3.11 Internal direct product
In this section \(G\) is a group. If \(S\subseteq G\), we set \(S^{-1} = \{ s^{-1}: s\in S\} \).
Suppose \(S\) is a nonempty subset of \(G\). Then the following are equivalent.
\(S\leq G\)
\(S S^{-1} \subseteq S\).
Moreover, if \(S\leq G\), we have \(SS = S = S^{-1}\).
Suppose \(H\) and \(K\) are subgroups of a group \(G\). Then \(HK\leq G\) if and only if \(HK = KH\). In this case, \(HK = \langle H, K\rangle \). In other words, \(HK\) is the group generated by \(H\) and \(K\).
(\(\Rightarrow \)) Suppose \(HK \leq G\). Then \(H\leq HK\) and \(K\leq HK\). So \(KH \subseteq HK\). But then \(HK = (KH)^{-1} \subseteq (HK)^{-1} = KH\).
(\(\Leftarrow \)) Suppose \(HK = KH\). Then \(HK (HK)^{-1} = HK K^{-1} H^{-1} = HK KH = HK H = H HK = HK\). So \(HK \leq G\) by Lemma 3.66.
Suppose that \(HK\) is a group. Then \(HK\leq \langle H, K\rangle \) as \(H, K \subseteq HK\). But \(HK\) is the intersection of all subgroups of \(G\) containing both \(H\) and \(K\). So, as \(\langle H, K \rangle \leq HK\).
Suppose \(H\), \(K\leq G\).
We say that \(H\) centralizes \(K\) if \(H\leq C_G(K)\), where \(C_G(K)\) denotes the centralizer of \(K\) in \(G\) (Definition 2.60).
We say that \(H\) normalizes \(K\) if \(H\leq N_G(K)\), where \(N_G(K)\) denotes the normalizer of \(K\) in \(G\) (Definition 3.31).
We say that \(H\) and \(K\) normalize each other if \(H\) normalizes \(K\) and \(K\) normalizes \(H\).
In the context of Definition 3.68, if \(H\) centralizes \(K\) then automatically \(K\) centralizes \(H\) (by Lemma 2.63). This is not true if we replace the word “centralizes" with the word “normalizes." If \(H\) is any subgroup of a group \(G\), then \(H\) clearly normalizes \(G\) itself, but \(G\) normalizes \(H\) if and only if \(H\) is normal.
Suppose \(H, K\leq G\). If \(H\) centralizes \(K\), then \(H\) normalizes \(K\).
Suppose \(H\) centralizes \(K\) and \(h\in H\). Then \(hK = Kh\). So \(h\) normalizes \(K\).
Suppose \(G\) is a group and \(H, K \leq G\) with \(H\) normalizing \(K\). Then \(HK\leq G\) and \(K\unlhd HK\).
Since \(H\) normalizes \(K\), for every \(h\in H\), \(hK = Kh\). So \(HK = \cup _{h\in H} hK = \cup _{h\in H} Kh = KH\). Therefore, by Theorem 3.67, we have \(HK\leq G\).
Suppose \((h,k)\in H\times K\). Then \(hkK = hKk = Khk\). So \(HK\) normalizes \(K\), and \(K\leq HK\) since \(1\in H\). So \(K\unlhd HK\).
Suppose \(H\), \(K\) and \(G\) are as in Proposition 3.71. Let \(i:H\to HK\) denote the inclusion homomorphism, and write \(\pi :HK \to HK/K\) for the quotient homomorphism (taking into account the fact that \(K\unlhd HK\) by Proposition 3.71. Set \(f = \pi \circ i: H\to HK/K\). Then \(f\) is onto with \(\ker f = H\cap K\). Hence, using Theorem 3.45, we get an isomorphism
Suppose \(hkK\in HK/K\). The \(hkK = hK = f(h)\). So \(f\) is onto.
Suppose \(h\in H\). Then \(f(h) = e \Leftrightarrow hK = K \leftrightarrow h\in K \Leftrightarrow h\in H\cap K\). So \(\ker f = H\cap K\). The rest follows from Corollary 3.50.
Suppose \(x,y\in G\). The commutator of \(x\) and \(y\) is the element \([x,y] = xyx^{-1}y^{-1}\).
We have \([x,y] = 1 \Leftrightarrow xy = yx\).
Exercise!
Suppose \(H\) and \(K\) are subgroups of a group \(G\) with \(H\cap K = \{ 1\} \). Then the following are equivalent:
\(H\) and \(K\) centralize each other.
\(H\) and \(K\) normalize each other.
1 \(\Rightarrow \) 2: Obvious by Proposition 3.70.
2 \(\Rightarrow \) 1: Suppose \(H\) and \(K\) normalize each other. Let \((h,k)\in H\times K\). Then \(hkh^{-1}\in K\). So \([h,k] = hkh^{-1} k^{-1}\in K\) as well. On the other hand, \(khk^{-1}\in H\). So, \([h,k] = hkh^{-1}k^{-1} \in H\). Therefore, \([h,k]\in H\cap K\), and, by our hypothesis, \(H\cap K = \{ 1\} \). So \([h,k] = 1\). But this proves that \(H\) and \(K\) centralize each other.
Suppose \(H, K\leq G\). Then the following are equivalent.
There exists a group isomorphism \(f:H\times K\to G\) with \(f(h,1) = h\) and \(f(1,k) = k\) for \((h,k)\in H\times K\).
\(H\unlhd G\), \(K\unlhd G\), \(H\cap K = \{ 1\} \) and \(HK = G\).
If these two equivalent conditions hold, then \(G\) is called the internal direct product of \(H\) and \(K\) Moreover, if \((h,k)\in H\times K\), we have \(hk = kh\), and \(f(h,k) = hk\).
1 \(\Rightarrow \) 2: Suppose \(f:H\times K\to G\) is an isomorphism of groups satisying \(f(h,1) = h\) and \(f(1,k) = k\). Then \(f(h,k) = f((h,1)(1,k)) = f(h,1) f(1,k) = hk\).
Since \(f\) is onto, \(HK = f(H\times K) = G\). On the other hand, if \(g\in H\cap K\), then \((g,g^{-1})\in H\times K\) and \(f(g,g^{-1}) = g g ^{-1} = 1\). So \((g,g^{-1})\in \ker f\). Therefore, as \((g,g^{-1}) = 1\). So \(g = 1\), which proves that \(H\cap K = \{ 1\} \).
Suppose \((h,k)\in H\times K\). Then \(hk = f(h,k) = f((1,k)f(h,1)) = f(1,k)f(h,1) = kh\). So \(hk=kh\). Therefore, \(H\) and \(K\) centralize each other. It follows from Proposition 3.76 that \(H\) and \(K\) normalize each other. But then \(H\unlhd HK = G\) and \(K\unlhd HK = G\).
2 \(\Rightarrow \) 1: Suppose the conditions of 2 hold. Then \(H\) and \(K\) centralize each other by Proposition 3.76. Therefore, by Theorem 3.64, the map \(f:H\times K\to G\) given by \(f(h,k) = hk\) is a group homomorphism. Suppose \((h,k)\in \ker f\). Then \(hk =1\). So \(h = k^{-1}\in H\cap K = \{ 1\} \). So \(h = k = 1\). Thus \((h,k) = (1,1)\), and we get that \(\ker f = \{ 1\} \). So \(f\) is one-one. On the other hand, \(f\) is onto because \(HK=G\). So \(f:H\times K\to G\) is an isomorphism of groups.
The rest is clear at this point.