Every group of given order is a permutable complement for symmetric groups

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Symbolic statement

Let G be a finite group of order n. Then the following are true:

  • Via the regular representation, we can view G as a subgroup of the symmetric group on n elements (this is the idea behind Cayley's theorem).
  • Under this representation, G is a permutable complement to the symmetric group on any n-1 of the elements

In particular this means that every isomorphism class of a group of order n occurs as a permutable complement of S_{n-1} in S_n. Since there can exist non-isomorphic groups of the same order, this shows that the isomorphism type of the permutable complement of S_{n-1} in S_n is not uniquely determined.


In fact, define the following equivalence relation on groups:

Two groups K_1 and K_2 are equivalent if there is a group G with embeddings of K_1 and K_2 in G, such that both K_1 and K_2 have a common permutable complement.

A priori, it is clear that two finite groups can be equivalent only if they have the same order. The above result shows that the converse is also true: any two finite groups that have the same order are equivalent.

Related facts

Facts used

  1. Cayley's theorem
  2. Group equals product of transitive subgroup and isotropy of a point (get more related information at proving product of subgroups)


Given: A group G of order n.

To prove: G is isomorphic to a permutable complement to the symmetric group on \{ 1,2,3, \dots, n-1 \} (which is the isotropy of the point \{ n \}, in the symmetric group on \{ 1,2,3, \dots, n \}.


  1. By Cayley's theorem, G embeds inside \operatorname{Sym}(G) -- the group of permutations on the underlying set of G, via the action by left multiplication.
  2. Under this embedding, every non-identity element of G acts as a fixed point-free permutation. In particular, the intersection of G with the isotropy subgroup of any point is trivial.
  3. Also, the action of G on itself by left multiplication is transitive. Thus, by fact (2), \operatorname{Sym}(G) is the product of the subgroup G and the isotropy of any point.
  4. Combining steps (2) and (3), we see that G is a permutable complement to the isotropy of any point in \operatorname{Sym}(G).
  5. A set-theoretic bijection between G and the set \{ 1,2,3,\dots,n \} thus shows that G is isomorphic to a permutable complement to the symmetric group on \{ 1,2,3,\dots,n-1\} in \{ 1,2,3,\dots,n \}.