- #1
joeblow
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For any prime p how do I show that there is a nonabelian group of order p^3?
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?
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?