Equivalent Order of Cosets

Theorem

Given a group \(G\) with subgroup \(H\), the order of all cosets of \(H\) in \(G\) (both left and right) are equal, which is in turn equal to the order of \(H\) itself. That is, for all \(g \in G\)

\[ |gH| = |H| = |Hg|\]

and hence

\[ |gH| = |g'H|.\]

Proof

This can be proven by showing that the map \(f : H \to gH\) given by

\[ h \mapsto gh\]

is bijective.

Here we prove this from scratch, although this fact should be obvious given that multiplication by \(g\) on the left is invertible through multiplication by \(g^{-1}\) on the left.

For injectivity we have

\[\begin{align*} & f(h) = f(h') \\ \iff& gh = gh' \\ \iff& g^{-1}gh = g^{-1}gh' \\ \iff& h = h' \\ \end{align*}\]

and for surjectivity we have for any \(gh \in gH\) we have

\[ f(h) = gh.\]