The quotient of a product by a product is isomorphic to the product of quotients

Let A and B be groups, with C \leq A and D \leq B normal. Prove that (C \times D) \leq (A \times B) is normal and that (A \times B)/(C \times D) \cong (A/C) \times (B/D).

We have already seen that C \times D is a subgroup. Now let (a,b) \in A \times B. we have (a,b)(C \times D) = aC \times bD = Ca \times Db = (C \times D)(a,b), hence C \times D is normal.

Define \varphi : A \times B \rightarrow (A/C) \times (B/D) by (a,b) \mapsto (aC,bD). Clearly \varphi is surjective. We now show that \mathsf{ker}\ \varphi = C \times D. If (a,b) \in \mathsf{ker}\ \varphi, we have (aC,bD) = (C,D), hence a \in C and b \in D. Thus (a,b) \in C \times D. If (a,b) \in C \times D, then (aC,bD) = (C,D). Thus \mathsf{ker}\ \varphi = C \times D, and by the First Isomorphism Theorem, (A \times B)/(C \times D) \cong (A/C) \times (B/D).

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  • Jer  On November 14, 2010 at 9:25 pm

    Where have we already seen that (C x D) is a subgroup. I know I did. I just don’t remember where. Thanks

    • nbloomf  On November 15, 2010 at 9:25 am

      I can’t recall which problem this was in, but it’s not so hard to see using the subgroup criterion: C \times D contains (1,1) and thus is not empty, and if (a_1,b_1),(a_2,b_2) \in C \times D, then (a_1,b_1)(a_2,b_2)^{-1} = (a_1a_2^{-1},b_1b_2^{-1}) \in C \times D.

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