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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 ...


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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 ...


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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 ...


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Using Carmichael Function, $$\lambda(100)=(\lambda(25),\lambda(4))=(20,2)=20$$ So, for any integer $x,(x,100)=1\iff(x,2)=(x,5)=1,$ $$x^{20}\equiv1\pmod{100}$$


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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 ...


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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 ...


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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.



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