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Let $n \in N$ and $S_n$ be the symmetric group on $n$.

(A) Let $\pi \in S_n$ and $z$ be the number of disjoint cycles of $\pi$ (here the 1-cycles are counted). Show that then ${\rm sgn} \pi = (-1)^{n-z}$

(B) Show that the subset To: = {π∈Sn|sgn (π) = 1} ⊆Sn is a subset of Sn.

(C) Determine the number of elements $|A_n|$ the subgroup $A_n$ from part (b).

Shaun
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ireallydontknowit
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  • For part (A) I'm partial to this argument. In other words, this question is also a duplicate. Askers are recommended to spend a while searching the site. Chances are that a standard question like this has already been covered. – Jyrki Lahtonen Dec 20 '18 at 19:12

1 Answers1

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For some $\pi\in S_n$, $\mathrm{sgn}(\pi) = (-1)^r$ where $\pi = \tau_1\tau_2\ldots\tau_r$ as a product of transpositions. This is well defined (although this is non-trivial to check).

(A) A cycle of length $k$ can be written as a product of $k-1$ transpositions - e.g. $(1 2 3 \ldots k) = (1 2)(2 3)\ldots(k-1\, k)$. If $\pi$ is a product of $z$ cycles of length $k_1,\ldots,k_z$ respectively then $\pi$ is a product of $$\sum_{i = 1}^z(k_i-1) = \left(\sum_1^zk_i\right) - z = n-z$$ transpositions and hence $\textrm{sgn}(\pi) = (-1)^{n-z}$.

(B) Check that $\sigma : S_n \rightarrow \{\pm 1\}$ where $\sigma(\pi) = \textrm{sgn}(\pi)$ is a well defined homomorphism. Then $A_n = \{\pi\in S_n : \textrm{sgn}(\pi) = 1\}$ is precisely the kernel of $\sigma$ and hence is a subgroup of $S_n$.

(C) Use the $\sigma$ from (B) and the first ismorphism theorem to show that $|A_n| = \frac{|S_n|}{2}$.

ODF
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