# Group actions

• Oct 8th 2011, 08:44 AM
dwsmith
Group actions
Let A be a nonempty set and $\displaystyle k\in\mathbb{N}$ with $\displaystyle k\leq |A|$. The symmetric group $\displaystyle S_A$ acts on B consisting of all subsets of cardinality k by $\displaystyle \sigma \{a_1,\cdots ,a_k\}=\{\sigma(a_1),\cdots , \sigma(a_k)\}$.

Prove that this is a group action.

$\displaystyle \sigma_{g_1g_2}(a)=(g_1g_2)(a)=g_1(g_2(a))=\sigma_ {g_1}(\sigma_{g_2}(a))$

Is this what it is asking for?
• Oct 8th 2011, 09:46 AM
Deveno
Re: Group actions
um,... what do you mean by "a"?
• Oct 8th 2011, 10:02 AM
dwsmith
Re: Group actions
Quote:

Originally Posted by Deveno
um,... what do you mean by "a"?

a I would assume is an element.
• Oct 8th 2011, 10:08 AM
emakarov
Re: Group actions
Quote:

Originally Posted by dwsmith
a I would assume is an element.

An element of what?
• Oct 8th 2011, 10:19 AM
dwsmith
Re: Group actions
Quote:

Originally Posted by emakarov
An element of what?

Probably of A.
• Oct 8th 2011, 10:26 AM
Deveno
Re: 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, 10:28 AM
dwsmith
Re: Group actions
Quote:

Originally Posted by Deveno
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).

This is word for word from Abstract Algebra by Dummit and Foote. I give you verbatim the text. I am not told where a is from either. But my guess is a would be in A.
• Oct 8th 2011, 10:46 AM
Deveno
Re: 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, 11:04 AM
emakarov
Re: Group actions
Quote:

Originally Posted by Deveno
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)!

By assumption, B consists of all subsets of A of cardinality k, so $\displaystyle |B| = {n\choose k}=n!/(k!(n-k)!=m$, and an element of A produces (through action) a permutation of B, i.e., an element of $\displaystyle S_m$.

Quote:

Originally Posted by Deveno
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").

I think we are moving too fast here. The question so far is whether $\displaystyle S_A$ (or $\displaystyle \mbox{Sym}_A$) acts on A or B, i.e., whether $\displaystyle a\in A$ or $\displaystyle a\in B$ in the equation $\displaystyle (\sigma_1\sigma_2)(a)=\sigma_1(\sigma_2(a))$.

The problem is clear that $\displaystyle S_A$ acts on B. Therefore, it asks to prove that

$\displaystyle (\sigma_1\sigma_2)(b)=\sigma_1(\sigma_2(b))$ for all $\displaystyle b\in B$

and

$\displaystyle e(b) = b$ for all $\displaystyle b\in B$.

Here $\displaystyle \sigma_1,\sigma_2\in S_A$, $\displaystyle \sigma_1\sigma_2$ is the result of the group operation on $\displaystyle \sigma_1$ and $\displaystyle \sigma_2$ and $\displaystyle \sigma_2(b)$ is the result of the group action of $\displaystyle \sigma_2$ on $\displaystyle b$. And, of course, $\displaystyle e$ is the unit of the group $\displaystyle S_A$.

Quote:

Originally Posted by Deveno