Tag Info

New answers tagged

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?


1

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 ...


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 ...



Top 50 recent answers are included