# Tag Info

4

The elements of $S_{E}$ are the bijections from $E$ to $E$. We can use the fact that compositions of bijective functions are bijective. In particular, we could define a map $\phi: S_{E}\rightarrow S_{F}$ by $\sigma\in S_{E}\mapsto \beta\circ\sigma\circ\beta^{-1}$. To show this is a homomorphism, let $\sigma_{1},\sigma_{2}\in S_{E}$. Then ...

4

Two permutations in $S_{n}$ are conjugate if and only if they have the same cycle type, so really the question is how many $m$-cycles there are in $S_{n}$. Well, choose your $m$ elements from $n$ in $n\choose{m}$ ways. But we can rearrange those $m$ elements in $m!$ ways. Finally, each of our cycles has been represented $m$ times by a cyclic rearrangement, ...

4

Let $\rho=\sigma_1\sigma_2\dots\sigma_r$ be the cyclic decomposition of $\rho$ in $S_n$. The $(\sigma_i)_{1\le i\le r}$ are cycles and are pairwise disjoint. Thus they commute with each others. Thus $\rho^3=\sigma_1^3\sigma_2^3\dots\sigma_r^3=I$. Since the cycles are disjoints, what can you say about $\sigma_1^3,\sigma_2^3,\dots,\sigma_r^3$? Then you should ...

3

The permutation $\rho$ has a decomposition as a product of disjoint, hence commuting, (non-trivial) cycles: $\;\rho=\gamma_1\dotsm\gamma_r$. The order of $\rho$ is the l.c.m. of the orders of the cycles, so each $\gamma_i$ has order $3$. As the order of a cycle is its length, this means each $\gamma_i$ is a $3$-cycle.

3

darij is incorrect; it's not even possible to write every even permutation as a square of some permutation, even or odd! As an exercise, show that a permutation has a square root (in $S_n$) iff it has an even number of cycles of each even length. (Derek Holt's example in the comments is the smallest example where a permutation has cycles of two different ...

3

You are asking whether the map $A_n \rightarrow A_n$ defined by $a \mapsto a^2$ is surjective. Since $A_n$ is a finite set that's the same asking whether the map is bijective. This is not true when $n \geq 4$, because then the map is not injective: for example, you can find $a \neq 1$ with $a^2 = 1$.

3

Functions $f : X \to X$ that satisfy $f\circ f = \mathrm{id}_X$ are called involutions. Notice that if $f : X \to X$ is a bijection, then we can partition $X$ according to the orbits of $f$, i.e., sets of the form $$\{ f^{\circ n}(x) : n \in \Bbb{Z} \}.$$ If $f$ is an involution, then the orbits are either a singleton $\{x\}$ when $f(x) = x$ or of the ...

3

I think you are confusing symmetric group and group of symmetry. The latter are more commonly called dihedral groups hence the notation $D_8$. Dihedral groups consist of rotations and flips of the polygon.

3

The rotational group of the cube certainly turns a (here always: main) diagonal into a diagonal, hence acts on the set of the four diagonals. This action gives us a homomorphism $\phi\colon G\to \operatorname{Sym}(4)$. On the other hand, $G$ certainly has order $24$: We can pick one of six faces as "ground" face and rotate it in steps of $90^\circ$, thus ...

3

Let $\sigma,\tau$ be a permutations on $\{1,\ldots, n\}$. If $\sigma$ is a cycle, say $\sigma=(i_1,\ldots i_k)$, then \begin{align} \tau\sigma\tau^{-1}(\tau(i_j)) & = \tau\sigma(i_j)\\ & = \tau(i_{j'})\\ \end{align} where $$j'=\begin{cases}j+1, & 1\le j<k\\1, & j=k\end{cases}.$$ So, ...

2

Since $S_n$ is generated by its transpositions, it is enough to show that $G$ contains all transpositions. Now, if $\tau\in G$, then $$\tau(12)\tau^{-1}=(\tau(1),\tau(2)).$$ Since $G$ acts transitively on $X$, for any $(i,j)\in X$, we can find $\tau\in G$ such that $\tau(1,2)=(i,j)$. Then, $\tau(1)=i$ and $\tau(2)=j$. This means $$\tau(12)\tau^{-1}=(ij)\in ... 2 Vectors in \Bbb{F}_2^2 are not the same as matrices in \operatorname{GL}(2,\Bbb{F}_2). The set of non-zero vectors in \Bbb{F}_2^2 is$$\left\{\binom10,\binom01,\binom11\right\}.$$Now check how each of the six elements of \operatorname{GL}(2,\Bbb{F}_2) you listed acts on this set. For example$$\begin{pmatrix}0&1\\1&0\end{pmatrix}\qquad\text{ ...

2

The Galois group $Gal(\mathbb{Q}(\alpha,\omega)/\mathbb{Q})$ contains the six automorphisms below (where each $\sigma_i$ fixes $\mathbb{Q}$): $$\begin{matrix} \sigma_1(\alpha)=\alpha & \sigma_1(\omega)=\omega \\ \sigma_2(\alpha)=\alpha\omega & \sigma_2(\omega)=\omega \\ \sigma_3(\alpha)=\alpha\omega^2 & \sigma_3(\omega)=\omega \\ ... 2 First of all I guess what you write as conjunction you mean conjugation (at least this is the name I know). Secondly your H' is the set of subgroups of S_3. (H' is not a group!!). So S_3 acts on H' by conjugation. As you mentioned correctly S_3 contains 6 subgroups, where of course S_3, the trivial subgroup \{1\} and the alternating group ... 2 As you suspect, your answer is not correct. The first thing you need to list all the subgroups of S_3. Now for each subgroup H \leq S_3 and for each g \in S_3, you need to compute gHg^{-1}. These conjugate subgroups are the elements of the orbit of H. For example, take H = \langle (1 \ 2) \rangle \leq S_3. Now we need to loop over all the g ... 2 If we avoid following two things, then answer is "interestingly" yes! x\neq 1 n\neq 4. Thus, we "state" the following theorem (Ref.- Problems in Group Theory: Dickson): Theorem [S. Picard]: If n\neq 4, then given any x\in S_n with x\neq 1, there exists y\in S_n such that x and y generate S_n. Sophie Piccard, Sur les Bases du ... 2 Consider the following elements in \Sigma_X where X = \mathbb{Z}$$ \sigma(i) = \begin{cases} i &: i \leq 0\\ i+1 &: i \in \{1,3,5,\ldots\} \\ i-1 &: i\in \{2,4,6,\ldots\} \end{cases}  \tau(j) = \begin{cases} i &: i\leq -1 \\ i-1 &: i\in \{1,3,5,\ldots\} \\ i+1 &: \{0,2,4,\ldots\} \end{cases} $$Then o(\sigma) = o(\tau) = ... 2 You are correct that since A_5 has no nontrivial normal subgroups, it can't be written as a nontrivial semidirect product. Now, for A_4, the copy of V=(\mathbb{Z}/2)^2 within the group is a nontrivial normal subgroup, so it is natural to try to use this. Now, you must consider A_4/V. By considering the order, this is a group of order 3. How ... 2 A permutation is conjugate if it has the same cycle structure. In this case you want to find the no. conjugates of a m cycle, so your question boils down to find the no. of m cycles in Sn. Which can be done by n(n-1)(n-2)...(n-m+1)/m ways as for first entry you have n options, for next entry you have (n-1) options and finally you divide n(n-1)...(n-m+1) by ... 2 By Burnside's lemma, 1/|G| \cdot \sum_{g \in G} f(g) is equal to the number of orbits under the action of G. Since G is transitive, the sum is 1. 1 By the orbit stabilizer theorem and transitivity, the number of cosets of the stabilizer of a point is p. In particular, as in a hint you received in a comment, the order of the group is divisible by p. By Cauchy's theorem there must be an element of order p. In general an element of the symmetric group of order p can only be shown to be a product of ... 1 Denote$$ \sigma=(a_{1}a_{2}\ldots a_{m}) $$There is a theorem that states that for every \tau\in S_{n}$$ \tau^{-1}\sigma\tau=(\tau(a_{1}),\tau(a_{2}),\ldots,\tau(a_{m})) $$thus, given any m elements of \{1,...,n\} there is some \tau s.t \tau^{-1}\sigma\tau will result in a cycle of those exactly m elements. For the above formula recall two ... 1 As stated in the comments above, you can always consider cycles (1,2, \cdots, k) . 1 The symmetry group (and not symmetric group, which involves permutations) of a polygon is the group of (geometric) actions (or transformations) which leave the polygon the same (invariant). For example, the symmetry group for a square (dihedral group), includes rotations of 90^o, reflections (or flips) along vertical/horizontal/diagonal axes and so on.. 1 Your derivation is correct but you can go one step further and write:$$x=(1\ 3\ 2\ 5\ 4), assuming that $fg(i)=f(g(i))$. For example for $3$ we have $3\to5$ because of $(5\ 3)$ and then $5\to2$ because of $(5\ 2)$, so $3\to2$.

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The symmetric group is even generated by all adjacent transpositions, hence in particular by all transpositions $(ij)$ with $i<j$. Furthermore, in the notes of Keith Conrad you can find all proofs for many different sets of generators of $S_n$, in particular, Theorem $2.1$. The proof is just this: every $n$-cycle is the product of transpositions as ...

1

Using the orbit-stabilizer theorem: first of all, we can see that we have "some" homomorphism: $G \to S_4$ by considering the fact that any such rotation of $G$ permutes the main diagonals. Now let's consider the action of $G$ on the set of faces of the cube, we have $6$ of these. $G$ clearly always takes a face to a face. Since it is possible, by just ...

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Hint: all those permutations have the following form $(ij)(kl).$

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Experiment with powers of cycles in order to see what happens. In particular, all new cycle types of the powers are obtained be considering only powers $\rho^d$ where $d$ is a divisor of $|\rho|$. Say $\rho=(1~2~3~4~5~6)$ within $S_6$ for instance. What are $\rho^2$ and $\rho^3$? In general, if $\rho$ is an $n$-cycle and we have a divisor $d\mid n$, can you ...

1

Consider the $N$ subspace $x_1 + \dots + x_{N+1} = 0$ which is orthogonal to the vector $\vec{1} =\underbrace{(1,\dots, 1)}_{N+1} \in \mathbb{R}^{N+1}$ Then $S_{N+1}$ acts on this subspace and embeds into $O(N+1)$ and yet $\vec{1}$ is preserved, so this action embeds into $O(N)$. This shows $n \geq N+1$. This is the same as @hardmath's answer. Showing ...

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