Homomorphisms
Table of Contents
1. Group Homomorphisms
A \(\textbf{\textit{homomorphisms}}\) between groups \((G,\cdot)\) and \((H, \circ)\) is a map \(\phi : G \rightarrow H\) such that \[\phi(g_1 \cdot g_2) = \phi(g_1) \circ \phi(g_2)\] for \(g_1, g_2 \in G\). The range of \(\phi\) in \(H\) is called the \(\textbf{\textit{homomorphic image}}\) of \(\phi\).
Let \(\phi: G_1 \rightarrow G_2\) be a homomorphism of groups. Then
- If e is the identity of \(G_1\), then \(\phi(e)\) is the identity of \(G_2\);
- For any element \(g \in G_1\), \(\phi(g^{-1}) = [\phi(g)]^{-1}\);
- If \(H_1\) is a subgroup of \(G_1\), then \(\phi(H_1)\) is a subgroup of \(G_2\);
- If \(H_2\) is a subgroup of \(G_2\), then \(\phi^{-1}(H_2) = \left\{ g \in G_1 : \phi(g) \in H_2 \right\}\) is a subgroup of \(G_1\). Furthermore, if \(H_2\) is normal in \(G_2\), then \(\phi^{-1}(H_2)\) is normal in \(G_1\).
Let \(\phi : G \rightarrow H\) be a group homomorphism and suppose that \(e\) is the identity of \(H\). \(\phi^{-1}(\left\{ e \right\})\) is a subgroup of \(G\). This subgroup is called the \(\textbf{\textit{kernel}}\) of \(\phi\) and will be denoted by \(\ker \phi\).
Let \(\phi : G \rightarrow H\) be a group homomorphism. Then the kernel of \(\phi\) is a normal subgroup of \(G\).
2. The Isomorphism Theorems
Let \(H\) be a normal subgroup of \(G\). Define the \(\textbf{\textit{natural}}\) or \(\textbf{\textit{canonical homomorphism}}\) \[\phi: G \rightarrow G/H\] by \[\phi(g) = gH.\]
- First Isomorphism Theorem
If \(\psi : G \rightarrow H\) is a group homomorphism with \(K = \ker \psi\), then \(K\) is normal in \(G\). Let \(\phi : G \rightarrow G/K\) be the canonical homomorphism. Then there exists a unique isomorphism \(\eta : G/K \rightarrow \psi(G)\) such that \(\psi = \eta\phi\).
- Second Isomorphism Theorem
Let \(H\) be a subgroup of a group \(G\)(not necessarily normal in \(G\)) and \(N\) a normal subgroup of \(G\). Then \(HN\) is a subgroup of \(G\), \(H \cap N \) is a normal subgroup of \(H\), and \[H/(H \cap N) \cong HN/N\]
- Correspondence Theorem
Let \(N\) be a normal subgroup of a group \(G\). Then \(H \mapsto H/N\) is a one-to-one correspondence between the set of subgroups \(H\) of \(G\) containing \(N\) and the set of subgroups of \(G/N\).
- Third Isomorphism Theorem
Let \(G\) be a group and \(N\) and \(H\) be normal subgroups of \(G\) with \(N \subset H\). Then \[G/H \cong \frac{G/N}{H/N}.\]
3. Exercises
- Let \(A\) be an \(m \times n\) matrix. Show that matrix multiplication, \(x \mapsto Ax\), defines a homomorphism \(\phi: \mathbb{R}^n \rightarrow \mathbb{R}^m\).
\(Proof.\) \(\phi(x+y) = A(x+y) = Ax + Ay = \phi (x) + \phi (y)\), so it defines a homomorphism.
\(\blacksquare\) - Describe all of the homomorphisms from \(\mathbb{Z}_{24}\) to \(\mathbb{Z}_{18}\).
\(Solution.\) Suppose \(\phi(1) = k \in \mathbb{Z}_{18}\),\(\phi(24) = 24\phi(1) = 24k \equiv 0 (\text{mod } 18)\) which is equivalent to \(3 \big | k\). So \(k = \{0, 3, 6, 9, 12, 15\}\).
\(\blacksquare\) - Describe all of the homomorphism from \(\mathbb{Z}\) to \(\mathbb{Z}_{12}\).
\(Solution.\) A homomorphism \(\phi\): \(\mathbb{Z} \rightarrow \mathbb{Z}_{12}\) is determinated by \(\phi(1) = k \in \mathbb{Z}_{12}\). Then \(\phi(n) = nk (\text{mod } 12)\). Hence there are exactly 12 Homomorphisms, one for each \(k \in \left\{ 1, 2, 3, \cdots, 11 \right\}\).
\(\blacksquare\) - If a group \(G\) has exactly one subgroup \(H\) of order \(k\), prove that \(H\) is normal in \(G\).
\(Proof.\) Define a homomorphism \(\phi\): \(G \rightarrow G/H\) where \(\phi(g) = gH\), then \(\ker \phi = H\). Thus, \(H\) is a normal subgroup of \(G\).
\(\blacksquare\) - Let \(G_1\) and \(G_2\) be groups, and let \(H_1\) and \(H_2\) be normal subgroups of \(G_1\) and \(G_2\) respectively. Let \(\phi\): \(G_1 \rightarrow G_2\) be a homomorphism. Show that \(\phi\) induces a homomorphism \(\overline{\phi}: (G_1/H_1) \rightarrow (G_2/H_2)\) if \(\phi(H_1) \subset H_2\).
\(Proof.\) Frist, we need to show that \(\overline{\phi}\) is well-defined. If \(g_1H_1 = g_2H_1\), then \(g_1=g_2h_1\) where \(h_1 \in H_1\). Because \(\phi(H_1) \subset H_2\), we have \[\overline{\phi}(g_1H_1) = \phi(g_1)H_2 = \phi(g_2h_1)H_2 = \phi(g_2)H_2 = \overline{\phi}(g_2H_1).\] Hence, \(\overline{\phi}\) is well-defined. It's obvious that \(\overline{\phi}\) is a homomorphism.
\(\blacksquare\) - If \(H\) and \(K\) are normal subgroups of \(G\) and \(H \cap K = \left\{ e \right\}\), prove that \(G\) is isomorphic to a subgroup of \(G/H \times G/K\).
\(Proof.\) Define a map \(\phi\): \(\phi(g) = \left( gH, gK \right)\). \(\phi(g_1g_2) = \left( g_1g_2H, g_1g_2K \right) = \left( g_1H, g_1K \right) \left( g_2H, g_2K \right) = \phi(g_1)\phi(g_2)\)
\(\blacksquare\) - Find all of the automorphisms of \(\mathbb{Z} \rightarrow \mathbb{Z}\). What is \(\text{Aut}(\mathbb{Z})\)?
\(Solution.\) \(1\) is the generator of \(\mathbb{Z}\), suppose \(\phi(1) = k\), then \(\phi(Z) = kZ\). Hence \(\left\{ \phi_k | \phi_k(1) = k, k \in \mathbb{Z}) \right\}\) are all of homomorphisms of \(\mathbb{Z}\). It's obvious that \(Aut(Z) = \left\{ \phi_1, \phi_{-1} \right\}\).
\(\blacksquare\) - Find all of the automorphisms of \(\mathbb{Z}_8\). Prove that \(\text{Aut}(\mathbb{Z}_8) = U(8)\).
\(Solution.\) \(Z_8\) is a cyclic group and \(1\) is one of its generator. The automorphism should keep the property of generators, so \(\text{Aut}(\mathbb{Z}_{8}) = \left\{ \phi_k | \phi_k(1) = k, \text{gcd}(k, 8) = 1 \right\} = \left\{ \phi_1, \phi_3, \phi_5, \phi_7 \right\}\). Define \(\psi\): \(U(8) \mapsto \text{Aut}(\mathbb{Z}_8)\) with \(\psi(k) = \phi_k\). If \(\psi(k) = \psi(m)\), then \(k = \phi_k(1) = \phi_m(1) = m\), thus \(\psi\) is injective. It's obvious that \(\psi\) is surjective. \(\psi(km)(x) = \phi_{km}(x) = (km)x = k(mx) = \phi_k\left( \phi_m(x) \right) = \psi(k)\psi(m)(x)\). Hence, \(\psi\) is an isomorphism.
\(\blacksquare\)