Let \(N\) and \(H\) be groups. Recall that the direct product is defined as:
\(N \times H = \{(n, h) \mid n \in N, h \in H\}\) with component-wise operation: \((n_1, h_1)(n_2, h_2) = (n_1n_2, h_1h_2)\) where the identity is \((e_N, e_H)\) and the inverse is \((n, h)^{-1} = (n^{-1}, h^{-1})\). This is also known as the external direct product of groups.
Notice how we are constructing a larger group (in terms of set cardinality) from two smaller groups using the external direct product. Informally, the internal direct product refers to the opposite process—splitting a larger group into two subgroups such that their direct product is isomorphic to the original group. Before we dive into the concept of the internal direct product, let us first define what it means for one subgroup to be a complement of another.
Definition
Let \(G\) be a group and \(N \leq G\). A subgroup \(H \leq G\) is called a complement of \(N\) in \(G\) if:
Remark
If \(H\) is a complement of \(N\) in \(G\), then \(N\) is a complement of \(H\) in \(G\) (i.e., \(NH = HN\) iff \(NH \leq G\))
Lemma
If \(N, H\) are complements in \(G\), then \(\phi: N \times H \to G\) with \(\phi(n, h) = nh\) is a bijection.
Proof
\(\phi\) is injective because
\(\begin{aligned} h_1 n_1 &= h_2 n_2 \\ \Rightarrow h_2^{-1} h_1 &= n_2 n_1^{-1} \\ \Rightarrow h_2^{-1} h_1 &= n_2 n_1^{-1} = e \in N \cap H = \{ e \} \\ \Rightarrow h_1 &= h_2, \quad n_1 = n_2 \end{aligned}\)
\(\phi\) is surjective because \(\text{Im}(\phi) = \{ nh \mid n \in N, h \in H \} = NH = G\).
Thus, \(\phi\) is a bijection. ∎
Now, let’s discuss the internal direct product.
Definition
A group \(G\) is the internal direct product of two subgroups \(N\) and \(H\) if:
We claim that \(G \cong N \times H\).
Proof
We know that \(\phi\) is a bijection. So, we only need to now show that \(\phi\) is a homomorphism.
First, notice that
\(h_1 n_2 = h_1 n_2 h_1^{-1} h_1 = (h_1 n_2 h_1^{-1}) h_1 = \mathbf{n} h_1 \quad (\text{since } h_1 n_2 h_1^{-1} \in N)\)
Then, notice
\(h_1 n_2 = n_2 n_2^{-1} h_1 n_2 = n_2 (n_2^{-1} h_1 n_2) = n_2 \mathbf{h} \quad (\text{since } n_2^{-1} h_1 n_2 \in H)\)
So, \(\mathbf{n} = n_2\), \(\mathbf{h} = h_1\) and \(h_1 n_2 = \textbf{nh} = n_2 h_1\).
Finally,
\(\begin{aligned} \phi((n_1, h_1)\cdot(n_2, h_2)) &= \phi((n_1n_2, h_1h_2)) \\ &= n_1n_2h_1h_2 \\ &= n_1 h_1 n_2 h_2 \quad \text{since } h_1n_2 = n_2h_1 \\ &= \phi(n_1, h_1) \cdot \phi(n_2, h_2) \end{aligned}\)
Therefore, \(G \cong N \times H\). ∎
Question. Can we still have an isomorphism $ G \cong N \ \text{(some operator)} \ H $ when \(H \leq G\) but not normal?
The answer is yes. We can define the group operation as
\((n_1, h_1)\cdot(n_2, h_2) = (n_1h_1n_2h_1^{-1}, h_1h_2),\)
which gives a structure known as the semi-direct product. Note that \(n_1(h_1n_2h_1^{-1}) \in N\) since \(N \trianglelefteq G\). Commonly, this is expressed as
\((n_1, h_1)\cdot(n_2, h_2) = (n_1 \varphi_{h_1}(n_2), h_1h_2)\)
where \(\varphi: H \to \text{Aut}(N)\) is defined by \(\varphi_h(n) = hnh^{-1}\). And this structure is denoted by the operator \(N \rtimes H\).
Definition
A group \(G\) is the internal semi-direct product of two subgroups \(N\) and \(H\) if:
We claim that \(G \cong N \rtimes H\).
Proof
We know that \(\phi\) is a bijection. So, we only need to now show that \(\phi\) is a homomorphism. Notice that,
Therefore, \(G \cong N \times H\). ∎
Remark
Notice how direct product is a special case of semi-direct product. To see this, let \(\varphi\) be trivial (i.e., \(\varphi_h(n) = hnh^{-1} = n\)), then the semi-direct product becomes a direct product since
\((n_1, h_1)\cdot(n_2, h_2) = (n_1 \varphi_{h_1}(n_2), h_1h_2) = (n_1n_2, h_1h_2).\)
Can we relax the requirements even further? The answer is yes.
If we only assume:
then \(G\) can be expressed as an internal Zappa–Szép product, written as \(G \cong N \bowtie H\).
We won’t go into the details here, but feel free to explore this generalization further.