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## New answers tagged symmetric-groups

2

In the future it will be worthwhile for you to know how to compute the order of an element of $\mathrm{Sym}(n)$, so I will detail it here. Recall that every permutation is a product of disjoint cycles, for example $\sigma=35412$ is $(134)(25)$. Disjoint cycles commute with each other, so exponentiating a permutation amounts to raising each disjoint cycle ...

1

Comment converted to an answer on OP's suggestion. This mathoverflow question: Is the following construction of the 0-Hecke monoid (well) known? might be relevant to your question. Let me also mention two possibly relevant references On the representation theory of finite J-trivial monoids by Tom Denton, Florent Hivert, Anne Schilling, Nicolas M. Thiéry A ...

0

In physics, the wave function is a mathematical function $\psi: \mathbb{R}^3 \to \mathbb{C}$. In the discussion of fermions and bosons we can talk about how the wave function behave under the interchange of two particles. There are two fundamental cases: $$\psi(x,y) = \pm \psi(y,x)$$ If there is a "+" we get bosons, in the case of "-" we get a fermion. ...

5

The answer is unfortunately not interesting. A permutation $\pi$ of the real line is called a translation if there exists a real number $a$ such that $\pi(x)=x+a$ for all $x$. The translations form a subgroup of the permutation group of $\mathbb{R}$, and this subgroup is isomorphic to the reals under addition.

1

Not sure this is what you mean by a parametrization, but $SO(3)$ is homeomorphic to a quotient of the ball $\{x\in R^3:|x|\le\pi\}$. The correspondence is, given $x$, there is the rotation of $R^3$ around $x$ by angle $|x|$. The quotient is to identify $x$ and $-x$ when $|x|=\pi$, since these give the same rotation. Then, since $SO(3)$ is a quotient of the ...

0

Let $T_r(n)$ denote the number of commuting $r$-tuples in $S_n$. The following formula for the exponential generating function of $T_r(n)$ is derived in [http://arxiv.org/abs/1304.2830]: $$\sum\limits_{n=0}^{\infty} T_r(n)\frac{u^n}{n!} = \prod\limits_{j=1}^{\infty} (1-u^j)^{-\lambda_{r-1}(j)}$$ where $$\lambda_r(n)= \sum\limits_{d_1d_2\cdots d_r=n}d_2 ... 1 Let G be a finite group. For each g\in G, define \varphi_g to be the conjugation by g (i.e., \varphi_g sends h\in G to ghg^{-1}). Let \tilde{G} be the set of conjugacy classes of G. For g\in G, let \Gamma_g be the conjugacy class of G containing g. We claim that the set$$T(G):=\left\{(g,h)\in G\times G\,\big|\,gh=hg\right\}$$... 3 S_n acts naturally on \mathbb Z^n by permuting the base vectors e_i. This action leaves the hyperplane orthogonal to e_1+\ldots+e_n invariant, which is spanned by then n-1 vectors e_1-e_n,\ldots, e_{n-1}-e_n. 0 Okay so A is a set being acted upon by a group G (as a group of permutations); the kernel of the action of G on A is defined to be the subgroup fixing A, that is the elements \phi such that a^{\phi} = a for all a \in A. Now, suppose that \sigma_{1}, \sigma_{2} \in G with a^{\sigma_{1}} = a^{\sigma_{2}} for all a \in A (this is what ... 4 Consider a permutation from S_n. First make a choice where on the available n positions you place the value one. All future elements to the left of this value will contribute an inversion. That gives the generating function$$q^{n-1}+q^{n-2}+\cdots+1.$$Having positioned one we position two and once again we have all remaining elements to the left of two ... 4 Let P_n(q) be the LHS polynomial and Q_n(q) be the RHS polynomial. Clearly P_1(q) = Q_1(q), and Q_n(q) satisfies the recurrence$$Q_{n+1}(q) = \frac{1-q^{n+1}}{1-q} Q_n(q) = (1+q+q^2+\cdots+q^n)Q_n(q).$$The question that remains is why P_n satisfies the same recurrence. To see this, define s_i \in S_n by the transposition s_i = (i \quad i+1), ... 1 Any permutation \pi\in{\cal S}_4\bigl([4]\bigr) acts on the elements of [4], and there is no simple formula describing the resulting action of \pi on the set P of pairings of [4]. Therefore I shall realize the pairings as edge colorings of the complete graph K_4 on the vertex set [4], as follows: Identifying an element of P, i.e., a pairing ... 1 Any element of S_4 sends any partition of \{1,2,3,4\} to a similar partition of \{1,2,3,4\}. For example, any \sigma \in S_4 sends a 3+1 partition of 4 to a 3+1 partition of 4: if \sigma = (1\ 3\ 4), for example, \sigma sends the 3+1 partition \{\{3\},\{1,2,4\}\} to the partition: \{\{\sigma(3)\},\{\sigma(1),\sigma(2),\sigma(4)\}\}= ... 1 The point is that there are exactly 3 possible pairings of 4 objects. Every permutation of the 4 objects will correspond to mapping one pairing to another, and so is essentially a permutation of the pairings. Composition of permutations will then also correspond to the composition of the corresponding mappings, and so it is a homomorphism from S_4 to ... 0 S_n is generated by q=(1 2) and p=(1 2\cdots n). In other words, every element can be written as a string of these two. Thus you only need check the two-and threefold compositions of these elements satisfy the equation you name: \phi(p^3)=\phi(p)^3 \phi(p^2q)=\phi(p^2)\phi(q) \phi(qp^2)=\phi(q)\phi(p^2) \phi(p^2)=\phi(p)^2 ... 3 A function satisfying the axioms you listed is called a length function. For S_n, the word length would be an example. The symmetric group S_n is generated by elements s_1, \ldots, s_{n-1}, where s_i is the transposition (i,i+1). Define the word length of an element w of S_n to be the smallest \ell for which there is a decomposition$$ ...

2

You can choose a permutation $a$ (others will work, too) such that: $a(1) = 5, a(2) = 6, a(3) = 1$ and $a(4) = 3$. It doesn't matter what we choose for $a(5),a(6)$, as long as we don't pick from the set $\{1,3,5,6\}$, since those values are already "taken". $a(5) = 2$, and $a(6) = 4$ will do. Thus $a = (1\ 5\ 2\ 6\ 4\ 3)$ is one possibility.

1

The key fact to note is that two elements of the symmetric group are conjugate if and only if they have the same number of $k$-cycles in their cycle decomposition for all $k$. In particular, every conjugate of a $3$-cycle is a $3$-cycle, which solves your second exercise. You have shown in a previous question that $$a(i_1i_2\cdots ... 4 The most direct way to solve this is to consider the complement of the given set - that is, the bijections that do have a fixed points.$$Y=\{f:S\rightarrow S\mid f\text{ is bijective and }f(x)=x\text{ for some } x\in S\}.$$Notice that if we define, for each s\in S the set$$Y_s=\{f:S\rightarrow S\mid f\text{ if bijective and }f(s)=s\} then we may ...

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