Classify groups of order \(21\) up to isomorphism.
Solution
Let \(|G|=21=3\cdot7\).
1. The Sylow \(7\)-subgroup is normal.
If \(n_7\) is the number of Sylow \(7\)-subgroups, then \[ n_7\equiv1\pmod7,\qquad n_7\mid3. \] Hence \(n_7=1\). Write this subgroup as \(P\cong C_7\).
2. Every such group is a semidirect product \(C_7\rtimes C_3\).
Let \(Q\) be a Sylow \(3\)-subgroup. Then \(Q\cong C_3\), \(P\cap Q=1\), and \(|PQ|=21\), so \[ G\cong P\rtimes Q. \] The action is a homomorphism \[ C_3\longrightarrow \operatorname{Aut}(C_7)\cong C_6. \]
3. There are exactly two isomorphism types.
The action is either trivial or has image the unique subgroup of order \(3\) in \(C_6\).
For the trivial action, \[ G\cong C_7\times C_3\cong C_{21}. \] For a nontrivial action, choose generators \(a\) of \(C_7\) and \(b\) of \(C_3\). Up to replacing \(b\) by \(b^{-1}\), the action is \[ bab^{-1}=a^2, \] since \(2\) has order \(3\) modulo \(7\). Thus the unique nonabelian isomorphism type has presentation \[ \langle a,b\mid a^7=b^3=1,\ bab^{-1}=a^2\rangle. \] The two possible nontrivial embeddings \(C_3\hookrightarrow C_6\) differ by an automorphism of \(C_3\), so they give isomorphic semidirect products.
Therefore exactly two groups of order \(21\) exist up to isomorphism: \(C_{21}\) and the nonabelian group \(C_7\rtimes C_3\).