I'm tasked with proving (for homework) any two nonabelian groups of order $4301 = (11)(17)(23)$ are isomorphic. I've proved that for any nonabelian group $G$, we have $G = P_{11}N$, with $P_{11} \cap N = \{1\}$, where $P_{11} \cong \mathbb Z_{11}$ is an 11-Sylow subgroup of $G$, and $N = P_{17}P_{23}$ is a normal subgroup of $G$ where $P_{17} \cong \mathbb Z_{17}$ and $P_{23} \cong \mathbb Z_{23}$ are the normal and unique 17-Sylow and 23-Sylow subgroups of $G$, where moreover these imply that $N \cong \mathbb Z_{17} \times \mathbb Z_{23} \cong \mathbb Z_{391}$ by the Chinese remainder theorem and some further computations.
I've also proved that the number of 11-Sylow subgroups must be 23 (or else $G$ is abelian, since the only other option is 1, but then this implies $P_{11}$ is normal and so we can write $G$ as a direct product of abelian normal subgroups), but I don't see how this fits in.
I'm not quite sure how to proceed from here, what I had in mind initially pursuing this direction was to show that I can determine $G$ up to isomorphism from how these subgroups fit together, but I don't see how to proceed.
I've read ahead in the textbook about semidirect products and automorphism groups a bit, but given that these haven't been covered in class, I'm not sure how I'm supposed to proceed (especially given that the homomorphisms from $\mathbb Z_{11} \to \mathrm{Aut}(\mathbb Z_{391})$ look really complicated, though I think this automorphism group is isomorphic to $\mathbb Z_{16} \times \mathbb Z_{22}$ which might simplify it a bit, though the proof I had in mind for this again veers away from what we have covered in class). Anything I should try (especially if there's something I'm missing not reliant on the knowledge not covered in my class)?
Best Answer
You've almost done it, just need to use one of the following lemmas:
$1$) Let $K$ be a finite group and $H$ a group of order $p$, where $p$ is a prime number. Assume that $\psi$ and $\phi$ are two homomorphisms from $H$ to $\mathrm{Aut}(K)$. If $\mathrm{Im}\psi$ and $\mathrm{Im}\phi$ are conjugate, then $K \times_\psi H \cong K \times_\phi H$.
$2$) Let $K$ be a finite group and $H$ a cyclic group. Assume that $\psi$ and $\phi$ are two monomorphisms from $H$ to $\mathrm{Aut}(K)$. If $\mathrm{Im}\psi=\mathrm{Im}\phi$, then $K \times_\psi H \cong K \times_\phi H$.
Here we have $\mathrm{Aut}(C_{391})$, which is isomorphic to $C_{16}\times C_{22}$. So image of any non-trivial homomorphism from $C_{11}$ to $\mathrm{Aut}(C_{391})$, equals the Sylow $11$-subgroup of $\mathrm{Aut}(C_{391})$. Hence both lemmas work here.