Question:
The generators of the set of integers $\mathbb{Z}$ under addition is:
The generators of the set of integers $\mathbb{Z}$ under addition is:
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An infinite cyclic group is always isomorphic to $(\mathbb{Z}, +)$ and possesses exactly 2 generators ($g$ and $g^{-1}$). A finite cyclic group $\mathbb{Z}_n$ has $\phi(n)$ generators, where $\phi$ is Euler's totient function.
Updated On: Jul 29, 2026
- Only 1.
- Only $-1$.
- Both 1 and $-1$.
- 1, 2 and $-1$.
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The Correct Option is C
Solution and Explanation
Step 1: Concept
A group $(G, +)$ is called cyclic if there exists an element $g \in G$ such that every element $x \in G$ can be written as an integer multiple of $g$, i.e., $x = n \cdot g$ for some $n \in \mathbb{Z}$.
The element $g$ is called a generator of $G$, and we write $G = \langle g \rangle$.
Step 2: Key Formulas and Approach
For an infinite cyclic group such as $(\mathbb{Z}, +)$, any generator $g$ must generate all positive integers, zero, and negative integers through additive repetition:
\[ \langle g \rangle = \{ n \cdot g \mid n \in \mathbb{Z} \} \] We need to identify all elements $g \in \mathbb{Z}$ for which $\langle g \rangle = \mathbb{Z}$.
Step 3: Step-by-step Explanation
• Let us evaluate $g = 1$:
$\langle 1 \rangle = \{ n \cdot 1 \mid n \in \mathbb{Z} \} = \{ \dots, -2, -1, 0, 1, 2, \dots \} = \mathbb{Z}$.
Thus, $1$ is a generator of $\mathbb{Z}$.
• Let us evaluate $g = -1$:
$\langle -1 \rangle = \{ n \cdot (-1) \mid n \in \mathbb{Z} \} = \{ \dots, 2, 1, 0, -1, -2, \dots \} = \mathbb{Z}$.
Thus, $-1$ is also a generator of $\mathbb{Z}$.
• Let us evaluate $g = 2$:
$\langle 2 \rangle = \{ n \cdot 2 \mid n \in \mathbb{Z} \} = \{ \dots, -4, -2, 0, 2, 4, \dots \} = 2\mathbb{Z} \neq \mathbb{Z}$.
Hence, $2$ is not a generator because odd integers cannot be produced.
• Generally, an infinite cyclic group has exactly two generators: if $g$ is a generator, then $g^{-1}$ is the only other generator. In $(\mathbb{Z}, +)$, the inverse of $1$ is $-1$. Therefore, the generators of $(\mathbb{Z}, +)$ are precisely $1$ and $-1$.
Step 4: Final Answer
The set of integers $\mathbb{Z}$ under addition has exactly two generators, namely $1$ and $-1$. Hence, Option (C) is correct.
A group $(G, +)$ is called cyclic if there exists an element $g \in G$ such that every element $x \in G$ can be written as an integer multiple of $g$, i.e., $x = n \cdot g$ for some $n \in \mathbb{Z}$.
The element $g$ is called a generator of $G$, and we write $G = \langle g \rangle$.
Step 2: Key Formulas and Approach
For an infinite cyclic group such as $(\mathbb{Z}, +)$, any generator $g$ must generate all positive integers, zero, and negative integers through additive repetition:
\[ \langle g \rangle = \{ n \cdot g \mid n \in \mathbb{Z} \} \] We need to identify all elements $g \in \mathbb{Z}$ for which $\langle g \rangle = \mathbb{Z}$.
Step 3: Step-by-step Explanation
• Let us evaluate $g = 1$:
$\langle 1 \rangle = \{ n \cdot 1 \mid n \in \mathbb{Z} \} = \{ \dots, -2, -1, 0, 1, 2, \dots \} = \mathbb{Z}$.
Thus, $1$ is a generator of $\mathbb{Z}$.
• Let us evaluate $g = -1$:
$\langle -1 \rangle = \{ n \cdot (-1) \mid n \in \mathbb{Z} \} = \{ \dots, 2, 1, 0, -1, -2, \dots \} = \mathbb{Z}$.
Thus, $-1$ is also a generator of $\mathbb{Z}$.
• Let us evaluate $g = 2$:
$\langle 2 \rangle = \{ n \cdot 2 \mid n \in \mathbb{Z} \} = \{ \dots, -4, -2, 0, 2, 4, \dots \} = 2\mathbb{Z} \neq \mathbb{Z}$.
Hence, $2$ is not a generator because odd integers cannot be produced.
• Generally, an infinite cyclic group has exactly two generators: if $g$ is a generator, then $g^{-1}$ is the only other generator. In $(\mathbb{Z}, +)$, the inverse of $1$ is $-1$. Therefore, the generators of $(\mathbb{Z}, +)$ are precisely $1$ and $-1$.
Step 4: Final Answer
The set of integers $\mathbb{Z}$ under addition has exactly two generators, namely $1$ and $-1$. Hence, Option (C) is correct.
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