Show that \(p\) groups have nontrivial centers.
Show that groups of order \(p^2\) are abelian.
Solution
\(p\dash\)groups have nontrivial centers:
Abusing notation by identifying sets with their cardinalities, the class equation says \(G = Z(G) + \sum_{g}' [G: Z(g)]\) where the terms in the sum are all bigger than 1.
Reducing mod \(p\) yields \(0 = Z(G) + 0\), since \(p\) must divide \([G:Z(g)]\) when \([G:Z(g)] > 1\) because \(G = [G:Z(g)]Z(g)\) and \(p\) divides the LHS.
So \(p\) divides \(Z(G)\), making \(Z(G)\) nontrivial.
\(p^2\) groups are abelian:
\(Z(G) = 1,p,p^2\), and by the first part \(Z(G)\neq 1\). If \(Z(G) = p^2\) then \(G = Z(G)\) is abelian, so assume \(Z(G) = p\).
Then \(G/Z(G) = p\) and groups of order \(p\) are cyclic. If \(G/Z(G)\) is cyclic then \(G\) is abelian, so \(G\) is abelian.