Homomorphism between groups of coprime order is trivial

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Suppose G,H are finite groups such that their orders are relatively prime. Then, the only homomorphism between G and H is the trivial map: the map sending every element of G to the identity element of H.

Related facts

  • Subgroup of least prime index is normal: This uses a similar idea -- except that the two groups in question are not of relatively prime orders; rather, the greatest common divisor of their orders is a prime number, which is forced to be the order of the image.
  • Any homomorphism from a simple group to any group is either trivial or injective.


  • Normal Hall implies order-unique: If H is a normal Hall subgroup of a finite group G, i.e., the order and index of H are relatively prime, then it is the unique subgroup of that order.

Facts used

  1. Order of quotient group divides order of group: Given a homomorphism \varphi:G \to H of groups, the order of the image \varphi(G) (which is a subgroup of H) divides the order of G.
  2. Lagrange's theorem: The order of any subgroup of a group divides the order of the group.


Given: Groups G and H of relatively prime order. A homomorphism \varphi:G \to H.

To prove: \varphi is the trivial map.


  1. The order of \varphi(G) divides the order of G: This follows from fact (1).
  2. The order of \varphi(G) divides the order of H: This follows from fact (2).
  3. The order of \varphi(G) is 1: This follows from the previous two steps, and the fact that the orders of G and H are relatively prime.
  4. \varphi(G) is the trivial subgroup and \varphi is the trivial map : Any subgroup of order one is trivial, so \varphi(G) is the trivial subgroup, i.e., it comprises the identity element of H. Thus, the map \varphi must send every element of G to the identity element of H.