Trouble with understanding classifying groups with semi direct products

Question

I'm trying to understand the following strategy on classifying groups of a particular order from Dummit & Foote's Abstract Algebra (p.181):

Let $G$ be a group of order $n$.

1. You find proper subgroups $H,K<G$ which satisfies the hypothesis of the Recognition Theorem for Semi-Direct Products. The theorem says:

If $G$ is a group with subgroups $H$ and $K$ such that $(1)\ G = HK,\ (2)\ H \cap K = \{e\},$ $(3)\ H \lhd G$, then $G \cong H \rtimes_{\phi} K$ with $\phi: K \to Aut(H)$ given by sending $k \mapsto \text{conjugation by }k$.

2. Now you find all isomorphism types for $H$ and $K$ and apply the Recognition Theorem for Semi-Direct Products to conclude there exist some homomorphism $\phi: K \to Aut(H)$ such that $G \cong H \rtimes_{\phi} K$. So now you just find all possible homomorphisms for pairs of $H$ and $K$ and you finish the classification.

Here's what confuses me:

1. Say in step 2 you find $H$ and $K$ satisfying the conditions for the Recognition Theorem. Doesn't the theorem only say that $\phi$ must be the homomorphism (i.e. permutation representation) associated with the group action of $K$ on $H$ by conjugation (not just any homomorphism as described in the strategy above)? How does constructing a group of order $n$ from semi-direct product of $H$ and $K$ relate to this Recognition Theorem?

Further: In the Recognition Theorem, the conjugation action $khk^{-1}$ is well-defined because we already know apriori what the group operation is in $G$. In case of the classification problem, however, when we find all isomorphism types of $H$ and $K$, what does it even mean by "$K$ acting on $H$" by conjugation?

2. Why is this the complete classification of groups of order $n$? (Again, I think the Recognition Theorem is relevant here but I failed to understand how it applies given the confusion described in 1.)

P.S: I went over many posts on MSE and many also use recognition theorem as part of their reasoning. Such as Find four groups of order 20 not isomorphic to each other.

Thanks in advance for anyone who would like to help !

Answer 1

You wrote

Doesn't the theorem only say that must be the homomorphism (i.e. permutation representation) associated with the group action of on by conjugation

but this misses the point that when you build a new group using (external) semidirect products you do not already have that conjugation action. Once you have a $\varphi : K \to {\rm Aut}(H)$ you can say that inside $H \rtimes_\varphi K$ the action using $\varphi$ can be interpreted as conjugation: $(1,k)(h,1)(1,k)^{-1} = (\varphi_k(h),1)$. But when you just start with two groups like $H = (\mathbf Z/(2))^2$ and $K = \mathbf Z/(3)$, you need to determine the possible homomorphisms $\varphi : K \to {\rm Aut}(H)$ even to have a group $H \rtimes_\varphi K$ and then you need to determine when different $\varphi$'s lead to isomorphic semidirect products. There is no such thing as a "conjugation" action of $\mathbf Z/(3)$ on $(\mathbf Z/(2))^2$. But once you write down an actual homomorphism $\varphi : \mathbf Z/(3) \to {\rm Aut}((\mathbf Z/(2))^2)$, inside $(\mathbf Z/(2))^2 \rtimes_\varphi \mathbf Z/(3)$ you could reinterpret $\varphi$ as being a conjugation action of the subgroup $\mathbf Z/(3)$ on the subgroup $(\mathbf Z/(2))^2$ inside that semidirect product determined by $\varphi$.

Lastly, not all groups are isomorphic to a semidirect product of smaller groups, such as cyclic $p$-groups with order greater than $p$ as mentioned in a comment above and the quaternion group $Q_8$ or even the generalized quaternion groups $Q_{2^n}$ where $n \geq 3$. (The groups $Q_{2^n}$ are quotient groups of a semidirect product, however.) Unless you have read or proved a result saying that every group of some particular order $n$ is isomorphic to a semidirect product of smaller groups that you can already classify up to isomorphism (that step uses the recognition theorem!), you usually aren't going to classify all groups of order $n$ just by forming semidirect products. In Sections 6 to 9 here, semidirect products are used to determine up to isomorphism all groups with order $pq$, $4q$, $1881$, and $p^3$, where $p$ and $q$ are distinct primes in $pq$ and $q$ is an odd greater than $3$ in $4q$. Groups with order $12$ are determined up to isomorphism using semidirect products here.

Answer 2

Addressing your (simple) second question. I will answer your second question first, for it is very simple...because no simple group has a nontrivial semidirect product decomposition. (Consider $\mathbb{Z}/3\mathbb{Z}$ for instance.)

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Addressing your first question. Consider two arbitrary groups $H,K$ (for instance, $H=\mathbb{Z}/5\mathbb{Z}$ and $K=\text{Mat}_n(\mathbb{F}_3)$...notice that these two groups are entirely arbitrary, i.e. we don't already know of a group which contains both $H,K$). Let's say we want to construct a group $G$ with underlying set $H\times K$ (Important! The product sign in this context will represent the Cartesian Product. This is distinct from the Direct Product, which involves equipping the set with a group operator.) in which (i) there exists subgroups $H_0,K_0<G$ such that $H\cong H_0,K\cong K_0$, (ii) these subgroups have trivial intersection $H_0\cap K_0=1$, (iii) $H_0\triangleleft G$, and (iv) there exists a natural isomorphism $\phi:G\to HK$ defined $\phi:(h,k)\mapsto hk$. Then, if we were to focus on the group homomorphism briefly, in order for such a homomorphism to be well-defined, observe:

$$\phi(h_1,k_1)\phi(h_2,k_2)=h_1k_1h_2k_2=(h_1h_2)(k_1k_2)=\phi(h_1h_2,k_1k_2).$$

Question! Noting how $H,K$ aren't already known to be subgroups of a given group, there exists no well-defined product between elements of $H$ and $K$. (e.g. The output of product $k_1h_2$ will be undefined.) In what ways could we define multiplication between these two elements so as to be well-defined?

Answer! In order for multiplication to be well-defined between these two elements, all that's needed is for the operation to represent an associative binary operator (i.e. well-defined function $HK\times HK\to HK$).

Looking now at how multiplication should be constructed in order for the group homomorphism to be well-defined, we only really need that our chosen multiplicative operator satisfies the relation $k_1h_2=h_2k_1$ (or rather, $k_1h_2k_1^{-1}=h_2$). Since $H_0\triangleleft G$, we know conjugation will define a group action on $H_0$, hence multiplication between elements in $K$ and $H$ should necessarily be defined via any group action, that is $kh=\phi_k(h)$ (where $\phi_{\_}:K\to\text{Aut}(H)$ denotes group homomorphism). And so, it is for this reason which we define $(h_1,k_1)(h_2,k_2)=(h_1\phi_{k_1}(h_2),$$k_1k_2).$ (And with this multiplication operator equipped on $G$, we arrive at our construction for the semi-direct product $G=H\rtimes_{\phi} K$.)