Let \(p\) be a prime number.
Show that the order of \(\widehat{1+p}\) in \((\mathbb{Z}/p^2\mathbb{Z})^\times\) is equal to \(p\).
Use (a) above to construct a non-abelian group of order \(p^3\).
Describe the non-abelian group you have constructed in (b) above via generators and relations.
Note. As usual, \((\mathbb{Z}/p^2\mathbb{Z})^\times\) denotes the multiplicative group consisting of all the congruence classes \(\hat{x}\in\mathbb{Z}/p^2\mathbb{Z}\) such that \(\gcd(x,p)=1\).
Solution
1. The class of \(1+p\) has order exactly \(p\) in \((\mathbb Z/p^2\mathbb Z)^\times\).
For every integer \(k\ge 0\), the binomial theorem gives \[ (1+p)^k =1+kp+\sum_{j=2}^k \binom{k}{j}p^j \equiv 1+kp\pmod{p^2}. \] In particular, \[ (1+p)^p\equiv 1+p^2\equiv1\pmod{p^2}, \] so the order divides \(p\).
If \(1\le k<p\), then \(p^2\nmid kp\), hence \[ (1+p)^k\equiv1+kp\not\equiv1\pmod{p^2}. \] Thus no positive exponent smaller than \(p\) gives the identity, and the order is exactly \(p\).
2. Let\[ A=\mathbb Z/p^2\mathbb Z \]with its additive group structure, and define\[ \varphi:A\longrightarrow A, \qquad \varphi(x)=(1+p)x. \]Then \(\varphi\) is an automorphism of order \(p\).
Since \(\gcd(1+p,p)=1\), multiplication by \(1+p\) is invertible modulo \(p^2\), so \(\varphi\in\operatorname{Aut}(A)\). Moreover \[ \varphi^k(x)=(1+p)^k x. \] Thus the order of \(\varphi\) equals the order of the unit \(\widehat{1+p}\), which is \(p\) by 1.
3. Let \(B=\langle b\rangle\cong C_p\) and let \(B\) act on \(A\) by sending the generator \(b\) to \(\varphi\). Then\[ G=A\rtimes_\varphi B \]has order \(p^3\).
2.As a set, the semidirect product is \(A\times B\), so\[ |G|=|A||B|=p^2p=p^3. \]
4. The group \(G\) is nonabelian.
Let \(a\) denote the class of \(1\) in the cyclic additive group \(A\), viewed as an element of \(G\). By the definition of the semidirect product, \[ bab^{-1}=a^{\,1+p}. \] Since \(1+p\not\equiv1\pmod{p^2}\), one has \(a^{1+p}\ne a\). Hence \(ba\ne ab\), so \(G\) is nonabelian.
5. The group constructed above has the presentation\[ G\cong \left\langle a,b\ \middle|\ a^{p^2}=1,\ b^p=1,\ bab^{-1}=a^{1+p}\right\rangle. \]
3.Every word can therefore be rewritten, using\[ ba=a^{1+p}b, \]in the form \(a^ib^j\) with \(0\le i<p^2\) and \(0\le j<p\).Thus the presented group has at most \(p^3\) elements, while the semidirect product \(G\) of 3 satisfies the relations and has exactly \(p^3\) elements.The induced surjection from the presented group onto \(G\) is therefore an isomorphism.