The center of a direct product is the direct product of the centers

Show that the center of a direct product is the direct product of the centers: Z(G_1 \times \cdots \times G_k) = Z(G_1) \times \cdots Z(G_k). Deduce that a direct product of groups is abelian if and only if each of the factors is abelian.


We proceed by induction on k.

For the base case k=2, let G and H be groups. (\subseteq) Let (g,h) \in Z(G \times H). Then for all (a,b) \in G \times H, (ga, hb) = (g,h)(a,b) = (a,b)(g,h) = (ag,bh). Thus ag = ga and bh = hb for all a \in G and b \in H; hence g \in Z(G) and h \in Z(H). Thus (g,h) \in Z(G) \times Z(H). (\supseteq) Let (g,h) \in Z(G) \times Z(H). Now for all a \in G and h \in H, (g,h)(a,b) = (ga,hb) = (ag,bh) = (a,b)(g,h). Thus (g,h) \in Z(G \times H). Thus Z(G \times H) = Z(G) \times Z(H).

Now suppose the conclusion holds for any product of k groups for some k \geq 2. Then Z(\times_{i=1}^{k+1} G_i) = Z(\times_{i=1}^k G_i \times G_{k+1}) = Z(\times_{i=1}^k G_i) \times Z(G_{k+1}) = \times_{i=1}^k Z(G_i) \times Z(G_{k+1}) = \times_{i=1}^{k+1} G_i. By induction, the result holds for any finite direct product.

Now let G = \times_{i=1}^k G_i be a finite direct product of groups.

If each factor G_i is abelian, then Z(G) = Z(\times_{i=1}^k G_i) = \times_{i=1}^k Z(G_i) = \times_{i=1}^k G_i = G; hence G is abelian.

If G is abelian, then \times_{i=1}^k Z(G_i) = Z(\times_{i=1}^k G_i) = Z(G) = G = \times_{i=1}^k G_i. If we let \pi_i denote the ith coordinate projection, we have G_i = \pi_i[G] = Z(G_i). Hence G_i is abelian for each i.

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