Semidirect product is not left-cancellative for finite groups

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Statement

Statement from the external semidirect product viewpoint

It is possible to have finite groups A,B,C, and semidirect products, such that:

A \rtimes B \cong A \rtimes C,

but B is not isomorphic to C.

Statement from the internal semidirect product viewpoint

We can construct a finite group P with subgroups H,K,L,M such that:

P = H \rtimes L = K \rtimes M,

where H is isomorphic to K but L is not isomorphic to M.

Related facts

Related facts about direct products and other notions of products and extensions

Other related facts about complements

Proof

Example involving the upper triangular matrices

Suppose p is any prime, and let G := U(5,p) be the group of upper-triangular unipotent 5 \times 5 matrices over the field of p elements. Let P be the subgroup of G comprising those matrices where the (12)^{th} entry is zero. Then, P is a group of order p^9.

By fact (1), we have that G has two subgroups that are Abelian of maximum order: the top right rectangle groups of dimensions 2 \times 3 and 3 \times 2 respectively. Call these subgroups H and K respectively. Then, observe that:

  • Both H and K are also Abelian subgroups of maximum order in P. Moreover, they are the only Abelian subgroups of maximum order in P since they are the only Abelian subgroups of maximum order in G.
  • H and K are isomorphic -- in fact, they are conjugate subgroups inside the bigger group GL(5,p). This conjugation restricts to an automorphism of G, but not of P.
  • Both H and K are normal in G, and hence in P.
  • The subgroup with nonzero entries in the (34),(35),(45) positions is a permutable complement to H in P. Call this subgroup L. Then P is an internal semidirect product of H and L. L is isomorphic to the prime-cube order group:U3p.
  • The subgroup with nonzero entries in the (13),(23),(45) positions is a permutable complement to K in G. Call this subgroup M. Then, P is an internal semidirect product of K and M. Note that M is isomorphic to the elementary Abelian group of order p^3.
  • Thus, we have P = H \rtimes L = K \rtimes M, with H \cong K but L not isomorphic to M.