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4

You are missing the identity element. It can be written as an "even" permutation: $$(12)(12)$$


3

The problem suggests that $H$ is isomorphic to the permutation group of some three elements. Find an element of order $3$ in $H$, an element of order $2$. Show that they must be, for example, $(123)$ and $(12)$, otherwise the order of $H$ is more than $6$.


2

elements of $H$ can be of order $1,2,3,6$ only. now if element is identity it is obvious. if order of an element is $2$ or $3$, then they are $2$ cycle or $3$ cycle resp. so will definitely leave $2$ or $1$ element fixed resp. Now if $a \in S_4$ and |$a$|=$6$ is not possible in $S_4$ as it will have to be either a $6$ cycle, absurd or product of disjoint ...


2

I do not think that this proposition allows a nice proof by induction. Note that you can't construct every subgroup of $S_n$ by adding an element to a subgroup of $S_{n-1}$. Instead, you can apply the pigeon-hole principle immediately: If $|H|>n$, there must exists $h_1,h_2 \in H$ with $h_1(1)=h_2(1)$. Then $1$ is a fixed point of $h_1^ {-1}h_2$.


2

The order of $A_n$ is always half the order of $S_n$, consider the bijective map from the even permutations to the odd permutations where $\varphi(\pi)=(12)\pi $. This is a bijection since the inverse is the map from the odd permutations to the even permutations $\varphi^{-1}(\pi)=(12)\pi$.


2

It's hard to help without knowing your background knowledge.Here is one approach. The transitive action of $A_5$ of degree $12$ has stabilizer of order $5$, which has four orbits, two of size $5$ and two of size $1$. So $A_5$ has four orbits on $I^2$ so, by a standard result on permutation characters (the fixed point formula applied to $A_5$ on $I^2$), we ...


1

The existence of an $n$-cycle shows that $H$ is transitive. Then the existence of an $(n-1)$-cycle shows that it is $2$-transitive. .... Prove by induction on $r$ that the existence of the $(n-r)$-cycle implies that $H$ is $(r+1)$-transitive.


1

"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 $\tau:=(b,c) \in A(S)$ . ...


1

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 ...


1

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 ...


1

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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