Composition factor-equivalent not implies embeddable as normal subgroups in a finite group with isomorphic quotient groups
We can have two groups and that are composition factor-equivalent: in other words, they have the same list of composition factors (possibly in a different order) but such that they are not embeddable as normal subgroups in a finite group with isomorphic quotient groups.
- Complete and composition factor-equivalent not implies isomorphic: There can exist finite complete groups and that are composition factor-equivalent but not isomorphic.
- Equivalence of definitions of complete group: A group is complete if and only if whenever it is embedded as a normal subgroup in another group, it is actually a direct factor of that group.
- Direct product is cancellative for finite groups: If for finite groups , then .
By Fact (1), we have that there exist finite non-isomorphic complete groups and such that and are composition factor-equivalent. We prove that these and work for our purpose, i.e., we will show that they cannot be embedded as normal subgroups in the same group with isomorphic quotient groups.
Proof by contradiction
Assumption from which we derive the contradiction: Suppose there exists a group with a normal subgroup isomorphic to and a normal subgroup isomorphic to , such that .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Expanation|
|1||There exist subgroups (possibly equal, possibly distinct) such that is an internal direct product of and (and hence is isomorphic to the external direct product ) and is also an internal direct product of and (and hence is isomorphic to the external direct product ).||Fact (2)|
|2||and .||Step (1)||Follows from basic properties of internal direct products.|
|3||.||Step (2)||Step-given direct.|
|4||is isomorphic both to and . Thus, .||Steps (1), (3)||Use that to replace by in the expression .|
|5||.||Fact (3)||is finite.||Step (4)||direct.|
|6||, the desired contradiction.||We use that .||Step (5)||direct.|