Brian Bi
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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 $$HK = KH$$.