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 ith 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.

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Comments

  • Gobi Ree  On December 30, 2011 at 10:18 pm

    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}.

    • nbloomf  On January 18, 2012 at 1:06 pm

      \overline{2} is still a generator (any nonidentity is), but I agree that \overline{1} is a better choice. Thanks!

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