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Generalized coordinate subgroups of a direct product


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.2

Solution:

(1) Define a mapping $\varphi : G_I \rightarrow \times_{i \in I} G_i$ by $(\varphi(g))_i = \pi_i(g)$. We need to show that $\varphi$ is a bijective homomorphism.

Homomorphism: Let $g = (g_i), h = (h_i) \in G_I$. Then for each $i \in I$, $$(\varphi(gh))_i = (\varphi((g_j)(h_j)))_i = (\varphi((g_jh_j)))_i = g_ih_i = (\varphi((g_j)))_i (\varphi((h_j)))_i = (\varphi(g) \varphi(h))_i.$$ Hence $\varphi(gh) = \varphi(g) \varphi(h)$, so that $\varphi$ is a homomorphism.

Surjective: Let $(g_i)_I \in \times_{i \in I} G_i$. Define $(a_i) \in G_I$ by $a_i = g_i$ if $i \in I$ and $a_i = 1$ otherwise. Clearly this is indeed an element of $G_I$. Moreover, $\varphi((a_j)) = (g_j)$. Because $(g_i)$ is arbitrary, $\varphi$ is surjective.

Injective: Let $g = (g_i), h = (h_i) \in G_I$. Suppose $\varphi(g) = \varphi(h)$. By definition, $g_j = h_j = 1$ for all $j \notin I$. Moreover, for $i \in I$, we have $$g_i = (\varphi(g))_i = (\varphi(h))_i = h_i.$$ Hence $g = h$, so that $\varphi$ is injective.

(2) Let $(g_i) \in G_I$ and $(x_i) \in G$. Note that $$(x_i)^{-1} (g_i) (x_i) = (x_i^{-1} g_i x_i).$$ For $i \in I$, we have $g_i = 1$, so that $x_i^{-1} g_i x_i = 1$. Thus $(x_i)^{-1} (g_i) (x_i) \in G_I$. Hence $G_I \leq G$ is normal.

Now define a mapping $\psi : G \rightarrow G_J$ as follows: $(\psi((g_i)))_j = g_j if j \in J$ and 1 otherwise. We wish to show that $\psi$ is a surjective homomorphism and that $\mathsf{ker}\ \psi = G_I$.

Homomorphism: Let $g = (g_i), h = (h_i) \in G$. If $i \in J$, then $$\psi(gh)_i = \psi((g_jh_j))_i = g_ih_i = \psi(g)_i \psi(h)_i.$$ If $i \notin J$, then $$\psi(gh)_i = 1 = 1 \cdot 1 = \psi(g)_i \psi(h)_i.$$ Thus $\psi$ is a homomorphism.

Surjective: It is clear that if $g \in G_J$, then $\psi(g) = g$.

Kernel: ($\subseteq$) Let $g = (g_i) \in \mathsf{ker}\ \psi$. Then $(\psi(g_j))_i = 1 $ for all $i$; in particular, $g_i = 1$ if $i \notin I$. Thus $g \in G_I$. ($\supseteq$) Let $g = (g_i) \in G_I$. Now if $j \in J$, $g_j = 1$. Clearly then $\psi(g) = 1$, so that $g \in \mathsf{ker}\ \psi$.

By the First Isomorphism Theorem, we have $G/G_I \cong G_J$.

(3) Define a mapping $\theta : G \rightarrow G_I \times G_J$ as follows. $(\theta(g)_1)_i = g_i$ for $i \in I$ and 1 otherwise, and $(\theta(g)_2)_j = g_j$ for $j \in J$ and 1 otherwise. We wish to show that $\theta$ is a bijective homomorphism.

Homomorphism: Let $g = (g_k), h = (h_k) \in G$. If $i \in I$, then $$(\theta((g_k)(h_k))_1)_i = (\theta((g_kh_k))_1)_i = g_ik_i = (\theta((g_k))_1)_i (\theta((h_k))_1)_i.$$ If $i \notin I$, then $$(\theta((g_k)(h_k))_1)_i = (\theta((g_kh_k))_1)_i = 1 = 1 \cdot 1 = (\theta((g_k))_1)_i (\theta((h_k))_1)_i.$$ Thus $\theta(gh)_1 = \theta(g)_1 \theta(h)_1$. Similarly, $$\theta(gh)_2 = \theta(g)_2 \theta(h)_2. Hence \theta(gh) = \theta(g) \theta(h),$$ so that $\theta$ is a homomorphism.

Surjective: Let $(a,b) \in G_I \times G_J$, with $a = (a_k)$ and $b = (b_k)$. Define $g = (g_k) \in G$ as follows: If $k \in I$ then $g_k = a_k$, and if $k \in J$ then $g_k = b_k$. Then by definition, we have $(\theta(g)_1)_k = a_k$ if $k \in I$ and 1 otherwise, so that $\theta(g)_1 = a$. Similarly, $\theta(g)_2 = b$. Hence $\theta(g) = (a,b)$, and $\theta$ is surjective.

Injective: Suppose $g = (g_k), h = (h_k) \in G$ such that $\theta(g) = \theta(h)$. For $k \in I$, we have $$g_k = (\theta(g)_1)_k = (\theta(h)_1)_k = h_k,$$ and for $k \in J$, we have $$g_k = (\theta(g)_2)_k = (\theta(h)_2)_k = h_k.$$ Thus $g = h$, and $\theta$ is injective.


Linearity

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