For any prime p how do I show that there is a nonabelian group of order p^3?(adsbygoogle = window.adsbygoogle || []).push({});

Since we are dealing with a p-group (call it G), its center is nontrivial (i.e., of order p,p^2, or p^3). Obviously, the center cannot have order p^3 (otherwise it's abelian). Also, if its center has order p^2, then [itex]|G/Z(G)|=p[/itex] implying that G is cyclic and thus abelian. Another no-go.

Thus, [itex]|Z(G)|=p[/itex] and [itex]Z(G) \cong \mathbb{Z}_p [/itex].

Now, I want G to be nonabelian, so I want a G containing Z_p such that [itex]G/\mathbb{Z}_p \cong \mathbb{Z}_p \times \mathbb{Z}_p [/itex] (which would be the only noncyclic group of order p^2).

From here, I have no idea how to recover the original G. Also, semidirect products are off-limits for this problem. (I'm going through an old algebra book and the semidirect product is not introduced till after the section from which this problem comes.)

Any ideas?

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# Nonabelian group of order p^3

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