Let \(H\leq G\) be a subgroup (not necessarily normal). Prove that any two cosets \(xH, yH\in G/H\) have the same cardinality.
Define a map \(m_g: G\to G\) where \(x\mapsto gx\), restrict to \(m_h:H\surjects gH\), inverse \(\inverseof{(m_g)} = m_{\inverseof{g}}\)
Solution
Theorem.
Any two left cosets of a subgroup have equal cardinality.
Proof.
Fix \(x,y\in G\). Define \(\tau:xH\to yH\) by \(\tau(xh)=yh\) for \(h\in H\).
\(\tau\) is well defined since multiplying \(h\) by an element of \(H\) gives another element of \(H\).
For \(u\in H\), define \(\tau^{-1}(yu)=xu\); then \(\tau^{-1}(yu)\in xH\) and \[ \tau(\tau^{-1}(yu))=y u,\qquad \tau^{-1}(\tau(xh))=x h. \] So \(\tau^{-1}\) is an inverse.
Hence \(\tau\) is a bijection and \(|xH|=|yH|\).