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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 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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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 ...


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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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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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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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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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"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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