I have a question regarding the classification of groups of "small" order; we'll take groups of order $102=2 \cdot 3 \cdot 17$ as an example.
Let G be a group of order 102, note that $P_{17}\in Syl_{17}(G)$ by Sylow theorems is unique, so $P_{17} \trianglelefteq G$. Now consider the surjective canonical projection map $\pi:G\twoheadrightarrow{}\frac{G}{P_{17}}$, then the subgroup $\pi(P_{3})$ is normal in $\frac{G}{P_{17}}$ (by Sylow theorems), so $P_{3} \trianglelefteq G$.
We can conclude that $G\cong (P_{17}\times P_{3})\rtimes_{\psi} P_{2}$, which is fully determined by the choice of $\psi:P_{2}\to Aut(P_{17}\times P_{3})$ with $Aut(P_{17}\times P_{3})\cong C_{16}\times C_{2}$. We have four cases (I believe, this is were my understanding get shaky); let x be non unit element in $P_2$, y be the unique element in $C_{16}$ with order two, and z be the unique element of $C_2$ with order two:
- trivial morphism, that correspond to $P_{17}\times P_{3}\times P_{2}=C_{102}$
- $\psi$ maps $x\longmapsto (y, 0)$
- $\psi$ maps $x\longmapsto (y, z)$
- $\psi$ maps $x\longmapsto (0, z)$
Intuitively, I think the groups are $D_{2\cdot 51}$, $C_{3}\times D_{2\cdot 17}$, and $C_{17}\times D_{2\cdot 6}$ (where D denotes the dihedral group), but the question I have is: how to map these groups to the cases 2), 3), and 4)?
Thanks in advance
In Case 2 the automorphism induced on $P_3$ is trivial, so $P_3 < Z(G)$. But the automorphism induced on $P_{17}$ is nontrivial, so $P_{17} \not< Z(G)$. The group is $D_{2.17} \times C_3$.
Similarly, in Case 4, the group is $C_{17} \times D_{2.3}$.
In Case 3, nontrivial automorphisms are induced on both $C_{17}$ and $C_3$, and the group is $D_{2.51}$.
This example is starightforward because all four of the options for $\psi$ lead to distinct (i.e. non-siomorphic) groups, and the groups have easily stated properties that can used to distinguish them (like haveing a central Sylow $p$-subgroup for some $p$). In general different maps $\psi$ can result in isomorphic groups, and deciding when that happens can be more difficult.