# Proof of Normal subgroup in symmetry group

Prove that $A_n$ is a normal subgroup of $S_n$

Proof

A subgroup is normal if its left cosets are the same as its right cosets.

That is, we want to show that for s in $S_n$, $A_ns$ = $sA_n$

Note that by Lagrange's theorem, $A_n$ has two cosets, one of which is $A_n$ itself.

Suppose that $sA_n$ = $A_n$: then $s$ is a member of $A_n$ which means that $A_ns$ = $A_n$ as well. Suppose that $sA_n \neq A_n$ then $s$ is not a member of $A_n$, which means that $A_ns$ cannot be $A_n$. Since neither $sA_n$ nor $A_ns$ are equal to $A_n$, they must both be the second coset. so, $A_n$ is normal

So I was reading this proof on the internet and if someone could please explain something to me. I've been going through my book and the only thing Lagrange's Theorem say is that the order of a subgroup divides the order of a finite group. Where does it say about cosets?

Also say what they say is true and that Lagrange Theorem says that the "group" ($A_n$ in this case) has two cosets, itself and the 'other', what exactly is this 'other'? Because the proof concludes that the right and left cosets are equal and therefore they are the 'other' cosets

• Read over the proof of Lagrange's Theorem or perhaps a remark just after it. Is there anything talking about "the index of $H$ in $G$"? If so, there should be some discussion there (a sentence or so) about how the standard proof relates to cosets. Namely, given $H \subset G$ subgroup, define an equivalence relation on $G$ by $a,b \in G$ are related if $ab' \in H$. Then the equivalence classes from this relation form a partition of $G$, each of which has $|H|$ elements. In particular, the equivalence classes are the cosets ($He$ etc). – Benjamin Dickman Nov 21 '12 at 8:44

For a finite group $G$, and $H \subseteq G$ a subgroup, the left cosets of $H$ in $G$ partition $G$ and $|aH| = |H|$ for all $a \in G$. Hence, $|G|/|H|$ gives you the number of left cosets.
The normality of $A_n$ in $S_n$ depends only on the fact that it has index $2$ (it has two left cosets). In general, if $|H|$ is coprime with $(|G|/|H| - 1)!$, then $H$ is normal.
Maybe the author of that proof is including in "Lagrange's Theorem" the fact that the number of cosets of a subgroup $H$ in a group $G$, is equal to $|G|/|H|$.