How does $A_n$ look in Aut$(X)$?

Let me phrase my question precisely:

Let $X=\{1,2,3,...,n\}$, $\ S_n=\mbox{Sym}\{1,2,3,...,n\}$ be symmetric group on $n-$letters. Let $\mbox{Aut}(X)$ denote the automorphism group of $X$. We can write $$S_n\cong \mbox{Aut}(X).$$ I want to understand what alternating group, $A_n$, corresponds to in $\mbox{Aut}(X)$. IOW, what is the subgroup of $S_n$ which $A_n$ is isomorphic to?

The only things I know about $A_n$ are

1. $A_n$ is the group of even permutations in $S_n$.
2. $A_n$ is the only normal subgroup of $S_n$ if $n\geq 5.$ (I know this as a fact; I haven't checked the proof.)
3. $\vert A_n \vert = \frac{n!}{2}.$

Using these information, how is it possible to construct a subgroup in $S_n$ s.t. $A_n$ is isomorphic to it? What if $n$ is not necessarily $\geq 5$? Any suggestions will be appreciated.

• $Aut(X)$ is defined when $X$ is a group. If you mean $S_{X}$ then this is identical to $S_{n}$ for your $X$. Aug 19 '14 at 18:22
• Normally the "automorphism" terminology would be used only where $X$ has some algebraic structure and we restrict attention to permutations of $X$ preserving that structure. Here $S_n = Sym(X)$ or $Perm(X)$ would suffice. Aug 19 '14 at 18:22
• @hardmath: Sets are algebraic structures. Their type is just empty. Universal algebra and category theory etc. also apply to sets. The automorphism group of a set makes sense. Aug 19 '14 at 19:10

I have often wondered why so many authors assume finite sets to be $\{1,2,\dotsc,n\}$. Although every finite set is isomorphic to such a set, a) the isomorphism is not canonical, b) in many applications there are finite sets (for example homogenous spaces) which are not of this form.

If $X$ is any finite set, one can consider the group $\mathrm{Aut}(X)$ of automorphisms of $X$. Because many people don't like category theory and would like to use old and classcial names, this group is usually denoted by $\mathrm{Sym}(X)$ or $\mathrm{Perm}(X)$.

The signature $\mathrm{sgn}(\sigma)$ of an automorphism $\sigma : X \to X$ is the rational function $\prod_{\{x,y\}} \frac{\sigma(x)-\sigma(y)}{x-y}$. The product is taken over all $2$-element subsets of $X$. Notice that the fraction only depends on $\{x,y\}$, so that $\mathrm{sgn}(\sigma)$ is well-defined. You can also view/define this as the determinent of the $k$-linear map $k^X \to k^X$ associated to $\sigma$ (for any commutative ring $k$). An easy calculation shows that $\sigma(x)^2=1$, so that in fact $\sigma(x)=\pm 1$. Another easy calculation shows that $\mathrm{sgn} : \mathrm{Aut}(X) \to \{\pm 1\}$ is a homomorphism. Then, $\mathrm{Alt}(X)$ is defined to be the kernel of $\mathrm{Sym}(X)$.

Equivalently, $\mathrm{Alt}(X)$ equals the commutator subgroup $\mathrm{Sym}(X)'$. This is best seen because both groups are generated by the $3$-cycles $(x \, y \, z)$. Again, this is true for every finite set $X$. In other words, you could also define $\mathrm{Alt}(X) := \mathrm{Sym}(X)'$ and then define $\mathrm{sgn}$ to be the Abelianization map (at least if $X$ has at least two elements) and then show that the Abelianization is $\cong \{\pm 1\}$.

• In the second paragraph you seem to be saying that $X$ can be any finite set. How do you calculate the rational function $\prod_{\{x,y\}} \frac{\sigma(x)-\sigma(y)}{x-y}$ when $X$ is an arbitrary finite set?
– MJD
Aug 20 '14 at 0:10
• One shows that $\sigma$ is a product of $2$-cycles and that $\mathrm{sgn}((x \, y))=-1$, as usual. Aug 20 '14 at 9:46
• What does $x-y$ mean when $x,y\in X$?
– MJD
Aug 20 '14 at 12:14
• The formula makes sense in the rational function field $\mathbb{Q}(X)$. It is just a quotient of two "formal" polynomials with variables in $X$. As I've said, it turns out to be $\pm 1$. For example, $\mathrm{sgn}(\sigma)=\frac{y-x}{x-y}=-1$. In any case, we don't need any ordering (or even subtraction) on $X$. Aug 20 '14 at 12:28
• I see now, thanks.
– MJD
Aug 20 '14 at 13:35