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To prove that every element in $A_n$ is a commutator, you would have to divide it into several cases and prove that each k-cycle can be written as a commutator. G.A. Miller's paper prove the statement: http://projecteuclid.org/euclid.bams/1183416038 as well as Noboru Ito's, which I can't find online. Hope it is helpful.


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Sometimes a little research can help. The formula you posted was discovered by Jon Perry in 2003. The generating function for this problem is: $$g(x) = \frac{1}{(1-x) \left(1-x^2\right) \left(1-x^3\right) \left(1-x^4\right)} $$ There does not seem to be something simple for your question but Michael Somos comes up with ...


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$H$ acts on $X$ by conjugation. This follows from $hxh^{-1} \in X$: it's easy to check that this implies $h^ix^jh^{-i} \in X$ for all $i, j$, i.e. $h'x'h'^{-1} \in X$ for all $h' \in H$, $x' \in X$. Checking that this action satisfies the other definitions of an axiom is also easy to check (it's the same as checking that any conjugation action is a valid ...


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By its very definition, $L$ acts (in fact, nontrivially) on the set $\{A,A'\}$ and $M$ is the kernel of this action.


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The standard way to see this goes as follows: the $i$th Jucys-Murphy-Young element is the orbit sum $$X_i=\sum_{1 \leq j < i} (ij)=\frac{1}{|S_{i-2}|}\sum_{w \in S_{i-1}} w(1i)w^{-1} $$ and hence commutes with $S_{i-1}$, and in particular with $X_1,\dots,X_{i-1}$.


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Well, this is one occasion, it might not be such a bad idea to do it computationally after all (that is, I am not aware of a better method). It is true that some transpositions don't commute, but we can pinpoint exactly which..: The arbitrary elements in the expansions $X_iX_j$ and $X_jX_i$ are $(ki)(mj)$ and $(mj)(ki)$. These are definitely equal when the ...


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I suggest avoiding the round-bracket notation for generated substructures at all costs, since they are already used for grouping expressions and delineating function inputs, and using it for anything else almost surely will result in unnecessary ambiguity. That said, in a group $G$ and given $g \in G$, "$\langle g \rangle$" denotes "$\{ g^k : k \in ...


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Let me try to shed some light on the different mathematical objects here. The Symmetric algebra $\operatorname{Sym}(V)$ over a vector space $V$ is a quotient of the Tensor algebra $T(V)$ of that space. If for instance, $V=\{v_1,v_2\}$, then $T(V)$ contains elements like $$v_2,\quad v_1\otimes v_2,\quad v_1\otimes v_2\otimes v_1+v_2\otimes v_2,\text{ etc... ...


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This is not a transposition, but in any case it is a permutation and hence is an element of $S_n$. The group of symmetries of a hypercube may be combinatorially realized as the hyperoctahedral group, or the group of signed permutations. These are permutations $f$ of $\{-n,-n+1,\ldots,-1,1,2,\ldots,n\}$ such that $f(-i)=-f(i)$ for all $i$ and are usually ...


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To check that a subset of a group is a subgroup, you need to check that it is closed under composition and inverses. In other words, you must check that if you are given two rotations $g$ and $g'$ fixing your plane $P$, then $gg'$ and $g^{-1}$ also fix $P$. [Note that this subgroup is indeed an isomorphic copy of SO(2) sitting inside of SO(3), but you have ...


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Perhaps it is worth noting the more general fact, that if $H$ is a group of order $n$, and $\psi : H \to S_{n}$ is the regular representation, then $N_{S_{n}}(\psi(H))$ is isomorphic to the holomorph $H \rtimes \operatorname{Aut}(H)$ of $H$.


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Consider the following bits/steps. I abbreviate $\alpha=(123\ldots n)$. If $g\in N_G(H)$, then $g\alpha g^{-1}\in H$, and $g\alpha g^{-1}$ must also have order $n$. Therefore $g\alpha g^{-1}=\alpha^k$ for some integer $k$, $\gcd(k,n)=1$. All the elements $\alpha^k$, $k$ as above, are conjugate in $S_n$. The elements $g$ such that $g\alpha g^{-1}=\alpha^k$, ...


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Fleshing out the comment a bit. Assume that the group $S_n$ has two non-trivial conjugacy classes of the same size, say $|[\sigma]|=|[\tau]|=C$ for some non-conjugate permutations $\sigma,\tau\in S_n$ and some positive integer $C\mid n!$. Define two class function $\psi_1$ and $\psi_2$ by declaring that $\psi_1(1)=Kn!=\psi_2(1)$, ...


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This is probably way too little too late, but someone just linked to this question from elsewhere on the site and seeing as it was never answered I thought I'd give an answer. Consider the natural action of $S_n$ on ($\mathbb{C}$-linear combinations of) $b$-element subsets of $\{1,2,\dots,n\}$. This is not quite the representation you want but its character ...


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Since $H$ is normal and all Sylow-$p$ subgroups are conjugate, once you show that $H$ contains one of them, you know that $H$ contains all of them. But since $H$ is transitive, the orbit of every element of $[p]$ is of size $p$ so $p$ divides the order of $H$, and so $H$ contains an element (and thus a subgroup) of order $p$. Thus, $H$ contains all Sylow-$p$ ...


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If $G\subseteq S_p$ acts transitively on $\{1,\dots,p\}$, the subgroup $G_1=\{g\in G:g1=1\}$ that fixes $1$ has index $p$, since the function $gG_1\in G/G_1\mapsto g1\in\{1,\dots,p\}$ is a bijection. It follows that $p$ divides the order of $G$ and, therefore, that the $p$-Sylow subgroup of $G$ is non-trivial. On the other hand, since $p^2$ does not divide ...


1

Sketch: Use the Orbit-Stabilizer theorem, Lagrange's theorem, Sylow's theorem, a little arithmetic.


1

Define $A = \langle (1\ 2\ 3\ 4)\rangle = \langle a\rangle$ and $B = \langle (1\ 3)\rangle = \langle b\rangle$. Note that $A \cap B = \{e\}$. Note as well that $ba = (1\ 3)(1\ 2\ 3\ 4) = (1\ 2)(3\ 4) = (1\ 4\ 3\ 2)(1\ 3) = a^{-1}b$. This, in turn, implies that $ba^i = a^{-i}b$, so $BA \subseteq AB$. Since $|AB| = |BA| = 8$, we conclude $AB$ is a subgroup ...


2

Let $g=(1\,2\,3\,4)$, and $h=(1\,3)$. So the order of $g$ is $4$ and the order of $h$ is $2$. We have $gh=(1\,4)(2\,3)$, so $(gh)^2=e$. Rather, $ghg=h$. That means that any word consisting of $g^i$s ($0<i<4$) and $h^1$s can be reduced to not contain the subword $ghg$. If you have an $h$ surrounded by $g$'s on either side, you can reduce the $ghg$ to ...


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Define a metric $d$ on $\{1,2,3,4\}$ as follows: $d(1,2)=d(2,3)=d(3,4)=d(4,1)=1$; $d(1,3)=d(2,4)=\sqrt2;\ d(x,y)=d(y,x);\ d(x,x)=0.$ An isometry of the metric space $\{1,2,3,4\}$ is a bijection $f:\{1,2,3,4\}\to\{1,2,3,4\}$ which preserves distances, i.e., $d(f(x),f(y))=d(x,y)$ for all $x,y\in\{1,2,3,4\}$. Observe that the set of all isometries of ...


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This can be solved by the following greedy strategy (where we assume each ribbon has a unique color and hence a unique target, as all cases may be reduced to this one). The main body of the post proves it solves things in at most $n+m-1$ moves. The addendum proves that it is optimal: Fill the grid from top to bottom (i.e. starting at the topmost cell, ...


2

For two elements $x,y \in A_n$ to be conjugate within $A_n$ they have to have the same cycle type. Suppose $axa^{-1}=y$ for some $a\in S_n$. Suppose $x$ contains a cycle $c$ which moves an even number of points (elements of the underlying set of size $n$), and is therefore an odd permutation. We have $cxc^{-1}=x$ so that $acxc^{-1}a^{-1}=axa^{-1}=y$. One of ...


1

Let $m$ and $n$ be positive integers. Denote the set $\{0, 1, \ldots, m - 1\}$ by $M.$ Let $p = m + n - 1,$ and denote the set $\{m, m + 1, \ldots, p\}$ by $N$. Let $\pi$ be any permutation of the set $P = M \cup N = \{0, 1, \ldots, p\}$ that satisfies the conditions: \begin{align*} i < m & \implies \pi(i) \leqslant i \text{ or } \pi(i) \geqslant m, ...


1

The proof relies on finding elements in $G$ of orders $4,5$ and $6$, therefor $|G|=lcm(4,5,6)=60$, so either $G=A_5$ or $G=S_5$, but $(a,b)\in G$ and $(a,b)\notin A_5$. Therefore $G=S_5$. Although it isn't true that $(a,c,d,e)(a,b)=(a,d)(b,c,e)$, we can still show that $(a,d)(b,c,e)\in G$ since $(a,b,c,d,e)(a,c,d,e)=(a,d)(b,c,e)$, hence $(a,d)(b,c,e)\in ...


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You are correct, $(a,c,d,e)(a,b)=(a,b,c,d,e)$. In fact, this is easy to see since $(a,b)$ is its own inverse, and therefore by what you found previously, $$(a,c,d,e)(a,b)=(a,b,c,d,e)(a,b)(a,b)=(a,b,c,d,e).$$


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For the first question. Let us consider the morphism : $$f_3:\mathbb{N}\rightarrow \mathbb{N} $$ $$3n\mapsto 3n $$ $$3n+1\mapsto 3n+2$$ $$3n+2\mapsto 3n+1$$ This is clearly a bijection of order $2$ whose support is $(3\mathbb{N})^c$ which is infinite. On the other hand could also consider : $$f_2:\mathbb{N}\rightarrow \mathbb{N} $$ $$2n\mapsto 2n+1 ...


2

Because $1$ is getting fixed so you can drop it. Any cycle of $1$ length i.e. (a) can be dropped , although you can write it too. It is just a way to make it look less complicated



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