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7

You are mixing up a few concepts here. The symmetric group $S_n$ is the group of permutations on $[n]=\{1,2,\dots,n\}$, i.e. $$S_n = \{ \varphi : [n]\to[n] \,|\, \text{$\varphi$ is bijective}\}.$$ The group operation on $S_n$ is given by composition of maps. The group $\mathbb Z/n\mathbb Z$ consists of cosets of $n\mathbb Z$ in $\mathbb Z$, i.e. $$\mathbb ...


6

You can't expect a purely group theoretic proof since the result is fundamentally about convergence. Some analytic element is thus necessary. In fact it is not at all a group-theoretic result. Moreover, I don't know which proof you refer to but the one I know (and I presume it's basically the only proof (up to irrelevant detail)) explains perfectly well ...


6

First: commutativity of finite sums. $\mathbb{R}$ is a field, and as such, addition is commutative. This means that for two elements $a,b\in\mathbb{R}$, $a+b=b+a$. However, it is possible to extend this result. Consider $a_1,\ldots,a_n\in\mathbb{R}$, and a permutation $\pi$ of $\{1,\ldots,n\}$. Then, by induction, one can easily show that $$ \sum_{i=1}^n ...


2

In any group $G$, any element $g \in G$ defines a function from $G$ to itself denoted $L_g : G \to G$, called "left multiplication": $$L_g(h) = gh $$ The group axioms show that this function is injective $$L_g(h) = L_g(h') \iff gh = gh' \iff h=h' $$ and that it is onto $$L_g(g^{-1} h) = g g^{-1} h = h $$ Therefore the formula $g \mapsto L_g$ defines a ...


2

Assuming that you know that $f$ fixes every $3$-cycle, it follows that $f$ also fixes every even permutation (because $A_4$ is generated by $3$-cycles). Now $A_n$ is the commutator subgroup of $S_n$ when $n \geq 3$ (proof: write a $3$-cycle as the commutator of two transpositions). So the parity of an permutation is invariant under an automorphism of $S_n$. ...


1

The intuation is that $$gG=G$$ for any $g\in G$ which means that every $g$ of permutes the elements of $G$. Surprisingly the corresponding permutations elements constitute a subgroup of $S_G$ which is isomorphic to $G$. If $G$ is simple, $G$ can be embeded into $A_n$ where $n$ is the smallest index of nontrivial subgroup of $G$.


1

Yup. $G$ is always a subgroup of $S_{|G|}$. On a related note, the question of the smallest $n$ such that $G\leq S_n$ is, in general, not known.


1

Come on, this is really not difficult. The left regular representation of $c$ is the matrix of the left translation by $c$ in the algebra $\mathbb{Z}S_3$. Thus, according to your table: $ca=f$, $cb=d$, $cc=e$, $cd=b$, $ce=c$, $cf=a$. In other words, $ca=0a + 0b + 0c + 0d + 0e + 1f$, $cb=0a + 0b + 0c + 1d + 0e + 0e$, $cc=0a + 0b + 0c + 0d + 1e + 0f$, ...



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