Left cosets are all in bijection

Let \(H\) be a subgroup of \(G\). Then for any two left cosets of \(H\) in \(G\), there is a bijective function between the two cosets.

Proof

Let \(aH, bH\) be two cosets. Define the function \(f: aH \to bH\) by \(x \mapsto b a^{-1} x\).

This has the correct codomain: if \(x \in aH\) (so \(x = ah\), say), then \(ba^{-1} a x = bx\) so \(f(x) \in bH\).

The function is injective: if \(b a^{-1} x = b a^{-1} y\) then (pre-multiplying both sides by \(a b^{-1}\)) we obtain \(x = y\).

The function is surjective: given \(b h \in b H \), we want to find \(x \in aH\) such that \(f(x) = bh\). Let \(x = a h\) to obtain \(f(x) = b a^{-1} a h = b h\), as required.

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