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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. ThiƩry 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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