Groups of order 75 (D&F 5.5.8)

Construct a non-abelian group of order 75. Classify all groups of order 75.

Let \(H = Z_5 \times Z_5\) and \(K = Z_3 = \langle x \rangle\). It is known that \(\text{Aut}(H)\) has order \(5(5-1)^2(5+1) = 2^5 \cdot 3 \cdot 5\), so it has a Sylow 3-subgroup \(H_3\ = \langle y \rangle\). Hence we can define a homomorphism \(\varphi : K \rightarrow \text{Aut}(H)\) by \(\varphi(x) = y\), inducing a semidirect product \(G = H \rtimes_{\varphi} K\) of order 75. Note that \(G\) is the unique (nontrivial) semidirect product of \(H\) and \(K\) by 5.5.6 since all Sylow 3-subgroups of \(\text{Aut}(H)\) are conjugate. Furthermore \(G\) is non-abelian, since \(\varphi\) is non-trivial meaning \(K \not\trianglelefteq G\).

Let \(G\) be an order \(75 = 3 \cdot 5^2\) group, and \(G_3\) and \(G_5\) be a Sylow 3- and 5-subgroup. \(n_3\) can either be 1 and 25. If \(n_3 = 25\), then the Sylow 3-subgroups contain \(2 \cdot 25 = 50\) nonidentity elements, forcing any Sylow 5-subgroup to be normal. If \(G_5 \cong Z_{25}\), then \(\text{Aut}(G_5) \cong Z_{20}\), so any homomorphism from \(G_3\) to \(\text{Aut}(G_5)\) must be trivial. Hence we can only construct the abelian group \(Z_{25} \times Z_3 \cong Z_{75}\). If \(G_5 \cong Z_5 \times Z_5\), then we effectively get the unique non-abelian group in the previous paragraph.

Suppose \(n_3 = 1\). Since \(\text{Aut}(G_3) \cong Z_2\), any homomorphism from \(G_5\) to \(\text{Aut}(G_3)\) must be trivial. Thus we only get one new abelian group \(Z_3 \times Z_5 \times Z_5\) (\(Z_3 \times Z_{25}\) was constructed in the previous paragraph).