**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.18**

Solution:

(1) Consider $\prod_{\mathbb{N}} \mathbb{Z}/(2)$; clearly this group is infinite, and moreover $$2(\prod x_i) = \prod 2x_i = 0$$ for all elements $\prod x_i$.

(2) Let $G$ be the restricted direct product of $G_i = \mathbb{Z}/(i)$. Note that $\prod x_i$ defined by $x_i = 1$ if $i = k$ and 0 otherwise is in $G$ and has order $k$, so that $G$ has elements of arbitrarily large order. However, if $\prod x_i \in G$ via the finite set $J$, then the integer $m = \prod_{j \in J} j$ exists and $m(\prod x_i) = 0$; thus every element of $G$ has finite order.

(3) Clearly $Z_2 \times \mathbb{Z}$ has this property.

(4) By Cayley’s theorem, we know that a finite group of order n is isomorphic to a subgroup of $S_n$. Moreover, $S_n$ is isomorphic to a subgroup of $S_\mathbb{N}$. Thus every finite group is isomorphic to a subgroup of $S_\mathbb{N}$.

(5) Let $G_i = G$ be a fixed group for each $i \in \mathbb{N}$, and define $$\theta : \prod_{\mathbb{N}} G_i \rightarrow (\prod_{\mathbb{N}} G_i) \times (\prod_{\mathbb{N}} G_i)$$ as follows: $((\theta(\prod g_i))_1)_k = g_{2k}$ and $((\theta(\prod g_i))_2)_k = g_{2k+1}$.

It is clear that $\theta$ is a homomorphism. Now if $\theta(\prod g_i) = 1$, then $g_k = 1$ for all $k \in \mathbb{N}$, so that in fact $\prod g_i = 1$; hence $\theta$ is injective. Finally, if $$(\prod g_i, \prod h_i) \in (\prod_{\mathbb{N}} G_i) \times (\prod_{\mathbb{N}} G_i),$$ then letting $k_i = g_{i/2}$ if $i$ is even and $k_i = h_{(i-1)/2}$ if $i$ is odd, we have $$\theta(\prod k_i) = (\prod g_i, \prod h_i).$$ Thus $\theta$ is surjective, hence an isomorphism.