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1

With the definition of sign in terms of transpositions, this is a one line proof. Namely, if $\phi$ is a product of an odd number of transpositions, then $\psi$ is the product of an even number of transpositions, and vice versa. Hence they have opposite signs. Less trivial is with the definition in terms of inversions (odd number of inversions means sign ...

4

Conjugacy classes in the symmetric group correspond to cycle-types. Thus, for $H = \langle (1234)\rangle$, the conjugacy class of $(1234)$ is just $\{(1234),(1243),(1324),(1342),(1423),(1432)\}$. We see that the cycles $(1234),(1432)$ generate the same group. Similarly, so do the cycles $(1243),(1342)$, and also the cycles $(1324),(1423)$. Thus, the orbit ...

3

The smallest $n$ is $6$: 1. $A$ is isomorphic to $\langle(1,2),(3,4),(5,6)\rangle$. 2. For $n=4,5$ the only subgroup of order $8$ which $S_n$ does contain is the dihedral group $D_4$ (and its conjugates, being a $2$-Sylow subgroup).

1

I denote the symmetric group on the set $G$ with $S_G$ Consider the following homomorphism $\varphi$: $G\rightarrow S_G\rightarrow Z_2$, where the first homomorphism is the left regular action (the homomorphism that sends $g$ to the permutation $f$ on $G$ defined by $f(a)=ga$ ) and the second homomorphism is the one that sends each permutation to its ...

1

A hint would be to take that element of order $2$, and decompose $\phi(g)$ as a permutation into disjoint cycles. If you know how many cycles and the length of each cycle, you can use parity to determine if that permutation is even or odd.

3

The quotient is actually $\mathbb{C}^n$ indeed. This is really an algebraic corollary and the geometric intuition here wouldn't be that helpful, mainly because we are talking about the homeomorphism type of the quotient, so we have quite a lot of flexibility. For that matter, we are in complex space, where the hyperplane of a reflection has topological ...

4

The class of $g \in S_n$ splits into two classes in $A_n$ if and only if all cycles of $g$ have odd length and their lengths are all distinct (this includes cycles of length $1$). In a class that splits, you get an element in the other class by conjugating by any odd permutation, such as $(1,2)$.

0

The problem is that you need to find $\tau\in A_n$ when your construction only guarantees $\tau\in S_n$. Start with $n\ge 5$. It suffices to show that any $3$-cycle $(a\,b\,c)$ is conjugate to $(1\,2\,3)$. We yould like to let $\tau(1)=a$, $\tau(2)=b$, $\tau(3)=c$ and otherwise $\tau$ is a bijecting between $\{4,\ldots, n\}$ and ...

1

For the second question, consider $A_4$ and the cycles $[123]$ and $[124]$. It is true that $[124]=[34][123][34]^{-1}$, but there is no even permutation $\tau$ for which $[124]=\tau[123]\tau^{-1}$.

2

$\newcommand{\Size}[1]{\lvert #1 \rvert}$First, a general fact. Suppose $H$ is a subgroup of index $2$ of the group $G$. Let $G$ act on a set $X$. Let $x \in X$. Then for the orbit $x H$ of $x$ under $H$, and writing $G_{x}, H_{x}$ for the stabilizers, we have  \Size{x H} = \Size{H : H_{x}} = \Size{H : G_{x} \cap H} = \Size{H G_{x} : G_{x}} = \frac{\Size{G ...

0

The idea is that a group of must be closed and the inverse element of it must be in it. By the group "generated" by some elements we mean the smallest such subgroup which contains all the elements first considered. You are right. $(123) = \{ (123), (132), I \}$ does not generate $S_3$. But the two elements $(123), (12)$ does. Let $G$ be the smallest ...

0

What you are dealing with is $\langle (123) \rangle = \{(\cdots,(123)^{-2},(123)^{-1},(123)^0=e,(123)^1,(123)^2,\cdots\}$ Which of course has elements repeating, because $(123)$ is of order $3$, and hence we obtain the elements $\{(123),(132),(1)(2)(3))\}=\langle(123)\rangle$. Why is this a subgroup? It has identity, $(1)(2)(3)$, that does nothing.

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