If you find any mistakes, please make a comment! Thank you.

Basic properties of the direct sum of groups

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.17


(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.


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu