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1

You need to define a composition law, not just a set, for this to be a group. For this to be a group (and a subgroup of $S_{\mathbb{R}})$ you probably wanted to define composition as $f_n,f_m \mapsto f_n \circ f_m$. Then $f_n, f_m \mapsto (x + m) + n = x + (m+n) = f_{m+n}$. This shows that $G \cong \mathbb{Z}^+$ so it must be cyclic. Alternatively, you ...

0

$x+m$ is not an element of $G$. Perhaps what you meant to say is that $f_m$ is a generator for any $m$. But that is not true either. For example, how can you get $f_1$ by composing $f_2$ with itself?

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If $n\ge 5$,for all $\tau=(i,j)(r,s)(i\not=j,r\not=s)$, we prove that this is in the subgroup generated by all $A_{n-1}^i$. This is because if $i=r,j=s$, $\tau=1$ is trivial, if $j=r,i\not=s$, $\tau=(jsi)=(i_1,i_2)(j,s)(s,i)(i_1,i_2)$, here $(i_1,i_2)(j,s)\in A^i_{n-1}$ and $(s,i)(i_1,i_2)\in A^j_{n-1}$($i_1,i_2$ are distinct and not the same as $i,j,s$) . ...

4

Hints: Write the permutation as a composite of disjoint cycles, some of which could have infinite length. It is then enough to solve the problem for the individual cycles. Any cycle of length greater than $2$ is a composite of two elements of order $2$. For example, $(1,2,3,4,5) = f \circ g$ with $f=(2,5)(3,4)$, $g=(1,5)(2,4)$. This also applies to cycles ...

1

We have $d=am+bn$ where $a$ and $b$ are integers. So, $f^d=f^{am}f^{bn}=\left(f^m\right)^a\left(f^n\right)^b=1$ because $f^n$ and $f^m$ are 1.

3

Apply the Bezout identity: $$\exists x,y\in\mathbb Z, xm+yn=d$$ $$f^d=f^{xm+yn}=(f^m)^x(f^n)^y=id$$

2

Regarding to @Sanath's post and that you already knew the structure of $S_3$'s structure, we may treat the group with the following presentation: $$\langle a,b\mid a^2=b^3=(ab)^2=1\rangle$$ It is good to know that $S_3=D_6$, the dihedral group of order $6$.

3

I have often wondered why so many authors assume finite sets to be $\{1,2,\dotsc,n\}$. Although every finite set is isomorphic to such a set, a) the isomorphism is not canonical, b) in many applications there are finite sets (for example homogenous spaces) which are not of this form. If $X$ is any finite set, one can consider the group $\mathrm{Aut}(X)$ of ...

2

As James noted in his comment, generating sets are not unique, since if $A$ is a set that generates the group, then any set containing $A$ will also be a generating set. However, I assume you are trying to find a smallest set of generators. If you allow an element $g$ to be a generator, then everything in the cyclic group $\langle g \rangle = ... 2 Any permutation in any$S_n$can be expressed as a product of transpositions (2-cycles), so they constitute a generating set. 2 There is a generalization to$S_n$. The generators$\alpha_1,\cdots,\alpha_{n-1}$, such that$\alpha_i^2 = 1$,$\alpha_i\alpha_j = \alpha_j\alpha_i$if$j \neq i\pm 1$,$\alpha_i\alpha_{i+1}\alpha_i = \alpha_{i+1}\alpha_i\alpha_{i+1}.\alpha_i$swaps the$i$th and$(i + 1)$-th position''. See Wikipedia, which says the following: Other popular ... 2$S_3$can be generated by a 2 cycle and a 3 cycle. For example$(12)$and$(1 2 3)$. 3 You can generate$S_3$with a rotation$(1\:2\:3)$and a flip$(1\:2)$, think geometrically. 0 Let$V = \Bbb C^2$with the standard basis, and let$\rho: S_3 \to GL_2(\Bbb C)$be given by:$\rho(e) = I$,$\rho((1\ 2\ 3)) = \begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&-\frac{1}{2} \end{bmatrix}$,$\rho((1\ 3\ 2)) = \begin{bmatrix}-\frac{1}{2}&\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&-\frac{1}{2} \end{bmatrix}$... 0 It is know that for any Young Tableaux$T_{\lambda}$, there exists$f_1 \in M_1$such that \begin{equation*} M_1=\mathbb{C}S_n e_{T_{\lambda}}f_1\end{equation*} Put$g_1=e_{T_{\lambda}}f_1$and observe that$\sigma g_1=g_1$for all$\sigma \in R_{T_{\lambda}}$. On the other hand, suppose$0\neq h \in M_1$is also stable under$R_{T_{\lambda}}$-action. ... 1 What I liked most about this problem is that it doesn't lend itself to be easily solved by counting (it might, but I haven't seen such a proof). Thus, we are forced to travel different paths to attack the problem from which we might learn something else interesting in its own right. For example, GA316's observation that$A_n$is the only normal subgroup of ... 1 The observation that$A$and$B$are simultaneously diagonalizable does help. Proffering the following route. Assume that a subgroup$G\le GL_2(\Bbb{R})$isomorphic to$A_4$would exist. Because we can replace$G$by any of its conjugates, we may as well assume that the matrices $$... 3 The derived subgroup is A_5, which has index 2. A homomorphism G\to H has abelian image if and only if G' is in the kernel. A linear character is a group homomorphism to the abelian group of units of the field. More generally, A_n is the derived subgroup of S_n, and for nice enough fields the group of linear characters of G is isomorphic to ... 1 Hint : For n \ge 5, A_n is the only normal subgroup of S_n 3 Let's count the number of permutations with a cycle of length \ge n/2. It can only have one such cycle, for obvious reasons. First, pick a natural n/2\le k\le n. Next, pick a k-cycle from S_n, and finally the rest of the cycle decomposition can be determined by choosing any permutation on the set of elements in \{1,\cdots,n\} not already present in ... 2 One attack is to use properties of orientation preserving orthogonal linear transformation of \Bbb{R}^3. Those have (assuming that the center of the icosahedron is at the origin) matrices of determinant 1 such that their transpose is also their inverse. A group of order 5 is necessarily cyclic. The generator g of a cyclic group \langle ... 2 Some things to consider: What does the stabilizer of a vertex v look like? N has order 5, so what kind of group is it? Consider the isometries of the icosahedron and consider which ones could possibly generate N. 3 Let G be a group of order 60 with no normal subgroups of order 2, 3, or 5. In each of the following cases, suppose that H \unlhd G were a normal subgroup of order n: 1) n = 10, 15, 20, 30: H has a normal Sylow 5-subgroup P_5, so P_5 \text{ char } H \unlhd G \implies P_5 \unlhd G. 2) n = 6: Same argument as in (1), considering a ... 1 Using Lagrange's theorem, we know that |N|=5,|G|=60 implies that |G/N|=\frac{60}{5}=12=|V|. The action of G on V is transitive. Hence$$\{g\in G:\mbox{order}(g)=5\}$$is the group of rotations around a vertex v\in V by angles of \frac{2\pi}{5},\frac{4\pi}{5},\frac{6\pi}{5},\frac{8\pi}{5},2\pi. Let this group G_v:=\{g\in G:g(v)=v\} be denoted by ... 0 The symmetries of a regular hexagon are those of dihedral group 6, D_6. Let R_n denote a rotation around the centre of an angle n \pi/ {\bf 3}, for n\in \{0,2,3,4,5\}. Let r_n denote a reflection about a line through the centre which is at angle of n \pi / {\bf 6} from the x-axis, for n\in\{0,1,2,3,4,5\}. The permutations are thus: ... 2 let \sigma and \kappa be the cycles suppose \sigma(x)\neq x , then \kappa(x)=x so \sigma(\kappa(x))\neq x so they are not inverses. 2 Say the first cycle, \sigma_1 has that$$ \sigma_1(a)= b\ne a$$all cycles have at least two symbols (or we don't write them in a cycle decomposition) so this is a valid assumption. Then after applying the second cycle,$\sigma_2$, which does not have the symbols$a$or$b\$, by definition of disjoint. We then have ...

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