## Basic properties of the direct sum of groups

Let $I$ be any nonempty set and let $G_i$ be a group for each $i \in I$. The restricted direct product or direct sum of the groups $G_i$ is the subset $H \leq G = \prod_I G_i$ of elements $\prod g_i$ such that $g_i = 1$ for all but a finite subset $J \subseteq I$.

1. Prove that $H$ is a subgroup of $G$.
2. Prove that $H$ is normal in $G$.
3. Let $I = \mathbb{N}^+$ and let $p_i$ be the $i$th integer prime. Show that if $G_i = \mathbb{Z}/(p_i)$ for all $i \in I$, then every element of the restricted direct product of the $G_i$ has finite order but that $\prod_I G_i$ has elements of infinite order. Show that in this example the restricted direct product is the torsion subgroup of the direct product. (See this previous exercise.)

1. Note that $\prod 1 \in H$, where we may take $J = \emptyset$. Thus $H$ is nonempty. Now suppose $\prod x_i$ and $\prod y_i$ are in $H$ via the finite subsets $J_x, J_y \subseteq I$. Note that $J_x \cup J_y$ is finite, and that if $k \notin J_x \cup J_y$, then $x_ky_k^{-1} = 1$. Thus $(\prod x_i)(\prod y_i)^{-1} \in H$ via $J_x \cup J_y$. By the Subgroup Criterion, $H \leq G$.
2. Let $\prod h_i \in H$ via the finite set $J$ and let $\prod g_i \in G$. Now $(\prod g_i)(\prod h_i)(\prod g_i)^{-1} = \prod g_ih_ig_i^{-1}$. Note that for $k \notin J$, $g_kh_kg_k^{-1} = 1$. Thus $\prod g_ih_ig_i^{-1} \in H$ via $J$; hence $H$ is normal in $G$.
3. Let $\prod g_i$ be in the restricted direct product via a finite set $J$. Define $k = \prod_J p_j$, where this product is simply multiplication in the integers. (Since $J$ is finite, this product exists.) Note that $g_ki = 0$ if $i \in J$ since $p_i$ divides $k$, and $kg_i = 0$ if $i \notin J$ since $g_i = 0$. Thus $k(\prod g_i) = 0$, and $\prod g_i$ has finite order.

On the other hand, consider $\prod g_i$ in the direct product defined by $g_i = \overline{1}$. Now for all odd primes $p_i$, $\overline{1}$ generates $\mathbb{Z}/(p_i)$, so that $\overline{1}$ has order $p_i$. Then $\prod \overline{1}$ cannot have finite order, since for all integers $k$, some component of $k(\prod \overline{1})$ is not 0; for instance, take any $i$ such that $p_i > k$.

Now the first part of this problem showed that the restricted direct product is contained in the torsion subgroup. Suppose now that $\prod g_i$ is in the torsion subgroup of $\prod_I G_i$. Then for some integer $k$, we have $k(\prod g_i) = \prod kg_i = 1$; hence $kg_i = 1$ for all $g_i$. Now if $i > k$ and $g_i \neq 0$, then $kg_i \neq 0$ since $p_i > k$ and $g_i$ has order $p_i$. Thus for all $i > k$, we have $g_i = 0$. In particular, $g_i = 0$ for all but finitely many $i$, namely the set $J = \{ 1 \leq i \leq k \}$. Thus $\prod g_i$ is in the restricted direct product, and in this case the restricted direct product and the torsion subgroup coincide.

Why did you choose $\overline{2}$ as an example? I thought that $\mathbb{Z}/(p_i)$ is a additive group, so the generator should be $\overline{1}$.
$\overline{2}$ is still a generator (any nonidentity is), but I agree that $\overline{1}$ is a better choice. Thanks!