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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.


4

I think the following is a counterexample. Let $G=S_{10}$. As expected let $A_7$ consist of the even permutations of $G$ that have $8,9,10$ as fixed points. Let $\alpha=(1234567)$. The elementary 2-abelian group $$ E=\langle(18)(24)(37)(56),(28)(14)(35)(67),(38)(46)(25)(17)\rangle $$ is stable under conjugation by $\alpha$. Unless I fumbled it, this is ...


4

In the case of $S_{2n}$, your argument almost goes through in most cases! Take $S_{2n-1} \subset S_{2n}$ (considered as the subgroup of permutations fixing $n$) and use your construction to obtain an element of order $(n)(n-1)$. In this case, $n(n-1) \leq 2n$ implies that $n^2 \leq 3n$, which fails for $n \geq 4$. This leaves just the case of $S_6$. By ...


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 $s_i = (i \quad i+1)$, ...


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 ...


3

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 ...


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 $e_1-e_n,\ldots, e_{n-1}-e_n$.


1

The space $F(p)$ is an open submanifold of $X^p$, so it has the same dimension*, namely $np$. The group $S_p$ acts freely and properly discontinuouly on $F(p)$ and is discrete, thus $B(p) = F(p) / S_p$ has the same dimension as $F(p)$, again $np$. * If you want to be convinced, each $x \in F(p)$ has an open neighborhood contained entirely in $F(p)$; $X^p$ ...


1

The commutator subgroup can be characterised as being the smallest normal subgroup such that the quotient is abelian; by this I mean that if $N$ is normal in $G$ with $G/N$ abelian then $N$ contains the commutator subgroup. The only normal subgroups of $S_3$ are $\{e\}$ and the $C_3$ subgroup generated by $(123)$ (being normal is a direct consequence of ...


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 $$T(G):=\left\{(g,h)\in G\times G\,\big|\,gh=hg\right\}$$ ...


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 ...


1

We wish to find all homomorphisms $\phi: \Bbb Z_4 \to \Bbb Z_9^*$. Now first of all it suffices to know what $\phi(1)$ is since for any other element $m \in \Bbb Z_4$ we have $\phi(m) = \phi(1 + 1 + .. + 1) = \phi(1) + ... + \phi(1) = m \phi(1)$. Hence $\phi(1)$ completely determines $\phi$. Now, $[\phi(1)]^4 = \phi(1) \cdot \phi(1) \cdot \phi(1) \cdot ...


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 ...



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