The inner automorphism group, denoted as $\text{Inn}(G)$, of a group $G$, consists of automorphisms of the form $f_g(x) = g x g^{-1}$, where $g \in G$. Now, let's analyze the given groups.

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1. $\text{Inn}(\mathbb{Z}_2)$:

• $\mathbb{Z}_2 = \{0, 1\}$ under addition modulo 2.
• The group is Abelian, so $g x g^{-1} = x$ for all $g, x \in G$.
• Thus, every element acts trivially under conjugation.
• Therefore, $\text{Inn}(\mathbb{Z}_2) = \{ \text{id} \}$, the trivial group.

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2. $\text{Inn}(\mathbb{Z}_3)$:

• $\mathbb{Z}_3 = \{0, 1, 2\}$ under addition modulo 3.
• Like $\mathbb{Z}_2$, $\mathbb{Z}_3$ is Abelian, so conjugation acts trivially: $g x g^{-1} = x$ for all $g, x \in G$.
• Thus, $\text{Inn}(\mathbb{Z}_3) = \{ \text{id} \}$, the trivial group.

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3. $\text{Inn}(\mathbb{Z})$:

• $\mathbb{Z}$, the group of integers under addition, is also Abelian.
• For any $g, x \in \mathbb{Z}$, conjugation is trivial: $g + x - g = x$.
• Therefore, $\text{Inn}(\mathbb{Z}) = \{ \text{id} \}$, the trivial group.

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4. $\text{Inn}(S_3)$:

• $S_3$, the symmetric group of degree 3, has $6$ elements and is non-Abelian.
• In $S_3$, conjugation is generally non-trivial, so $\text{Inn}(S_3)$ is non-trivial.
• The inner automorphism group of $S_3$ is isomorphic to $\text{S}_3 / Z(S_3)$, where $Z(S_3)$ (the center of $S_3$) is trivial ($Z(S_3) = \{ \text{id} \}$).
• Hence, $\text{Inn}(S_3) \cong S_3$.

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Final Results:

• $\text{Inn}(\mathbb{Z}_2) = \{ \text{id} \}$ (trivial group).
• $\text{Inn}(\mathbb{Z}_3) = \{ \text{id} \}$ (trivial group).
• $\text{Inn}(\mathbb{Z}) = \{ \text{id} \}$ (trivial group).
• $\text{Inn}(S_3) \cong S_3$ (non-trivial, isomorphic to $S_3$).

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5 Related Questions:

1. What is the difference between inner and outer automorphisms?
2. Why is $\text{Inn}(G)$ trivial for Abelian groups?
3. How can $\text{Inn}(S_n)$ be computed for $n > 3$?
4. What is the role of the center $Z(G)$ in determining $\text{Inn}(G)$?
5. Can a non-trivial group have a trivial $\text{Inn}(G)$?

1 Tip:

For a group $G$, if it is Abelian, $\text{Inn}(G)$ is always trivial because all elements commute, rendering conjugation ineffective.