Groups of order 57

Since \[ 57=3\cdot 19, \] let \(G\) be a group of order \(57\). By Sylow’s theorem, the number \(n_{19}\) of Sylow \(19\)-subgroups satisfies \[ n_{19}\mid3,\qquad n_{19}\equiv1\pmod{19}. \] Hence \(n_{19}=1\). Thus the Sylow \(19\)-subgroup \(N\cong C_{19}\) is normal.

Let \(P\) be a Sylow \(3\)-subgroup. Then \(P\cong C_3\), \(N\cap P=1\), and \[ |NP|=\frac{|N||P|}{|N\cap P|}=57, \] so \(G=NP\). Therefore \[ G\cong C_{19}\rtimes_\varphi C_3 \] for a homomorphism \[ \varphi:C_3\longrightarrow \operatorname{Aut}(C_{19}). \] Now \[ \operatorname{Aut}(C_{19})\cong (\mathbb Z/19\mathbb Z)^\times\cong C_{18}. \] Since \(C_{18}\) has a unique subgroup of order \(3\), there are, up to automorphisms of \(C_3\) and \(C_{19}\), exactly two possible actions:

1. the trivial action, giving \[ C_{19}\times C_3\cong C_{57}; \]

2. the unique nontrivial action, whose image is the unique subgroup of order \(3\) in \(C_{18}\), giving one nonabelian semidirect product.

For example, since \(7^3\equiv1\pmod{19}\) and \(7\not\equiv1\pmod{19}\), the nonabelian group can be presented as \[ \langle a,b\mid a^{19}=b^3=1,\;bab^{-1}=a^7\rangle. \] The two groups are not isomorphic because one is abelian and the other is not.

Hence, up to isomorphism, there are exactly two groups of order \(57\): \[ \boxed{C_{57}} \qquad\text{and}\qquad \boxed{C_{19}\rtimes C_3}. \]