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

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.

• I think it is useful to consider the center of the group. In your example, if $|Z(G)|=1$, then $G=D_{15}$; if $|Z(G)|=30$, then $G=Z_{30}$; $|Z(G)|=3$, then $G=Z_3\times D_{5}$; $|Z(G)|=5$, then $G=Z_5\times D_{3}$ and so on. Commented Mar 6, 2022 at 4:15
• @kabenyuk I agree that that is helpful once you already know the possible isomorphism types, for example analyzing the centers is how I would go about showing that none of the four groups are pairwise isomorphic. However, if you don't have the list of possible groups ahead of time, I don't see how this helps. Am I missing something? Commented Mar 6, 2022 at 14:41

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 $$G=P_q\times (P_p\rtimes P_2)$$.