Let \(G\) be a group (not necessarily finite), and suppose that \(H\) is a subgroup of index \(n\). Show that there is a normal subgroup \(N\) of \(G\) with \(n!\ge [G:N]\ge n\).
Question
Math 6310 • Algebra
Summer 2016 • Question 1
Question
Summer 2016 • Question 1
Let \(G\) be a group (not necessarily finite), and suppose that \(H\) is a subgroup of index \(n\). Show that there is a normal subgroup \(N\) of \(G\) with \(n!\ge [G:N]\ge n\).