Strategy for classifying some groups of order $pqr$ - recognizing direct factors

Question

I've been reviewing some of my notes from an abstract algebra class that I took and have been thinking about/redoing some of the examples we did classifying groups of smallish order $pqr$. In classifying these kinds of groups, we usually leverage the Sylow theorems to find a normal subgroup and a complement to get a semidirect product $H \rtimes K$ and then consider all possible homomorphisms $K \to \mathrm{Aut}(H)$ and determine which are isomorphic. I feel pretty comfortable with this overall strategy, but my question is as follows. In quite a few of the examples once we pick a homomorphism $\phi:K\to \mathrm{Aut}(H)$ we can determine a direct factor of the group, but I feel like the process for finding these factors seems a little haphazard.

For example in classifying groups of order $30$, we can find a subgroup $H$ of order $15$ and then classify groups of order $30$ by considering maps $\mathbb{Z}_2\to \mathrm{Aut}(\mathbb{Z}_{15}) \cong \mathrm{Aut}(\mathbb{Z}_5) \times \mathrm{Aut}(\mathbb{Z}_3)$. There are four such homomorphisms, which corresponds to sending neither, one, or both of the generators of $\mathbb{Z}_5,\mathbb{Z}_3$ to their inverses. The resulting groups then are $\mathbb{Z}_{30}, \mathbb{Z}_5 \times D_3, \mathbb{Z}_3 \times D_5, D_{15}$.

My question is:

What is the best way to recognize the direct factor in those middle two cases?

For example in the case of $\mathbb{Z}_5 \times D_3$ which corresponds to the homomorphism fixing the generator of $\mathbb{Z}_5$ and sending the generator of $\mathbb{Z}_3$ to its inverse, I explicitly constructed an element of order $5$ in $H \rtimes K$, showed it generated a normal subgroup by showing it was closed under conjugation, and then constructed $D_3$ as the kernel of the homomorphism $H\rtimes K \to \mathbb{Z}_5$. This worked, but felt really messy and inefficient, I'm hoping there is a cleaner more general approach. I had a similar experience working through groups of order $182$. Any insight for a better way to approach this problem would be appreciated.

Answer 1

Let $P_2$, $P_3$ and $P_5$ be Sylow subgroups of order $2$, $3$ and $5$, respectively.

Let $\{p,q\}=\{3,5\}$.

You already have that $P_p$ and $P_q$ are normal in $G$. This means that $P_2P_p=P_p\rtimes P_2$ is a subgroup of order $2p$, and thus is a complement to $P_q$.

So $P_q$ will be a direct factor if and only if it is central. Since it is central in $P_pP_q=P_p\times P_q$, this is if and only if it is centralised by $P_2=K$, in other words, if $\phi$ fixes the generator. In this case, we have $G=P_q\times (P_p\rtimes P_2)$.