Order of 1+p in (Z/p²Z)* and construction of a non-abelian group of order p³

1.1. By the Binomial Theorem:\[ (1 + p)^p = \sum_{k=0}^p \binom{p}{k} p^k = 1 + \binom{p}{1} p + \sum_{k=2}^p \binom{p}{k} p^k = 1 + p^2 + \sum_{k=2}^p \binom{p}{k} p^k. \]

1.2. For every \(k \ge 2\), the power \(p^k\) is a multiple of \(p^2\), so:\[ (1 + p)^p \equiv 1 + p^2 \equiv 1 \pmod{p^2}. \]

1.3. Since \(p \ge 2\), \(1 + p \not\equiv 1 \pmod{p^2}\), so the order of \(1 + p\) is strictly greater than 1.Because the order of \(1 + p\) divides the prime \(p\), the order must be exactly \(p\):\[ o(1 + p) = p \quad \text{in } (\mathbb{Z}/p^2\mathbb{Z})^\times. \]

2.1. Consider the cyclic group \(N = \mathbb{Z}/p^2\mathbb{Z}\) and \(H = \mathbb{Z}/p\mathbb{Z}\).The automorphism group of \(N\) is \(\operatorname{Aut}(N) \cong (\mathbb{Z}/p^2\mathbb{Z})^\times\).

2.2. By Part (a), the element \(1 + p\) has order \(p\) in \((\mathbb{Z}/p^2\mathbb{Z})^\times\).Therefore, there exists a non-trivial group homomorphism \(\theta: H \to \operatorname{Aut}(N)\) defined on the generator \(1 \in \mathbb{Z}/p\mathbb{Z}\) by:\[ \theta(1)(x) = (1 + p)x \pmod{p^2}. \]

2.3. Define \(G\) as the semidirect product \(G = N \rtimes_\theta H = (\mathbb{Z}/p^2\mathbb{Z}) \rtimes_\theta (\mathbb{Z}/p\mathbb{Z})\).The order of \(G\) is:\[ |G| = |N| \cdot |H| = p^2 \cdot p = p^3. \]Since \(\theta\) is non-trivial (\(1 + p \not\equiv 1 \pmod{p^2}\)), \(G\) is a non-abelian group of order \(p^3\).

3.1. Let \(a = (1, 0) \in G\) and \(b = (0, 1) \in G\).Then \(a\) generates \(N \cong \mathbb{Z}/p^2\mathbb{Z}\) (so \(a^{p^2} = e\)) and \(b\) generates \(H \cong \mathbb{Z}/p\mathbb{Z}\) (so \(b^p = e\)).The semidirect product action gives \(b a b^{-1} = \theta(1)(a) = a^{1 + p}\).

3.2. Thus \(G\) has the presentation:\[ G = \langle a, b \mid a^{p^2} = e, \; b^p = e, \; b a b^{-1} = a^{1 + p} \rangle. \]