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Hint: It is easy to show that $e^{i\theta_1}\cdot e^{i\theta_2}=e^{i(\theta_1+\theta_2)\pmod{2\pi}}$ (Since, $e^{2i\pi}=1$ ). This would show that the group operation of $S_2$ is binary. If $e^{i\theta}$ is identity, then $e^{i\theta_1}\cdot e^{i\theta}=e^{i\theta_1}=e^{i(\theta_1+\theta)\pmod{2\pi}}$, can you find what $\theta$ is? Similarly, you can ...

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Let's think about $S(p, q)$ and $S_p \times S_q$. $S(p, q)$ has $\binom{p+q}{p}$ elements. Think of $n$ adjacent slots: we have to choose $p$ of them to put the first $p$ in order, and afterwards the $q$ remaining elements must go in the consecutive remaining slots. $S_p \times S_q$ has $p!q!$ many elements: $p!$ permutations of the first $p$ elements and ...

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It looks like you want the permutations corresponding to all cycles of length $d$. We can characterize these permutations as follows: Let $\sigma_d$ denote the permutation $i \mapsto i+1 \pmod d$, corresponding to the matrix $$\pmatrix{ &&&&1\\ 1\\ &1\\ &&\ddots\\ &&&1 }$$ The cycles of length $d$ on $d$ elements ...

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The point is that the cycle-type of an element of order $2$ is $1,1,2,2$. Now I know I can conjugate this element to any other element of the same cycle-type within $S_6$ - because conjugacy classes in $S_n$ are defined by cycle-types for all $n$. Now suppose I conjugate some permutation $p$ of this type to another permutation $q$. If I am conjugating using ...

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Trace what happens to each element: in $(ab)(cd)$ the element $d$ goes to $c$ which then is not changed; in $(dac)(abd)$ the element $d$ goes to $a$ which then goes to $c$. So $d$ ends up at $c$ in both cases. Check $a,b,c$ in the same way. And any element $x$ other than $a,b,c,d$ is not affected by either permutation, so ends up as $x$ itself in both ...

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The cycle $(abd)$ sends $a$ to $b$, $b$ to $d$, and $d$ to $a$, leaving $c$ alone. The cycle $(dac)$ sends $d$ to $a$, $a$ to $c$, and $c$ to $d$, leaving $b$ alone. What happens when you apply first $(abd)$ and then $(dac)$? \begin{align*} &a\mapsto b\mapsto b\\ &b\mapsto d\mapsto a\\ &c\mapsto c\mapsto d\\ &d\mapsto a\mapsto c ... 0 So you want to see that\sigma = (ab)(cd) = (dac)(abd)= \tau$$You can simply try to see what happens to each "number". Note that$$ \sigma(a) = b\\ \sigma(b) = a\\ \sigma(c) = d\\ \sigma(d) = c\\ \\ \tau(a) = b\\ \tau(b) = a\\ \dots  For example, $\tau$ applied to $d$ first sends $d$ to $a$ and then $a$ to $c$. Since $\sigma(x) = \tau(x)$ for $X$ ...

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For any partition (n) or (k,1^n-k) alias (k,1,1,..,1) with n-k '1's, the resulting tableaux are always () alias empty tableau or ((1),(2), ...,(n-k)) alias a vertical one-column tableau. Bur for any partition (k,2,1,1,..,1) with n-k-2 '1's, the beheaded tableau can be either ((1,2),(3),..,(n-k-1)) or ((1,3),(2),..,(n-k-1)) depending on T. So, beyond these ...

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Hint: For the first, if the two top sides are the same you do not have a different triangle to pair it with, so count them all. If the top two sides are different, you have another triangle to pair it with, so divide these by $2$. How many of the $k^3$ have the top two sides the same? Similarly for the second, if all three sides are the same you don't ...

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In fact, every element of $S_p$ is, for example, if $2< p$, $[2]=\{n \in \Bbb Z: n\mod(p) =2\}.$ All the numers of $[2]$ are viewed as the same element of $S_p$, i.e., $[2]$.

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I think a conjugation of a glide reflection like rg$r^{-1}$, can be seen as rRt$r^{-1}$=rR$r^{-1}$rt$r^{-1}$, as the conjugation of reflection is still reflection, so does reflection, and conjugate will not change the direction, thus rR$r^{-1}$rt$r^{-1}$ is still a glide reflection.

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The following idea is super-simple. However, it only works if $n$ is odd (so it works precisely half the time!). But as it is so simple I thought it would be nice to record it here. Suppose $n$ is odd. Note that $S_n$ can be generated by the elements $\alpha:=(1, 2)$ and $\beta:=(1, 2, \ldots, n)$. Now, every subgroup $H$ of index 2 is the kernel of some ...

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Your proof looks simple because you assumed not so simple result that $A_n$ is simple for $n\geq 5$... Actually something more is true... Suppose that $H\leq S_n$ of index $m$ with $m< n$ then we have homomorphism $\eta: S_n \rightarrow S_m$. As $Ker(\eta)$ is a normal subgroup of $S_n$ we should have $Ker(\eta)=1$ or $Ker(\eta)=A_n$. Suppose ...

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As suggested, I am writing the complete answer. First we use the not so simple result: Theorem: Let $n = 3$ or $n\geq 5$. Then $A_n$ is simple. Statement: Now for $n=3$ or $n \geq 5$ let's show that $\{id\}, A_n$ and $S_n$ are the only normal subgroups of $S_n$.In particular, $A_n$ is the only sugbroup of $S_n$ of index $2$. Proof: It is clear that ...

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"Let $S$ be a set having more than two elements , then the center of $A(S)$ , the permutation group of $S$ , is trivial" . Let $id \ne \sigma \in A(S)$ , then $\exists a \in S$ such that $b:=\sigma(a)\ne a$ . As $S$ has more than two elements , $\exists c \in S$ such that $c \ne a , c \ne b$ . Now consider the transposition $\tau:=(b,c) \in A(S)$ . ...

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As you correctly calculated, the order of$A$ is $24$ and the order of $B$ is $6$. Therefore the quotient group is of order $4$. If you restrict the homomorphism $A\to A/B$ to the first factor, you obtain a subgroup of $A/B$ isomorphic to $\mathbb{Z}_2$, and the same is true for the second factor, which yields a different subgroup of order 2. Thus the ...

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