- #1
Lee33
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Homework Statement
If ##|G| = p^2## where ##p## is a prime, then ##G## is abelian. 2. The attempt at a solution
The book my proof gives is confusing at the very last part.
Suppose ##|Z(G)| = p##. Let ##a\in G##, ##a\notin Z(G)##. Thus ##|N(a)| > p##, yet by Lagrange ##|N(a)| \ | \ |G| = p^2##.
Why is ##|N(a)| \ | \ |G| = p^2##? Since ##|G| = p^2## and ##|N(a)|> p ## how does that make ##|N(a)| \ | \ |G| = p^2##?