(March 2013) Let \(G\) be a group, let \(N\subseteq G\) be a normal subgroup, and suppose that \([G:N]=42\), where \([G:N]\) denotes the index of \(N\) in \(G\). Prove that \(x^{42}\in N\) for every \(x\in G\). (Suggestion: Consider the quotient group \(G/N\).)
Solution
Proof. The quotient groupG/N has order 42 since|G/N| = [G :N] = 42. Given any x∈G, consider the coset Nx∈ G/N. By Lagrange’s Theorem applied to the group G/N and the element Nx, we have (Nx)42 =Ne, since Ne is the identity element of G/N. Now (Nx)42 =N(x42) [by definition of the group operation in G/N], and therefore Nx 42 =Ne, and hence x42 =x42e−1∈N by the coset relation. QED