Direct product of groups is essentially commutative

Let A and B be groups. Prove that A \times B \cong B \times A.


We know from set theory that the mapping \varphi : A \times B \rightarrow B \times A given by \varphi((a,b)) = (b,a) is a bijection with inverse \psi : B \times A \rightarrow A \times B given by \psi((b,a)) = (a,b). Also \varphi is a homomorphism, as we show.

Let a_1, a_2 \in A and b_1, b_2 \in B. Then \varphi((a_1,b_1) \cdot (a_2,b_2)) = \varphi((a_1a_2, b_1b_2)) = (b_1b_2, a_1a_2) = (b_1,a_1) \cdot (b_2,a_2) = \varphi((a_1,b_1)) \cdot \varphi((a_2,b_2)).

Hence A \times B \cong B \times A.

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Comments

  • muhammad izhar  On January 4, 2011 at 11:09 pm

    great sir!!!!!!!! i learnt many a things here
    really great………

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