I have some trouble in understanding properly the concept of abelianization in the case of the symmetric group $S_n$. More specifically, it is known that the commutator group of $S_n$ is $A_n$, the group of all even permutations in $S_n$. Now, assume that we are given two elements $[\rho_1], [\rho_2] \in S_n / A_n$ such that $[\rho_1], [\rho_2] \neq e$ in $S_n / A_n$ ($e$ is the neutral element). This means that $\rho_1$ and $\rho_2$ are odd permutations. At the same time, however, we get $[\rho_1] \cdot [\rho_2] = [\rho_1 \circ \rho_2] = e$ because the composition of two odd permutations results in an even permutation (which is an element of $A_n$). Since the inverse of a group element must be unique, we thus infer that $S_n / A_n$ is not a group, which makes not much of a sense.
What am I missing?
Another question that I have is: is it true that $|S_n / A_n| \leq \binom{n}{2} + 1$? (The argumentation would go like that: since any $[\rho] \in S_n / A_n$ with $[\rho] \neq e$ must be such that $\rho$ is odd, we can write $\rho$ as a product of transpositions $\tau_1 \circ \tau_2 \circ \ldots \circ \tau_k$ with $k$ being odd. Noting that $\tau_2 \circ \ldots \circ \tau_k$ is in $A_n$, we infer that $[\tau_1] = [\rho]$. The estimation follows by observing that there are $\binom{n}{2}$ transpositions.)
Thanks a lot.
Best Answer
Your argument shows that any $[\rho_1],[\rho_2]\neq e$ are inverses to each other. You are right that inverses are unique, but this isn't a contradiction: it just proves that if $[\rho_1],[\rho_2]\neq e$ then $[\rho_1]=[\rho_2]$ (since they are both inverse to $[\rho_1]$). That is, $S_n/A_n$ contains only one element other than the identity. Explicitly, $S_n/A_n$ has two elements: the set of all even permutations, and the set of all odd permutations. (That is, assuming $n>1$: if $n\leq 1$, then there are no odd permutations and $S_n/A_n$ only has one element!)