course the full image of f and g are conjugate (even identical), it is just the conjugating element doesn't match up the same elements of Sym(6). This happens precisely because Sym(6) has an outer automorphism (and no other finite symmetric group does). In the language of permutation representations, the images are permutation isomorphic, but the representations are not.
- This would be a counterexample to "For finite sets X, every permutation representation of Sym(X) is isomorphic, so it makes sense to talk about the permutation representation of Sym(X)."
- Metawiki: if the discussion drifts too far or too long, the reference desk is a better place to discuss with the community. However, a short discussion might reveal a way to add something nice to the article (or one of its relatives), and might help future editors understand our current point of view. JackSchmidt (talk) 02:50, 8 March 2009 (UTC)
