give answer in 2 step with explanation at the end of each step and final answer at the end:Show that:
a) Ifg:G; — G, andy: G, — G3 are group homomorphisms, then o ¢: G; — Gy is
a group homomorphism.

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:Show that:
a) Ifg:G; — G, andy: G, — G3 are group homomorphisms, then o ¢: G; — Gy is
a group homomorphism.
b) For any group G, the set:
Aut(G) :={¢:G — G | ¢ is an isomorphism of groups}
Itis a group under composition known as the automorphism group of G.
c) Calculate the groups Aut(Z,) and Aut(Z,).

VIPSolutions · Accepted Answer

Let's solve each part step by step, with explanations after each step.

---

### (a) **Composition of Homomorphisms is a Homomorphism**

**Step 1:**  
Let $\varphi: G_1 \to G_2$ and $\psi: G_2 \to G_3$ be group homomorphisms. Consider the composition $\psi \circ \varphi: G_1 \to G_3$ defined by $(\psi \circ \varphi)(g) = \psi(\varphi(g))$ for $g \in G_1$.

**Explanation:**  
We need to show that $\psi \circ \varphi$ preserves the group operation, i.e.,  
$$
(\psi \circ \varphi)(g_1g_2) = (\psi \circ \varphi)(g_1) \, (\psi \circ \varphi)(g_2) \quad \forall g_1, g_2 \in G_1.
$$

**Step 2:**  
Calculate:
$$
(\psi \circ \varphi)(g_1g_2) = \psi(\varphi(g_1g_2))
$$
Since $\varphi$ is a homomorphism:
$$
= \psi(\varphi(g_1)\varphi(g_2))
$$
Since $\psi$ is a homomorphism:
$$
= \psi(\varphi(g_1)) \psi(\varphi(g_2)) = (\psi \circ \varphi)(g_1) (\psi \circ \varphi)(g_2)
$$

**Explanation:**  
We've shown that the composition preserves the group operation, so $\psi \circ \varphi$ is a homomorphism.

---

### (b) **Automorphism Group is a Group**

**Step 1:**  
Define $\text{Aut}(G) = \{ \phi: G \to G \mid \phi \text{ is an isomorphism} \}$ and consider composition as the operation.

**Explanation:**  
We need to show that:  
1. The composition of two automorphisms is an automorphism.  
2. The identity map is an automorphism.  
3. The inverse of an automorphism is an automorphism.

**Step 2:**  
- The composition of two isomorphisms is an isomorphism (proved in (a)).
- The identity map $id_G: G \to G$ is an isomorphism.
- The inverse of an isomorphism is an isomorphism.

**Explanation:**  
Thus, $\text{Aut}(G)$ with composition forms a group (it is closed, associative, has identity and inverses).

---

### (c) **Calculate $\text{Aut}(\mathbb{Z}_2)$ and $\text{Aut}(\mathbb{Z}_3)$**

**Step 1:**  
Recall:  
- $\mathbb{Z}_2 = \{0, 1\}$ under addition mod 2.  
- $\mathbb{Z}_3 = \{0, 1, 2\}$ under addition mod 3.  
Automorphisms of a finite cyclic group $\mathbb{Z}_n$ correspond to isomorphisms, i.e., generators sent to generators.

**Explanation:**  
For $\mathbb{Z}_n$, automorphisms are determined by where $1$ is sent, which must be a generator (i.e., an element coprime to $n$).

**Step 2:**  
- In $\mathbb{Z}_2$, the only generator is $1$. So the only automorphism is the identity: $\text{Aut}(\mathbb{Z}_2) = \{id\}$ (trivial group).
- In $\mathbb{Z}_3$, the generators are $1$ and $2$ (since both are coprime to $3$). Map $1 \mapsto 1$ (identity) or $1 \mapsto 2$ (which gives the automorphism $x \mapsto 2x \mod 3$). So $\text{Aut}(\mathbb{Z}_3) \cong \mathbb{Z}_2$.

**Explanation:**  
- $\text{Aut}(\mathbb{Z}_2)$ is the trivial group.
- $\text{Aut}(\mathbb{Z}_3)$ is the cyclic group of order $2$, i.e., $\mathbb{Z}_2$.

---

### **Final Answers**

**a)** The composition of group homomorphisms is a group homomorphism.  
**b)** The set of automorphisms of a group $G$, with composition, forms a group called the automorphism group $\text{Aut}(G)$.  
**c)** $\text{Aut}(\mathbb{Z}_2)$ is the trivial group, and $\text{Aut}(\mathbb{Z}_3) \cong \mathbb{Z}_2$.

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