Equivalence Relation on Symmetric Group by Image of n is Congruence Modulo Subgroup

Theorem
Let $S_n$ denote the symmetric group on $n$ letters $\set {1, \dots, n}$.

Let $\sim$ be the relation on $S_n$ defined as:
 * $\forall \pi, \tau \in S_n: \pi \sim \tau \iff \map \pi n = \map \tau n$

Then $\sim$ is an equivalence relation which is congruence modulo a subgroup.

Proof
We claim that $\sim$ is left congruence modulo $S_{n - 1}$, the symmetric group on $n - 1$ letters $\set {1, \dots, n - 1}$.

Notice that every element of $S_{n - 1}$ fixes $n$.

For all $\pi, \tau \in S_n$ such that $\pi \sim \tau$:

so $\pi^{-1} \circ \tau$ fixes $n$ as well.

This shows that $\pi^{-1} \circ \tau \in S_{n - 1}$.

By definition of Left Congruence Modulo Subgroup:
 * $\pi \equiv^l \tau \pmod {S_{n - 1} }$

Now we show the converse.

Suppose $\pi \equiv^l \tau \pmod {S_{n - 1} }$.

Then $\pi^{-1} \circ \tau \in S_{n - 1}$.

Hence $\map {\paren {\pi^{-1} \circ \tau} } n = n$.

Then:

Therefore $\sim$ and left congruence modulo $S_{n - 1}$ are equivalent.

The fact that $\sim$ is an equivalence relation follows from Left Congruence Modulo Subgroup is Equivalence Relation.