Let's break the problem into **two steps** as requested: --- ### **Step 1: Show that \( \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \) is not a finitely generated abelian group** #### **Explanation:** - **Definition:** A group is **finitely generated** if there exists a finite set of elements such that every element of the group can be written as a combination (product) of these generators. - The group \( \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \) is the direct product of countably infinite copies of \( \mathbb{Z}_2 \). - If we try to generate all elements with a finite set, any element outside the coordinates indexed by the generators would be unreachable. - For example, if you select \( n \) generators, you can only affect the first \( n \) coordinates nontrivially. But the group has infinitely many coordinates, so elements with 1 in the \( (n+1) \)-th coordinate cannot be generated. - **Thus,** it cannot be finitely generated. #### **Summary for Step 1:** **The infinite direct product \( \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \) is not finitely generated because any finite set of generators only covers finitely many coordinates, but the group has infinitely many.** --- ### **Step 2: Given \( G = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \), show \(\{e\} \times G\) is a normal subgroup of \( (\mathbb{Z}_2)^n \times G \), and \((\mathbb{Z}_2)^n \times G / \{e\} \times G \cong (\mathbb{Z}_2)^n\)** #### **Explanation:** - Let \( H = (\mathbb{Z}_2)^n \) and \( K = G \). - The group \( H \times K \) consists of pairs \( (h, k) \) with \( h \in H \), \( k \in K \). - The subset \( \{e\} \times K = \{ (e, k) \mid k \in K \} \) is a subgroup. - **Normality:** Since \( H \times K \) is abelian (both \( H \) and \( K \) are abelian), every subgroup is normal. - Now, consider the quotient group \( (H \times K)/(\{e\} \times K) \): - By the **First Isomorphism Theorem** and the **definition of quotient groups**, the cosets are of the form \( (h, k) (\{e\} \times K) = \{ (h, k k') \mid k' \in K \} \). - Each coset is uniquely determined by \( h \), regardless of \( k \). - Thus, the set of cosets is in bijection with \( H \), and the group operation matches. - **Therefore,** \( (H \times K)/(\{e\} \times K) \cong H \). #### **Summary for Step 2:** **The subgroup \( \{e\} \times G \) is normal in \( (\mathbb{Z}_2)^n \times G \), and the quotient \( (\mathbb{Z}_2)^n \times G / (\{e\} \times G) \) is isomorphic to \( (\mathbb{Z}_2)^n \).** --- ## **Final Answer:** 1. **\( \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \) is not finitely generated because any finite set of generators only covers finitely many coordinates.** 2. **For \( G = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \), the subgroup \( \{e\} \times G \) is normal in \( (\mathbb{Z}_2)^n \times G \), and the quotient \( (\mathbb{Z}_2)^n \times G / (\{e\} \times G) \cong (\mathbb{Z}_2)^n \).**