Basic properties of n-ary direct products of groups

Let I be a set and let G_i be a group for each i. Let G = \prod_I G_i.

  1. For each k \in I, define \iota_k : G_i \rightarrow G by (\iota_k(g))_i = g if i=k and 1 otherwise. Then \iota_k is an injective homomorphism, \mathsf{im}\ \iota_k \leq G is normal, and G/\mathsf{im}\ \iota_k \cong \prod_{I \setminus \{k\}} G_i.
  2. For each k \in I, define \pi_k : G \rightarrow G_k by \pi_k(\prod g_i) = g_k. Then \pi_k is a surjective homomorphism with \mathsf{ker}\ \pi_k \cong \prod_{I \setminus \{k\}} G_i.
  3. If x \in G_i and y \in G_j for some i \neq j, then \iota_i(x) \iota_j(y) = \iota_j(y) \iota_i(x).

  1. First we show that \iota_k is an injective homomorphism. If g,h \in G_k, then (\iota_k(gh))_i = gh if i=k and 1 otherwise. In either case, we have (\iota_k(gh))_i = (\iota_k(g))_i (\iota_k(h))_i. Thus \iota_k is a homomorphism. Now suppose g \in \mathsf{ker}\ \iota_k. Then \iota_k(g) = \prod 1_i, so that in particular g = 1. Thus \mathsf{ker}\ \iota_k = 1, hence \iota_k is injective.

    Now suppose (\prod g_i) \in G and (\prod a_i) \in \mathsf{im}\ \iota_k. Now (\prod g_i)(\prod a_i)(\prod g_i)^{-1} = \prod (g_ia_ig_i^{-1}), and if i \neq k, we have g_ia_ig_i^{-1} = 1. Then (\prod g_i)(\prod a_i)(\prod g_i)^{-1} = \iota_k(g_ka_kg_k^{-1}), hence \mathsf{im}\ \iota_k is normal in G.

    Now define a mapping \theta : G \rightarrow \prod_{I \setminus \{k\}} G_i by (\theta(\prod g_i))_j = g_j. Now for all j \in I \setminus \{k\}, (\theta((\prod g_i)(\prod h_i)))_j = (\theta(\prod g_ih_i))_j = g_jh_j = (\theta(\prod g_i))_j (\theta(\prod h_i))_j, so that \theta is a group homomorphism. If \prod h_i \in \prod_{I \setminus \{k\}} G_i, then \prod g_i \in G defined by g_i = h_i if i \neq k and g_k = 1 maps to \prod h_i under \theta; hence \theta is surjective. Finally, we see that \mathsf{ker}\ \theta = \mathsf{im}\ \iota_k as follows. (\subseteq) If \theta(\prod g_i) = 1, then g_i = 1 for all i \neq k. Thus \prod g_i = \iota_k(g_k). (\supseteq) If g \in G_k, then (\theta(\iota_k(g)))_j = 1 for all j \neq k; hence \iota_k(g) \in \mathsf{ker}\ \theta. Finally, by the First Isomorphism Theorem, we have G/\mathsf{im}\ \iota_k \cong \prod_{I \setminus \{k\}} G_i.

  2. First we show that \pi_k is a surjective homomorphism. We have \pi_k((\prod g_i)(\prod h_i)) = \pi_k(\prod g_ih_i) = g_kh_k = \pi_k(\prod g_i) \pi_k(\prod h_i). Thus \pi_k is a homomorphism. Now let h \in G_i, and define \prod g_i \in G by g_i = h if i = k and 1 otherwise. Clearly then \pi_k(\prod g_i) = h; hence \pi_k is surjective.

    Now define \tau : \mathsf{ker}\ \pi_k \rightarrow \prod_{I \setminus \{k\}} G_i as follows: (\tau(\prod g_i))_j = g_j. This \tau is clearly a group homomorphism. Now if \prod h_i \in \prod_{I \setminus \{k\}} G_i, then \prod g_i \in G defined by g_i = 1 if i=k and h_i otherwise is in \mathsf{ker}\ \pi_k, and moreover \tau(\prod g_i) = \prod h_i. Thus \tau is surjective. Finally, if \prod g_i \in \mathsf{ker}\ \tau$, then g_i = 1 for all i \neq k, and (by definition) g_k = 1. Thus \prod g_i = 1, so that \tau is injective. Thus \mathsf{ker}\ \pi_k \cong \prod_{I \setminus\{k\}} G_i.

  3. Let g \in G_i and h \in G_j where i \neq j. Then (\iota_i(g) \iota_j(h))_k is g if k=i, h if k=j, and is 1 otherwise. Similarly for (\iota_j(h) \iota_i(g))_k; thus these elements are equal.
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