Theorem
If \(G\) is a non-abelian group, then show that \(\operatorname{Aut}(G)\) cannot be cyclic.
Proof
Step 1
Let us consider the inner automorphism group
Recall that the inner automorphism group \(\operatorname{Inn}(G)\) is defined as:
\[
\operatorname{Inn}(G) = \{\theta_g \mid \theta_g(x) = gxg^{-1}, \, g \in G\}.
\]
Moreover, we have an isomorphism:
\[
\operatorname{Inn}(G) \cong G/Z(G), \tag{1}
\]
where \(Z(G)\) is the centre of \(G\).
Step 2
Subgroup of a Cyclic Group is Cyclic
If \(\operatorname{Aut}(G)\) were cyclic, then every subgroup of \(\operatorname{Aut}(G)\) would also be cyclic. In particular, \(\operatorname{Inn}(G)\) would be cyclic.
Step 3
Cyclic \(G/Z(G)\) Implies \(G\) is Abelian
It is a standard result that if \(G/Z(G)\) is cyclic, then \(G\) must be abelian. In detail, suppose:
\begin{align}
G/Z(G) = \langle gZ(G) \rangle \nonumber
\end{align}
Then for any \(x \in G\), we can write \(xZ(G) = (gZ(G))^k\) for some integer \(k\), which implies that every element of \(G\) is in the coset of a power of \(g\). This forces the commutator subgroup of \(G\) to be trivial, and hence \(G\) is abelian.
Step 4
Deriving a Contradiction
Now assume, for contradiction, that \(\operatorname{Aut}(G)\) is cyclic. Then \(\operatorname{Inn}(G)\), being a subgroup of \(\operatorname{Aut}(G)\), is cyclic. By the isomorphism in (1), \(G/Z(G)\) is cyclic, which implies that \(G\) is abelian. This contradicts the hypothesis that \(G\) is non-abelian.
