# Tag Info

## Hot answers tagged group-theory

2

The idea here is that we want to find an invariant subspace, that is, a subspace $W$ of $V$ such that if $w \in W$ we have that $gw \in W$ for all $g \in S_3$. As Tobias mentioned, we want to pick a subspace where no matter how we permute the basis vectors, we get the same subspace. The subspace $W = <(1,1,1)>$ is invariant, and it's complement is a ...

1

If we take a complete sets of left cosets of $\;H_i,\,i=1,2\;$ in $\;G\;$ , say $$\left\{\,y_j H_2\,\right\}_{j\in J}\;,\;\;\left\{\,x_i H_1\,\right\}_{i\in I}\;,\;\;|J|,\,\,|I|<\infty$$ then for any two indexes $\;i\in I,\,j\in J\;$ we have that either $\;x_iH_1\cap y_jH_2=\emptyset\;$ or else the $\;x_iH_1\cap y_jH_2\;$ is a coset in $\;H_1\cap ... 1 Since$U_{100} \thickapprox \mathbb Z_2\oplus\,\mathbb Z_{20}$, then there exists a group isomorphism$f : U_{100} \to \mathbb Z_2\oplus\,\mathbb Z_{20}$. Note that$U_{100}$is a multiplicative group while$\mathbb Z_2$and$\mathbb Z_{20}$are additive groups. So the identity element of$U_{100}$is$1$while the identity element of$\mathbb Z_2$and ... 1 Using Carmichael Function, $$\lambda(100)=(\lambda(25),\lambda(4))=(20,2)=20$$ So, for any integer$x,(x,100)=1\iff(x,2)=(x,5)=1,$$$x^{20}\equiv1\pmod{100}$$ 1 Put$\;k=2^n\;for simplicity, and observe that \begin{align*}&\left(2^k-1\right)^2=2^{2k}-2^{k+1}+1=1\pmod{2^k}\\{}\\&\left(2^{k-1}-1\right)^2=2^{2k-2}-2^k+1=1\pmod{2^k}\end{align*} The last equivalence because\;2k-2\ge k\iff k\ge 2\;$, which is the case we're interested in. Thus, since$\;2^{k-1}-1\neq 2^k-1=-1\pmod{2^k}\;$, there are ... 1 Each permutation$g \in S_n$can be represented by an$n \times n$permutation matrix$P(g)$defined by$P(g)_{ij} = 1$iff$g$maps$j$to$i$, and$P(g)_{ij}=0$otherwise. Let$A$denote the adjacency matrix of the graph$G$. It can be verified that a permutation$g$is an automorphism of the graph$G$if and only if$P(g)^{-1}AP(g) = A$. In other ... 1 Your$\sigma$is not an automorphism. It does not map the identity element to the identity element. As you say in the definition you only require that$\sigma(H) = H$for all automorphisms$\sigma\$. Not for all bijections or all functions or all homomorphisms.

Only top voted, non community-wiki answers of a minimum length are eligible