# Tag Info

6

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 $g\alpha g^{-1}=\alpha^k$, ...

3

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 ...

2

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 $H$.

2

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 $X_1,\dots,X_{i-1}$.

2

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 "$\{ g^k : k \in ... 2 To check that a subset of a group is a subgroup, you need to check that it is closed under composition and inverses. In other words, you must check that if you are given two rotations$g$and$g'$fixing your plane$P$, then$gg'$and$g^{-1}$also fix$P$. [Note that this subgroup is indeed an isomorphic copy of SO(2) sitting inside of SO(3), but you have ... 1 Define$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$is a subgroup ... 1$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$. Checking that this action satisfies the other definitions of an axiom is also easy to check (it's the same as checking that any conjugation action is a valid ... 1 Fleshing out the comment a bit. Assume that the group$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)$, ... 1 Well, this is one occasion, it might not be such a bad idea to do it computationally after all (that is, I am not aware of a better method). It is true that some transpositions don't commute, but we can pinpoint exactly which..: The arbitrary elements in the expansions$X_iX_j$and$X_jX_i$are$(ki)(mj)$and$(mj)(ki)$. These are definitely equal when the ... 1 Sketch: Use the Orbit-Stabilizer theorem, Lagrange's theorem, Sylow's theorem, a little arithmetic. 1 This is not a transposition, but in any case it is a permutation and hence is an element of$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$and are usually ... 1 To prove that every element in$A_n$is a commutator, you would have to divide it into several cases and prove that each k-cycle can be written as a commutator. G.A. Miller's paper prove the statement: http://projecteuclid.org/euclid.bams/1183416038 as well as Noboru Ito's, which I can't find online. Hope it is helpful. 1 Let$n=|\sigma|$, then$n|p$so$n=1$or$p$. If$\sigma\ne e$, then$|\sigma|=p$. Now since$1\le r<p$,$r$and$p$are coprime and so$|\sigma^r|=p$. Finally let$i\in\{1,\ldots, p\}$. The orbit of$i$consists of$i,\sigma(i),\sigma^2(i),\ldots,\sigma^{p-1}(i)$, therefore$\sigma$is a$p\$-cycle.

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