- Show that the order of \(1+p\) in \(\unitsof{\qty{\ZZ/p^2\ZZ}}\) is equal to \(p\). Use this to construct a non-abelian group of order \(p^3\).
Solution
1. The class of \(1+p\) in \((\ZZ/p^2\ZZ)^\times\) has order \(p\).
For \(1\le j<p\), the binomial theorem gives \[ (1+p)^j=1+jp+\binom j2p^2+\cdots\equiv1+jp\pmod{p^2}. \] Since \(p\nmid j\), the residue \(jp\) is not divisible by \(p^2\), so \((1+p)^j\not\equiv1\pmod{p^2}\). On the other hand, \[ (1+p)^p\equiv1+p^2\equiv1\pmod{p^2}; \] for \(p=2\) this is the same computation, \((1+2)^2\equiv1\pmod4\). Hence the order is exactly \(p\).
2. Multiplication by \(1+p\) defines an automorphism \(\varphi\) of the cyclic group \(C_{p^2}\) of order \(p\).
Identify \(C_{p^2}\) with the additive group \(\ZZ/p^2\ZZ\). Multiplication by the unit \(1+p\) is an automorphism. Its order as an automorphism is the multiplicative order of \(1+p\) in \((\ZZ/p^2\ZZ)^\times\), which is \(p\) by 1.
3. Therefore the semidirect product\[ G=C_{p^2}\rtimes_{\varphi} C_p \]has order \(p^3\) and is nonabelian.
The order is \[ |G|=p^2p=p^3. \] Let \(a\) generate \(C_{p^2}\) and \(b\) generate \(C_p\). The defining action gives \[ bab^{-1}=a^{1+p}. \] Since \(a^p\ne1\) in \(C_{p^2}\), one has \(a^{1+p}\ne a\). Thus \(a\) and \(b\) do not commute, so \(G\) is nonabelian.