Group Theory Problem-Solving: If |G| = p^2, how do we prove |Z(G)| ≠ p (p is prime)?

In Group Theory from Abstract Algebra, if p is prime and a group G has order p^2, then the center Z(G) of G cannot have order p. This is a consequence of the G/Z Theorem and Lagrange's Theorem. To prove this, suppose to the contrary that |Z(G)|=p. Then Lagrange's Theorem implies that the order of the factor group G/Z(G) is p as well, since |G/Z(G)| = |G|/|Z(G)| = p^2/p = p. But this implies, also by Lagrange's Theorem, that G/Z(G) is a cyclic group. The G/Z Theorem then allows us to conclude that G is Abelian, which means that Z(G)=G and |Z(G)|=p^2, a contradiction. In the end, it turns out if |G|=p^2, then G is Abelian so Z(G)=G. In fact, G is isomorphic either to the cyclic group ℤ_{p^2} or to the (non-cyclic) direct product ℤp ⊕ ℤp.
#GroupTheory #AbstractAlgebra #LagrangeTheorem
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[@nominativ1941]

Thanks, great explaination!

[@billkinneymath]

Thanks! Glad it was helpful for you!