Lemma. Let \(f: G \to H\) and \(g: H \to K\) be homomorphisms; assume \(f\) is surjective. If \(x_0 \in G\), and if \(\ker g\) is the least normal subgroup of \(H\) containing \(f(x_0)\), then \(\ker(g \circ f)\) is the least normal subgroup \(N\) of \(G\) containing \(\ker f\) and \(x_0\).
Proof. Show that \(f(N)\) is normal; conclude that \(\ker(g \circ f) = f^{-1}(\ker g) \subset f^{-1}f(N) = N\).
Solution
Goal: Prove that if \(f: G \to H\) is a surjective homomorphism and \(\ker g = \langle\langle f(x_0) \rangle\rangle_H\), then \(\ker(g \circ f)\) is the least normal subgroup \(N = \langle\langle \ker f \cup \{x_0\} \rangle\rangle_G\).
1. Inclusion \(N \subseteq \ker(g \circ f)\):Proof:
1.1. \(\ker(g \circ f) = \{x \in G \mid g(f(x)) = 1_K\} = f^{-1}(\ker g)\).
1.2. For any \(k \in \ker f\), \((g \circ f)(k) = g(f(k)) = g(1_H) = 1_K\), so \(\ker f \subseteq \ker(g \circ f)\).
1.3. For \(x_0\), \((g \circ f)(x_0) = g(f(x_0)) = 1_K\) because \(f(x_0) \in \ker g\), so \(x_0 \in \ker(g \circ f)\).
1.4. Because \(\ker(g \circ f)\) is the kernel of a homomorphism, it is a normal subgroup of \(G\).
1.5. Since \(N\) is defined as the least normal subgroup of \(G\) containing \(\ker f\) and \(x_0\), we have \(N \subseteq \ker(g \circ f)\).
2. Normality of the image \(f(N)\) in \(H\):Proof:
2.1. Let \(h \in H\) and \(y \in f(N)\).
2.2. Since \(f\) is surjective, choose \(g \in G\) such that \(f(g) = h\).
2.3. Choose \(n \in N\) such that \(f(n) = y\).
2.4. Then \(h y h^{-1} = f(g) f(n) f(g)^{-1} = f(g n g^{-1})\).
2.5. Because \(N \trianglelefteq G\), \(g n g^{-1} \in N\), so \(f(g n g^{-1}) \in f(N)\).
2.6. Thus \(f(N)\) is a normal subgroup of \(H\).
3. Inclusion \(\ker g \subseteq f(N)\):Proof:
3.1. Since \(x_0 \in N\), \(f(x_0) \in f(N)\).
3.2. By hypothesis, \(\ker g\) is the least normal subgroup of \(H\) containing \(f(x_0)\).
3.3. Since \(f(N)\) is a normal subgroup of \(H\) containing \(f(x_0)\), it follows that \(\ker g \subseteq f(N)\).
4. Inclusion \(\ker(g \circ f) \subseteq N\):Proof:
4.1. Taking preimages under \(f\) gives:\[\ker(g \circ f) = f^{-1}(\ker g) \subseteq f^{-1}(f(N)).\]
4.2. We claim \(f^{-1}(f(N)) = N\):- We have \(N \subseteq f^{-1}(f(N))\): for any \(n \in N\), \(f(n) \in f(N)\), so \(n \in f^{-1}(f(N))\) by definition of the preimage.- If \(x \in f^{-1}(f(N))\), then \(f(x) = f(n)\) for some \(n \in N\).- Then \(f(x n^{-1}) = f(x) f(n)^{-1} = 1_H\), which means \(x n^{-1} \in \ker f\).- Since \(\ker f \subseteq N\), \(x = (x n^{-1}) n \in N \cdot N = N\).
4.3. Thus \(f^{-1}(f(N)) = N\), which proves \(\ker(g \circ f) \subseteq N\).
5. Conclusion:\(\ker(g \circ f) = N\), so \(\ker(g \circ f)\) is the least normal subgroup of \(G\) containing \(\ker f\) and \(x_0\). Q.E.D.