Let p be a fixed prime. Show that there exist a noncommutative ring (with identity) of order $p^3$.
RemarkI was able to $p = 2$: $U_n(\mathbb{Z}_2)$ - the set of $n \times n$ matrices with entries from $\mathbb{Z}_2$. But where $p>2$ is not worth.
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1Hint for (a): Show if the $G/Z(G)$ is cyclic, then $G$ is abelian. – Viktor Vaughn Mar 25 '15 at 13:38
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1See Ring of order $p^2$ is commutative. and Noncommutative ring of order $np^2$ – J.-E. Pin Mar 25 '15 at 13:40
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Take the ring $$R=\{A\in M_{2\times2}(\mathbf{F}_p)\mid A_{21}=0\}$$ consisting of matrices with $(2,1)$ entry zero. As sum and product of upper tringular matrices is again upper triangular, and as $I\in R$ this is a subring. As there are no other conditions on the entries it indeed has $p^3$ elements. Definitely we can find two such matrices not commuting with each other.
P Vanchinathan
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