Let \(H \subseteq G\).
(a) Show that \(\left|N_G(H) : H\right|\) is equal to the number of left cosets of \(H\) in \(G\) that are invariant under left multiplication of \(H\).
(b) Suppose that \(\left|H\right|\) is a power of the prime \(p\) and that \(\left|G : H\right|\) is divisible by \(p\). Show that \(\left|N_G(H) : H\right|\) is divisible by \(p\).
(a) \(h(aH) = aH\) for all \(h \in H\) \(\iff\) \(a^{-1}ha \in H\) \(\iff\) \(a^{-1}\), hence \(a\) in \(N_G(H)\) \(\iff\) \(aH \subseteq N_G(H)\)
(b) Let \(H\) act on the left cosets of \(H\) by left multiplication. Then the orbit of the coset \(1H\) has size 1. But all orbit sizes must be of some \(p\)-power (by the condition imposed on \(\left|H\right|\)) and sum up to \(\left|G : H\right|\) which is itself a \(p\)-multiple, so the number of size-1 orbits must be divisible by \(p\). The elements in these orbits are precisely the left cosets invariant under left multiplication by \(H\), so by (a) the conclusion follows.
No comments:
Post a Comment