Prove that every finite group of order at least \(3\) has a nontrivial automorphism.
Solution
Let \(G\) be a finite group with \(\abs{G}\geq3\).
1. If \(G\) is nonabelian, then \(G\) has a nontrivial automorphism.
Since \(G\) is nonabelian, its center is a proper subgroup: \[ Z(G)\neq G. \] Choose \(g\in G\setminus Z(G)\). Conjugation by \(g\), \[ c_g:G\longrightarrow G, \qquad c_g(x)=gxg^{-1}, \] is an automorphism. Because \(g\) is not central, there is some \(x\in G\) with \[ gxg^{-1}\neq x. \] Thus \(c_g\) is not the identity automorphism.
2. Suppose \(G\) is abelian. Then inversion\[ \iota:G\longrightarrow G, \qquad \iota(x)=x^{-1}, \]is an automorphism.
For \(x,y\in G\), commutativity gives \[ \iota(xy) =(xy)^{-1} =x^{-1}y^{-1} =\iota(x)\iota(y). \] Also \(\iota^2=\operatorname{id}_G\), so \(\iota\) is bijective.
3. If the inversion automorphism in step 2 is nontrivial, then\(G\) has a nontrivial automorphism.
This is immediate from step 2.
4. If the inversion automorphism is the identity, then every elementof \(G\) has order at most \(2\).
If \(\iota=\operatorname{id}_G\), then \[ x=x^{-1} \] for every \(x\in G\). Hence \[ x^2=e \] for every \(x\in G\).
5. Under the hypotheses of step 4, the group \(G\) is a vectorspace over \(\FF_2\) of dimension at least \(2\).
Because \(G\) is abelian and every element has order dividing \(2\), its group law makes it an elementary abelian \(2\)-group, equivalently an \(\FF_2\)-vector space. Since \(G\) is finite, \[ \abs{G}=2^{\dim_{\FF_2}G}. \] The hypothesis \(\abs{G}\geq3\) rules out dimensions \(0\) and \(1\). Therefore \[ \dim_{\FF_2}G\geq2. \]
6. Under the hypotheses of step 5, \(G\) has a nontrivialautomorphism.
Choose a basis \[ e_1,e_2,e_3,\ldots,e_r \] with \(r\geq2\). The linear map that interchanges \(e_1\) and \(e_2\) and fixes every other basis vector is an invertible \(\FF_2\)-linear map. Hence it is a group automorphism of \(G\), and it is nontrivial because it sends \(e_1\) to \(e_2\neq e_1\).
7. Every finite group of order at least \(3\) has a nontrivialautomorphism.
1. If \(G\) is abelian, eitherinversion is nontrivial, in which case step 3 applies, or inversionis the identity, in which case step 6 applies.
8. Q.E.D.
Step 7 is the required conclusion.