# Tag Info

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I see that you want to know how to deal with this kind of problems . How about splitting your question into two paths? if S4 is normal then you know what to do , if not then you have to use this fact "S4 is not normal in A(n)" in order to complete your answer . As you know the answer to the question already, you can't say that "an argument along these lines ...

4

Not sure how to "convince" you, but I suspect you are worried because some of the 2-cycles do the job and others do not. The trick is in picking the right 4-cycle to go with your 2-cycle or vice versa. (1,2) works with (1,2,3,4) because 1 and 2 are adjacent in (1,2,3,4) whereas 1,3 is not. But (1,3) will work with (1,3,2,4). For more info: ...

2

Hint Let $X \subseteq S_n$ be the set of all permutations having $1$ and $2$ in separate cycles and $Y \subseteq S_n$ the set of all permutations having $1$ and $2$ in the same cycle. Show that the mapping $f : X \to Y$ which "glues together" the two cycles containing $1$ and $2$ (cycles written down such that they start with $1$ and $2$, resp.) is a ...

2

Note that if $\ker(f) \neq \mathbb{Z}$, then $f(1) \neq e$ ($e$ referring to the identity of $S_3$). Why? With the above established, we have two possibilities: either the order of $f(1)$ is $2$, or the order of $f(1)$ is $3$. In the first case, we have $f(2k) = f(2)^k = e^k = e$, so that $\ker(f) = 2\mathbb{Z}$. Similarly, in the second case, we have ...

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Hint : $\mathbb{Z}_2$ and $\mathbb{Z}_3$ are subgroups of $S_3$ (of course upto isomorphism) .

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As a counterexample, you may consider an irrational rotation on $\mathbb{R}^2$.

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A subgroup $G$ of $E(n)$ is called discrete if for each point $x\in \mathbb{R}^n$ the family $\{g.x\mid g\in G\}$ is locally finite. Since this family contains each point with multiplicity equal to the order of the stabiliser $G_x$ of $x$, the family is locally finite if and only if the orbit $Gx$ is discrete and the stabilizer $G_x$ is finite. This shows ...

1

These permutations are called involutions. The counting function for involutions on $n$ elements is documented at OEIS here. You'll be able to find explicit formulas, recurrence relations, asymptotics, and generating functions there, along with some references. OEIS (Online Encyclopedia of Integer Sequences) is a pretty nice resource in general for ...

2

The way I was taught it, there are $$\frac{n(n-1)}{2}$$ 2-cycles, $$\frac{n(n-1)(n-2)(n-3)}{2^2 \cdot 2}$$ products of two disjoint 2-cycles, and in general $$\frac{n(n-1) \cdot \dots \cdot (n-2k+2)(n - 2 k +1)}{2^k \cdot k!}$$ products of $k$ disjoint 2-cycles, provided $2 k \le n$.

5

Can you prove that every element of $S_n$ is equal to a product of transpositions? If so, you just need to show that each of those generating sets contains all transpositions. Edit: To make this more direct, this task becomes straight forward once you understand how conjugation works in $S_n$.

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The definition means that $S$ is discrete if for each point $x\in \mathbb{R}^n$ the family $Sx$ is locally finite, i.e., for each point there is a neighbourhood intersecting only finitely many subsets of this family. This is equivalent to the fact each compact set intersects only finitely many subsets of this family.

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We will use Pieri's formula which is a special case of the Littlewood-Richardson-rule. It states: Let $a+b = n$ be natural numbers and $\alpha \vdash a$ a partition of $a$. Then the induced representation $\mathrm{Ind}_{S_a\times S_b}^{S_n} V_\alpha \otimes V_{(b)}$ (where $V_\alpha$ is the irreducible representation corresponding to $\alpha$ and ...

0

In both cases, you could use the fact that if $H$ and $K$ are finite subgroups of a group $G$, then $|HK|=|H|~|K| / |H \cap K|$. To show $S_4=A_3 D_4$, observe that the nonidentity elements of $A_3$ have order 3, whereas the symmetries of a square $D_4$ is a group of order 8, and 3 does not divide 8, whence $A_3 \cap D_4=1$. Thus, $|A_3 D_4|=3 \times 8 ... 1 Consider the restriction of$\operatorname{sign}$to$H$. Since$H$has an odd permutation, this restriction is still surjective. Hence,$\ker( \operatorname{sign}\mid H)$has index$2$in$H$. 2 Let's count. There is a well-known formula for the number of ways of choosing an ordered set of$n$linearly independent 1-dimensional subspaces of an$n$-dimensional vector space over$\mathbb{F}_q$, namely: $$\frac{q^n - 1}{q - 1} \frac{q^n - q}{q - 1} \cdots \frac{q^n - q^{n-1}}{q - 1} = q^{\frac{1}{2} (n - 1) n} \frac{q^n - 1}{q - 1} \frac{q^{n-1} - 1}{q ... 0 If G contains an injective map then it suffices to show that the identity of G must be the identity of the monoid X^X, i.e. the identity map. Let e be the identity of G and f an injective function in G, then f(e(x)) = e(x) for all x \in X. Hence, e is also injective. This is sufficient to show that e(x) = x for all x \in G since ... 2 Let \pi=\langle p_1,p_2,\ldots,p_n\rangle be a permutation of [n]=\{1,\ldots,n\}. If \pi has no inversions, it’s the identity permutation, which is even, so assume that \pi has at least one inversion. Suppose that 1\le j<k\le n, and p_j>p_k. Let \rho=(p_j,p_k)\pi (where I apply the transposition first and then \pi); if \rho=\langle ... 1 I wrote up a general classification for the centers of D_n, (the dihedral group of order 2n, not n) just the other week. Perhaps it will be useful to read: If n=1,2, then D_n is of order 2 or 4, hence abelian, and Z(D_n)=D_n. Suppose n\geq 3. We have the presentation$$ D_n=\langle x,y:x^2=y^n=1,\; xyx=y^{-1}\rangle.$$Then$yx=xy^{-1}$... 0 As an alternative to Hagen von Eitzen's answer for the$k>3case, we can avoid using Bertrand's postulate. Write the equation as \begin{align*} \underbrace{k(k-1)\cdots 2}_{k-1} \cdot \underbrace{2\cdot2\cdots 2}_{k-1} = \underbrace{(n-2)(n-3)\cdots(n-2k+1)}_{2k-2} \end{align*} and compare terms pairwise (i.e. comparek$and$n-2$, then ... 2 I think the difference is that you are thinking of group actions of$G$on$S$as mapping elements of$G$to elements of$S$, thus inducing a homomorphism of$G$into$S$. What is actually happening is that elements of$G$are being mapped to permutations of S. The following should help understand this difference. You can think of finite group actions as ... 1 If you want to define an action of a group$G$on another group$H$, you need a homomorphism from$G$to the automorphism group of$H$. The automorphism group of the cyclic group of order 5 is the cyclic group of order 4, so you can define a nontrivial action since there are maps from$S_3$to$\mathbb{Z}/4$. 1 You have defined a successful group action on the set$\{0,1,2,3,4\}$, but this does not give a homomorphism to$\mathbb Z_5$. To do this, you would need to assign an element of$\mathbb Z_5$to each element of$S_3$. So for example if your homomorphism is called$\phi\colon \mathbb S_3\to \mathbb Z_5$, then$\phi((12))$would have to be an element of ... 1 Conjugation affects cycles like so:$\sigma(a_1~a_2~\cdots~a_r)\sigma^{-1}=(\sigma(a_1)~\sigma(a_2)~\cdots~\sigma(a_r))\$. So, how does conjugation affect a permutation's disjoint cycle representation?

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