Let's answer each part step by step, with explanations after each step and a final answer at the end. --- ## 1. Show that \( \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \) is **not** a finitely generated abelian group ### Step 1: Assume it is finitely generated and seek a contradiction Suppose \( G = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \) (countably infinite product) **is finitely generated**. Let \( S = \{g_1, g_2, ..., g_n\} \) be a finite generating set for \( G \). #### Explanation: Every element of \( G \) is a tuple of 0s and 1s, with each coordinate from a copy of \( \mathbb{Z}_2 \). If \( G \) is generated by \( n \) elements, then every element is a finite combination of these generators. --- ### Step 2: Show there exists an element that cannot be generated by \( S \) Since each \( g_i \) has only finitely many nonzero entries, their span can only produce elements that are nonzero in at most \( n \) positions. But in \( G \), we can define an element \( h \) that is 1 in the \( (n+1) \)-th position and 0 elsewhere, which **cannot** be written as a combination of the \( g_i \). #### Explanation: No combination of the finite set \( S \) can produce an element that is nonzero in a position where all the \( g_i \) are zero. Hence, \( G \) cannot be finitely generated. --- **Final answer for 1:** \( \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \) is **not** finitely generated, because any finite subset of generators can only control finitely many coordinates, but \( G \) has infinitely many. --- ## 2. Let \( G = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \). Show \( \{e\} \times G \) is normal in \( (\mathbb{Z}_2)^n \times G \) and \( (\mathbb{Z}_2)^n \times G / (\{e\} \times G) \cong (\mathbb{Z}_2)^n \) ### Step 1: Show \( \{e\} \times G \) is a normal subgroup of \( (\mathbb{Z}_2)^n \times G \) Let \( H = \{e\} \times G \). Any element of \( (\mathbb{Z}_2)^n \times G \) is of the form \( (a, g) \), and any \( h \in H \) is of the form \( (e, h) \). Since the group is abelian, for any \( (a, g) \) and \( (e, h) \): \[ (a, g)(e, h)(a, g)^{-1} = (e, h) \] #### Explanation: All subgroups of an abelian group are normal by definition, so \( H \) is normal in \( (\mathbb{Z}_2)^n \times G \). --- ### Step 2: Show the quotient \( (\mathbb{Z}_2)^n \times G / (\{e\} \times G) \cong (\mathbb{Z}_2)^n \) Define the map: \[ \phi: (\mathbb{Z}_2)^n \times G \to (\mathbb{Z}_2)^n, \quad \phi(a, g) = a \] This is a surjective homomorphism with kernel \( \{(e, g) \mid g \in G\} = H \). By the First Isomorphism Theorem: \[ (\mathbb{Z}_2)^n \times G / H \cong (\mathbb{Z}_2)^n \] #### Explanation: The quotient group collapses the \( G \) part, leaving only the \( (\mathbb{Z}_2)^n \) part. --- **Final answer for 2:** - \( \{e\} \times G \) is a normal subgroup of \( (\mathbb{Z}_2)^n \times G \) since the group is abelian. - The quotient \( (\mathbb{Z}_2)^n \times G / (\{e\} \times G) \cong (\mathbb{Z}_2)^n \) by the First Isomorphism Theorem.