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Can the structures of \(N\) and \(H\) be used to determine the structure of \(NH\)? First consider that both subgroups are normal.
Let \(N\) and \(H\) be two normal subgroups of \(G\). Suppose \(N\cap H = \{e\}\), then \(nh = hn\) for all \(h\in H, n\in N\) and \(NH \cong N \times H\).
Proof
Since \(H\) is normal, we have that for all \(n\in N\), \(nhn^{-1} \in H\). This also holds for \(N\) and all \(h \in H\). Then the element
\[nhn^{-1}h^{-1} = (nhn^{-1})h^{-1} = n(hn^{-1}h^{-1})\]is contained in both \(N\) and \(H\). Therefore, \(nhn^{-1}h^{-1} = e\), which implies \(nh = hn\).
Now consider the map \(\rho:N\times H\to NH\) given by \(\rho(n, h) = nh\). We can prove that \(\rho\) is an isomorphism. (Here we only show that \(\text{ker}(\rho) = \{(e, e) \}\).)
Suppose \((n,h)\in\text{ker}(\rho)\), i.e., \(\rho(n, h) = nh = e\). This implies that \(n = h^{-1} \in N\cap H\). Therefore, \(n=h=e\). ◼
Then, consider one subgroup is normal and the other is arbitrary.
Now, suppose \(N\) is normal and \(N\cap H = \{e\}\). By this theorem, the elements in \(NH\) can be uniquely represented as \(nh\) for some \(n\in N\) and \(h\in H\). Moreover. the product of two elements \(n_1h_1\) and \(n_2h_2\) can be written as
\[n_1h_1n_2h_2 = n_1(h_1n_2h_1^{-1})h_1h_2 = n_1\rho_{h_1}(n_2)h_1h_2,\]where \(\rho_{h_1}(x) := h_1xh_1^{-1}\) is an inner automorphism on \(N\).
\(\rho: H\to \text{Aut}(N)\), given by \(h_1\mapsto \rho_{h_1}\).
Then, consider the binary structure \(N \rtimes_{\rho} H\) defined on \(N\times H\) given by
\[(n_1, h_1)*(n_2, h_2) = (n_1, h_1)(n_2, h_2) := (n_1\rho_{h_1}(n_2), h_1h_2).\]這是定義在 \(N\rtimes_{\rho} H\) 上的乘法，而 \(N\rtimes_{\rho} H\) 的 elements 和 \(N\times H\) 一模一樣！
Let’s first prove that \((N\rtimes_{\rho}H, *)\) is a group:
associativity
Expand everything.
identity
\((e_N, e_H)\).
inverse
Let \((n_2, h_2) = (n_1, h_1)^{-1}\). Then \((n_1, h_1)(n_2, h_2) = (e, e)\), which means that \(n_1h_1n_2h_1^{-1} = e\) and \(h_1h_2 = e\). Thus we have
\[(n_1, h_1)^{-1} = (n_2, h_2) = (h_1^{-1}n_1h_1, h_1^{-1}). \tag*{$\blacksquare$}\]Then prove that \((N\rtimes_{\rho}H, *) \cong (NH, \cdot)\). Define \(\phi: NH\to N\rtimes_{\rho}H\) given by \(nh\mapsto (n, h)\). Here we only show that \(\phi\) is a homomorphism:
\[\begin{align*} \phi(n_1h_1n_2h_2) &= \phi(n_1\rho_{h_1}(n_2)h_1h_2) \\ &= (n_1\rho_{h_1}(n_2), h_1h_2) \\ &= \phi(n_1h_1)\phi(n_2h_2). \tag*{$\blacksquare$} \end{align*}\]Especially, the group \(N\rtimes_{\rho}H\) is called the semi-direct product of \(N\) and \(H\) with respect to \(\rho\).
因為 \(NH\) 是 group，所以證明 \(NH\) 和 \(N\rtimes_{\rho}H\) 同構也就證明了 \(N\rtimes_{\rho}H\) 是 group！
Let \(H\) and \(N\) be two subgroups of a finite group \(G\). Suppose \(N\) is normal and \(H\cap N = \{e\}\). Then there exists some group homomorphism \(\rho:H\to \text{Aut}(N)\) such that \(NH \cong N\rtimes_{\rho}H\).
當其一 normal 且兩者交集只有 identity！
The binary structure \((N\rtimes_{\rho}H, *)\) only involves the group operation of \(N\), the group operation of \(H\), and the group homomorphism \(\rho\).
因為 \(NH \leq G\)，所以如果 \(\vert NH\vert = \vert G\vert\)，則 \(NH=G\)。參考 Three Isomorphism Theorems。
Let \(H\) and \(N\) be two groups and \(\rho\) is a group homomorphism from \(H\) to \(\text{Aut}(N)\). Then \(G_{\rho}=N\rtimes_{\rho}H\) is a group satisfying the following conditions:
\[h'n'h'^{-1} = (\rho_h(n), e_H),\]
- \(N'= N\times \{e_H\} \cong N\), which is a normal subgroup of \(G_\rho\).
- \(H'= \{e_N\}\times H\cong H\), which is a normal subgroup of \(G_\rho\).
- \(N' \cap H' = \{(e, e)\}\) and \(G_{\rho} = N'H'\).
- For all \(n' = (n, e_H)\in N', h'=(e_N, h)\in H'\),
where \(\rho_h(n) = hnh^{-1}\).
Let \(H\) and \(N\) be two subgroups of a finite group \(G\). Suppose
- \(N\) is normal;
- \(\vert H\cap N\vert = 1\);
- \(\vert G\vert = \vert N\vert \vert H\vert\),
then
\[\{N\rtimes_\rho H \mid \rho: H\to \text{Aut}(N)\}\]consists all possible group structures of \(G\).
Let \(G\) be a group with a normal subgroup \(N=\langle a\rangle\) of order \(n\) and a subgroup \(H=\langle b\rangle\) of order \(m\), where \(\gcd(n, m) = 1\) and \(\vert G\vert=nm\). Then \(G\) is isomorphic to \(N\rtimes_\rho H\) where \(\rho_b(a) = a^k\) for some \(k\) satisfying \(k^m \equiv 1 \pmod n\).
Proof (?)
這裡的 \(\rho\) 一樣是 \(\rho: H\to \text{Aut}(N)\)。因為 \(\rho\) 是 group homomorphism，所以一旦 \(\rho_b(a) = a^k\) 決定了，\(\rho_{b^r}(a)\) 必然是 \({a^k}^r\)，於是 \(\rho_b\) 唯一決定了 \(\rho\)（用 generator 就夠）。
又因為 \(b\) 的 order 是 \(m\)，於是 \(\rho_{b^m}\) 是 identity map，亦即 \(\rho_{b^m}(a) = a = {a^k}^m\)、對於某些滿足 \(k^m \equiv 1 \pmod n\) 的 \(k\)。
然後：
\[N\rtimes_\rho H = \langle a,b\mid a^n=b^m=e, bab^{-1}=a^k\rangle.\]想想 Dihedral groups！
這個定理的意義在於，如果 \(G\) 可以找到兩個 cyclic subgroups（其中一個 normal），我們可以用只牽涉到該 subgroups 的 order 的同餘式，來明確的找出所有 \(G\) 可能的結構！（用到這個定理。）
Check this.
Let \(G\) be a group of order \(39\) with elements \(a\) and \(b\) of order \(13\) and \(3\) respectively. Find all non-isomorphic group structures of \(G\).
Solution
By the above theorem, we know that \(G\cong \langle a\rangle \rtimes_\rho \langle b\rangle\), where \(\rho_b(a) = bab^{-1} = a^k\) for some \(k\) satisfies \(k^3 \equiv 1 \pmod{ 13}\).
Note that \(\Bbb Z_{13}^\times \cong \Bbb Z_{12}\), which means that the number of elements of order divisible by \(3\) is the same in the two groups. Then, there \(3\) elements, \(\{0, 4, 8\}\) of order divisible by \(3\), which implies that \(k\) has \(3\) choices. After some calculation, we obtain
\[k = 1, 3, 3^2 = 9.\]When \(k=1\), we have \(G_1 \cong \Bbb Z_3 \times \Bbb Z_{13}\) for it is abelian.
When \(k=3\) or \(k=9\), we have \(bab^{-1}=a^3\) and \(bab^{-1}=a^9\) respectively. We need to check whether these two groups are isomorphic. First observe that \(b^{-1} = b^2\), and thus let \(c=b^2\). Then
\[c^{-1}ac = a^3 \implies ca^3c^{-1} = a \\ \implies (ca^3c^{-1})^9 = ca^{27}c^{-1} = cac^{-1} = a^9.\]Therefore, \(G_3\cong G_9\). ◼
\(G\cong \langle a,b\mid a^{13}=b^3=e, bab^{-1}=a^k\rangle\).
Let \(G\) be a group of order \(2p\) where \(p\) is an odd prime. Then \(G\cong \Bbb Z_{2p}\) or \(D_p\).
Proof
By Cauchy’s theorem, \(G\) contains a subgroup \(N=\langle a\rangle\) of order \(p\) and a subgroup \(H=\langle b\rangle\) of order \(2\). Since \([G:N]=2\), \(N\) is normal. Moreover, \(\vert G\vert =\vert N\vert \vert H\vert\) and \(\vert N\cap H\vert =1\). Therefore by this theorem, \(G\) is isomorphic to \(N\rtimes_\rho H\) where \(\rho_b(a) = a^k\) for some \(k\) with \(k^2\equiv 1\pmod p\).
Since \(\Bbb Z_p\) is a field, \(x^2=1\) has at most \(2\) distinct zeros; here, \(1\) and \(-1\).
When \(k=1\), \(\rho\) is the trivial homomorphism, which implies that
\[N\rtimes_\rho H \cong N\times H\cong \Bbb Z_p\times \Bbb Z_2 \cong \Bbb Z_{2p}.\]When \(k=-1\), \(bab^{-1} = \rho_b(a) = a^{-1}\), which implies that
\[N\rtimes_\rho H = \langle a, b\mid a^p=b^2=e, bab^{-1}=a^{-1}\rangle \cong D_p. \tag*{$\blacksquare$}\]