Let \(\phi: \mathbb{Z}_m \to \operatorname{Aut}(\mathbb{Z}_n)\) be a homomorphism, for some integers \(m, n \geq 2\). Let \(G\) be the semidirect product \(\mathbb{Z}_m \rtimes_\phi \mathbb{Z}_n\). Find a presentation for \(G\) by generators and relations and prove carefully that your presented group is isomorphic to \(G\).
Solution
1. Setup of the semidirect product and candidate presentation:
1.1. Let \(G = \mathbb{Z}_n \rtimes_\phi \mathbb{Z}_m\), where the action \(\phi: \mathbb{Z}_m \to \operatorname{Aut}(\mathbb{Z}_n) \cong (\mathbb{Z}/n\mathbb{Z})^\times\) is determined by an integer \(k\) such that \(\phi(1)(a) = a^k\), with \(\gcd(k, n) = 1\) and \(k^m \equiv 1 \pmod n\).
classification of automorphisms of cyclic groups.
1.2. We claim that \(G\) has the presentation:\[ \Gamma = \langle x, y \mid x^n = 1, \, y^m = 1, \, y x y^{-1} = x^k \rangle. \]
candidate presentation for semidirect products of cyclic groups.
2. Existence of a surjective homomorphism \(\Phi: \Gamma \to G\):
2.1. In \(G = \mathbb{Z}_n \rtimes_\phi \mathbb{Z}_m\), define the elements \(\bar{x} = (1, 0)\) and \(\bar{y} = (0, 1)\).Compute the group relations in \(G\):\[ \bar{x}^n = (n \cdot 1, 0) = (0, 0) = e_G, \]\[ \bar{y}^m = (0, m \cdot 1) = (0, 0) = e_G, \]\[ \bar{y}\bar{x}\bar{y}^{-1} = (0, 1)(1, 0)(0, -1) = (\phi(1)(1), 1)(0, -1) = (k, 1)(0, -1) = (k, 0) = \bar{x}^k. \]
definition of the semidirect product multiplication \((a_1, b_1)(a_2, b_2) = (a_1 + \phi(b_1)(a_2), b_1 + b_2)\).
2.2. By the universal property of group presentations, there exists a unique group homomorphism \(\Phi: \Gamma \to G\) such that:\[ \Phi(x) = \bar{x} = (1, 0) \quad \text{and} \quad \Phi(y) = \bar{y} = (0, 1). \]
universal property of presentations.
2.3. Every element \((u, v) \in G\) satisfies \((u, v) = (u, 0)(0, v) = \bar{x}^u \bar{y}^v = \Phi(x^u y^v)\), so \(\Phi\) is surjective.Thus \(|\Gamma| \ge |G| = mn\).
generation of \(G\) by \(\bar{x}\) and \(\bar{y}\).
3. Upper bound on the order of \(\Gamma\):
3.1. The conjugation relation \(y x y^{-1} = x^k\) can be written as the commutation relation \(y x = x^k y\).By induction, \(y^j x^i = x^{i k^j} y^j\) for all \(i, j \ge 0\).
induction on word length.
3.2. Using \(y x = x^k y\) and the relations \(x^n = 1\) and \(y^m = 1\), every element of \(\Gamma\) can be written in the normal form:\[ x^i y^j \quad \text{for } 0 \le i < n \text{ and } 0 \le j < m. \]Thus \(\Gamma = \{x^i y^j \mid 0 \le i < n, \, 0 \le j < m\}\).
moving all powers of \(x\) to the left of powers of \(y\).
3.3. Therefore \(|\Gamma| \le mn\).
at most \(mn\) distinct normal form words.
4. Bijectivity and Conclusion:Since \(|\Gamma| \le mn = |G|\) and \(\Phi: \Gamma \to G\) is surjective, \(\Phi\) is an isomorphism.Thus \(G \cong \langle x, y \mid x^n = 1, \, y^m = 1, \, y x y^{-1} = x^k \rangle\). Q.E.D.
2 and 3.