Show that if \(H < G\) is a proper subgroup, then \(\Union_{g\in G} gH\inverseof{g} \subset G\) is a proper subset.
Hint: consider the intersection and count. Try Orbit-stabilizer?
Erratum: the statement needs \(G\) finite. Every matrix in \(\mathrm{GL}_2(\CC)\) is conjugate to an upper-triangular matrix, so the conjugates of the proper subgroup of invertible upper-triangular matrices cover \(\mathrm{GL}_2(\CC)\).
Solution
Let \(G\) be finite, \(m=\# H\), and \(n'=[G:H]>1\).
1. \(H\) has \(n=[G:N_G(H)]\le n'\) distinct conjugates.
By orbit-stabilizer for the conjugation action on subgroups, \(n=[G:N_G(H)]\), and \(H\le N_G(H)\) gives \(n\le[G:H]\).
2. \(\# \Union_{g\in G} gH\inverseof{g}\le 1+n(m-1)\).
Each of the \(n\) conjugates has \(m\) elements, one of which is the identity, and the identity is common to all of them.
3. Q.E.D.
By steps 1 and 2 and Lagrange’s theorem \(n'm=\# G\), \[\# \Union_{g\in G} gH\inverseof{g}\le 1+n'(m-1)=\# G-(n'-1)<\# G .\]