Show there are exactly two groups of order 21 up to isomorphism.
Solution
1. Let \(G\) be a group of order\[ 21=3\cdot7. \]Then \(G\) has a unique Sylow \(7\)-subgroup \(N\).
If \(n_7\) denotes the number of Sylow \(7\)-subgroups, the Sylow theorems give \[ n_7\equiv1\pmod7, \qquad n_7\mid3. \] The only possibility is \(n_7=1\). Hence \(N\trianglelefteq G\), and since \(|N|=7\), we have \[ N\cong C_7. \]
2. Let \(H\) be a Sylow \(3\)-subgroup. Then\[ H\cong C_3, \qquad N\cap H=1, \qquad G=NH. \]
Since \(|H|=3\), the group \(H\) is cyclic. The intersection \(N\cap H\) has order dividing both \(7\) and \(3\), hence is trivial. Therefore \[ |NH|=\frac{|N||H|}{|N\cap H|}=21=|G|, \] so \(NH=G\).
3. Thus every group of order \(21\) is a semidirect product\[ G\cong C_7\rtimes_\varphi C_3 \]for some homomorphism\[ \varphi:C_3\longrightarrow\operatorname{Aut}(C_7). \]
By 1, \(N\) is normal, and by 2, \(G=NH\) with \(N\cap H=1\). Conjugation by \(H\) on \(N\) gives the homomorphism \(\varphi\), and these data identify \(G\) with the corresponding internal semidirect product.
4. If \(\varphi\) is trivial, then\[ G\cong C_7\times C_3\cong C_{21}. \]
A trivial conjugation action means every element of \(H\) commutes with every element of \(N\), so the semidirect product is direct. Since \(3\) and \(7\) are coprime, the direct product of cyclic groups \(C_7\times C_3\) is cyclic of order \(21\).
5. Up to isomorphism, there is exactly one nontrivial semidirect product \(C_7\rtimes C_3\).
We have \[ \operatorname{Aut}(C_7)\cong(\mathbb Z/7\mathbb Z)^\times\cong C_6. \] A nontrivial homomorphism \(C_3\to C_6\) must be injective, and its image must be a subgroup of order \(3\). A cyclic group has a unique subgroup of each order dividing its order, so \(C_6\) has exactly one subgroup of order \(3\).
There are two injections \(C_3\to C_6\) with that image, according to which of the two generators of the image is assigned to a chosen generator of \(C_3\). They differ by the automorphism of \(C_3\) sending a generator to its inverse, and precomposing a semidirect-product action by an automorphism of the acting group yields an isomorphic semidirect product. Hence there is exactly one isomorphism class arising from a nontrivial action.
Explicitly, if \[ C_7=\langle a\rangle, \qquad C_3=\langle b\rangle, \] then multiplication by \(2\) modulo \(7\) has order \(3\), because \[ 2^3\equiv1\pmod7, \qquad 2\not\equiv1\pmod7. \] Thus the unique nonabelian group may be presented as \[ \langle a,b\mid a^7=b^3=1,\;bab^{-1}=a^2\rangle. \]
6. The two groups in 4 and 5 are not isomorphic.
The group in 4 is abelian. In the group from 5, \[ bab^{-1}=a^2\ne a, \] so \(a\) and \(b\) do not commute; hence it is nonabelian.
7. Therefore there are exactly two groups of order \(21\) up to isomorphism:\[ \boxed{C_{21}} \qquad\text{and}\qquad \boxed{C_7\rtimes C_3\text{ with nontrivial action}}. \]