Brian Bi
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Return to table of contents for Brian's unofficial solutions to Artin's Algebra

Section 2.11. Product Groups

Exercise 2.11.1 \((x, y)^n = 1\) \(\Leftrightarrow\) (\(x^n = 1\) in \(G\) and \(y^n = 1\) in \(H\)) \(\Leftrightarrow\) (\(n\) is a multiple of both \(r\) and \(s\)) \(\Leftrightarrow\) \(n\) is a multiple of the LCM or \(r\) and \(s\). Therefore the order of \((x, y)\) is exactly the LCM of \(r\) and \(s\).

Exercise 2.11.3 Let \(G\) and \(H\) be infinite cyclic groups. Then \(G\) and \(H\) are each isomorphic to the additive group of integers and their product is isomorphic to \(\mathbb{Z}^2\) with elementwise addition. This group is not cyclic, since each nonzero element generates a subgroup of \(\mathbb{Z}^2\) consisting only of those lattice points that lie on the line joining the generator to the origin. So \(G \times H\) is not cyclic either.

Exercise 2.11.4

  1. Yes. \(H \cap K = \{1\}\), \(HK = G\), \(H\) and \(K\) are closed under multiplication and under inverses, making them subgroups of \(G\), and they are normal subgroups since \(G\) is abelian. Therefore the conditions of Proposition 2.11.4(d) are satisifed.
  2. No. The condition of Proposition 2.11.4(b) is violated. For example, the matrices \(\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\) and \(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\), which lie in \(H\) and \(K\), respectively, don't commute with each other.
  3. Yes. \(H \cap K = \{1\}\), \(HK = G\) since a nonzero complex number \(z\) can always be written as \((z/|z|)|z|\), and \(H\) and \(K\) are again subgroups of \(G\), which are automatically normal as \(G\) is abelian. So again Proposition 2.11.4(d) applies.

Exercise 2.11.5 \((z_1, z_2) \in Z_1 \times Z_2\) \(\Leftrightarrow\) \(\forall (g_1, g_2) \in G_1 \times G_2, (z_1, z_2)(g_1, g_2) = (z_1 g_1, z_2 g_2) = (g_1 z_1, g_2 z_2) = (g_1, g_2)(z_1, z_2)\) \(\Leftrightarrow\) (\((z_1, z_2)\) is central). Therefore, \(Z_1 \times Z_2\) is the centre of \(G_1 \times G_2\).

Exercise 2.11.6 A subgroup of order 3 must be cyclic; we can write it in the form \(H = \{1, x, x^2\}\) for some \(x\). Likewise a subgroup of order 5 must take the form \(K = \{1, y, y^2, y^3, y^4\}\). The intersection of these subgroups is necessarily \(\{1\}\) since all non-identity elements of the former have order 3 and all non-identity elements of the latter have order 5. According to Proposition 2.11.4(c), \(HK\) is a subgroup of \(G\). Since \(H\) and \(K\) are normal in \(G\), they are also normal in \(HK\). By Proposition 2.11.4(d), \(HK \cong H \times K \cong C_3 \times C_5\) and therefore contains elements of order 15.

Exercise 2.11.7 We know that \(H \cap N = \{1\}\), so Proposition 2.11.4(a) applies and the map is injective; also since \(N\) is a normal subgroup of \(G\), Proposition 2.11.4(c) tells us that the image is a subgroup of \(G\). However, the product map is not necessarily a homomorphism since elements of \(H\) may not commute with elements of \(N\).

Exercise 2.11.8 If \(\varphi, \varphi'\) are homomorphisms \(H \to G\) and \(H \to G'\) respectively, then the pair \((\varphi, \varphi')\) is mapped to the homomorphism \(\Phi(h) = (\varphi(h), \varphi'(h))\). We can easily verify that \(\Phi\) is a homomorphism whenever \(\varphi\) and \(\varphi'\) are. Conversely, if \(\Phi : H \to G \times G'\) is a homomorphism, then there is exactly one pair of functions \(\varphi : H \to G, \varphi' : H \to G'\) such that \(\Phi(h) = (\varphi(h), \varphi'(h))\), namely, \(\varphi(h)\) must be the first element of \(\Phi(h)\) and \(\varphi'(h)\) the second; the properties of homomorphisms as applied to \(\Phi\) then imply that these properties are satisfied for \(\varphi\) and \(\varphi'\) as well.

Exercise 2.11.9 Suppose first that \(HK = KH\). Obviously \(1 \in HK\). Also for all \(h \in H, k \in K\), we have that \((hk)^{-1} = k^{-1} h^{-1} \in KH = HK\), so \(HK\) is closed under inverses; also, for all \(h_1, h_2 \in H, k_1, k_2 \in K\), we have that \((h_1 k_1)(h_2 k_2) \in h_1 KH k_2 = h_1 HK k_2 \subseteq HK\), so \(HK\) is closed under the operation of \(G\); therefore \(HK\) is a subgroup.

In the other direction, suppose \(HK\) is a subgroup of \(G\). Then for all \(h \in H, k \in K\), we have \(kh = (h^{-1} k^{-1})^{-1} \in (HK)^{-1} = HK\); therefore \(KH \subseteq HK\). Also for all \(x \in HK\), we have that \(x^{-1} \in HK\), so \(x^{-1} = hk\) for some \(h \in H, k \in K\), whence \(x = k^{-1} h^{-1} \in KH\); therefore \(HK \subseteq KH\). We conclude that \(HK = KH\).