Normality is not transitive for any pair of nontrivial quotient groups

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Statement

Suppose A and B are (possibly equal) nontrivial groups. Then, there exist groups H \le K \le G such that all the following conditions are satisfied:

  • H is a Normal subgroup (?) of K and the quotient group K/H is isomorphic to A.
  • K is a normal subgroup of G and the quotient group is isomorphic to B.
  • H is not a normal subgroup of G.

Related facts

Similar facts

Generalizations

Proof

The construction is as follows. Let G be the wreath product of A and B for the regular group action of B. Let K be the subgroup A^B, i.e., the normal subgroup that forms the base of the semidirect product, and let H be the subgroup of K where a particular coordinate is the identity element. (If we are thinking of A^B as functions from B to A, then H can be taken as the subgroup comprising those functions that send the identity element of B to the identity element of A -- here, the particular coordinate becomes the coordinate corresponding to the identity element of B).

Thus, K is isomorphic to H \times A. Then:

  • H is normal in K and K/H is isomorphic to A: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
  • K is normal in G and G/K is isomorphic to B: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
  • H is not normal in G: The action of B on the coordinates in K = A^B is transitive on the coordinates, so H is not preserved under this action.