Let A be a nonempty set and with . The symmetric group acts on B consisting of all subsets of cardinality k by .

Prove that this is a group action.

Is this what it is asking for?

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- Oct 8th 2011, 09:44 AMdwsmithGroup actions
Let A be a nonempty set and with . The symmetric group acts on B consisting of all subsets of cardinality k by .

Prove that this is a group action.

Is this what it is asking for? - Oct 8th 2011, 10:46 AMDevenoRe: Group actions
um,... what do you mean by "a"?

- Oct 8th 2011, 11:02 AMdwsmithRe: Group actions
- Oct 8th 2011, 11:08 AMemakarovRe: Group actions
- Oct 8th 2011, 11:19 AMdwsmithRe: Group actions
- Oct 8th 2011, 11:26 AMDevenoRe: Group actions
when discussing math, it never hurts to define your terms, so as to be perfectly clear what you mean. at worst, you will take up more space than necessary, saying things that are already understood. but you avoid vagueness, and lessen the risk of misunderstanding.

i'm not in your course, i'm not reading your notes, or privy to the notational shorthand used therein. and contrary to public belief, there is no one "standard way" of expressing mathematical concepts (although certain conventions are wide-spread).

let's look at an example of the construction, here.

suppose A = {a1,a2,a3,a4}, then Sym(A) is a group isomorphic to S4. let's say that k = 2, so:

B = {{a1,a2},{a1,a3},{a1,a4},{a2,a3},{a2,a4},{a3,a4}} = {b1,b2,b3,b4,b5,b6}

so let's say σ is the element (a1 a2) in Sym(A). what is the associated mapping b-->σb in B?

well, {σ(a1),σ(a2)} = {a2,a1} = {a1,a2}. so we have b1-->b1 that is:

σb1 = b1. similarly {σ(a1), σ(a3)} = {a2,a3} = b4, so σb2 = b4. so σ inducees the permutation on B:

b1-->b1

b2-->b4

b3-->b5

b4-->b2

b5-->b3

b6-->b6, which is the permutation (b2 b4)(b3 b5).

do you see what is happening here? - Oct 8th 2011, 11:28 AMdwsmithRe: Group actions
- Oct 8th 2011, 11:46 AMDevenoRe: Group actions
suppose |A| = n. then we start with a group isomorphic to Sn. if |B| = k, we end up with an element in Sm, where m = n!/(k!(n-k)!

you can do one of 2 things: show the subgroup of Sm we end up with is a quotient group (homomorphic image) of Sn (this is the "abstract approach")

or show that the "G-set multiplication" is associative, with identity id_A (the "concrete approach"). what's your poison? - Oct 8th 2011, 12:04 PMemakarovRe: Group actions
By assumption, B consists of all subsets of A of cardinality k, so , and an element of A produces (through action) a permutation of B, i.e., an element of .

I think we are moving too fast here. The question so far is whether (or ) acts on A or B, i.e., whether or in the equation .

The problem is clear that acts on B. Therefore, it asks to prove that

for all

and

for all .

Here , is the result of the group operation on and and is the result of the group action of on . And, of course, is the unit of the group .

Oh, no... (Rofl) - Oct 8th 2011, 12:29 PMDevenoRe: Group actions
i've never read Dummit and Foote, some people love it, some people hate it. from what i gather, it takes the "abstract approach" rather hard-core, which is probably the source of dwsmith's bewilderment. i'm not trying to "move too fast", the way i would prefer to handle group actions (especially on a "first-look") is by seeing it as a way to introduce group multiplication into a set, which appears to be at odds with the way that his text is dealing with it.

which is what emarkov is doing in his post, good job :)