- Show that a Sylow \(p_1\)-subgroup and a Sylow \(p_2\)-subgroup intersect trivially when \(p_1\ne p_2\).
- Show that every nontrivial normal subgroup of a finite \(p\)-group intersects the center nontrivially.
Solution
1. Let \(P_1\) and \(P_2\) be \(p_1\)- and \(p_2\)-subgroups with \(p_1\ne p_2\).
The order of \(P_1\cap P_2\) divides both \(|P_1|\), a power of \(p_1\), and \(|P_2|\), a power of \(p_2\). Hence \[ |P_1\cap P_2|=1, \] so \(P_1\cap P_2=\{e\}\).
2. Let \(G\) be a finite \(p\)-group and let \(1\ne N\trianglelefteq G\).
Let \(G\) act on \(N\) by conjugation. Because \(N\) is normal, this action is well-defined. Its fixed points are exactly \[ N\cap Z(G). \] Every non-fixed orbit has size \[ [G:C_G(x)], \] a positive power of \(p\), hence is divisible by \(p\). Therefore the class equation for this action gives \[ |N|\equiv |N\cap Z(G)|\pmod p. \] Since \(N\) is a nontrivial subgroup of a \(p\)-group, \(p\mid |N|\). Hence \[ p\mid |N\cap Z(G)|. \] Thus \(N\cap Z(G)\) contains more than the identity.