# Composition factor-equivalent not implies embeddable as normal subgroups in a finite group with isomorphic quotient groups

## Statement

We can have two groups $G$ and $H$ 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.

## Facts used

1. Complete and composition factor-equivalent not implies isomorphic: There can exist finite complete groups $G$ and $H$ that are composition factor-equivalent but not isomorphic.
2. 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.
3. Direct product is cancellative for finite groups: If $X \times Y \cong X \times Z$ for finite groups $X,Y,Z$, then $Y \cong Z$.

## Proof

### Construction

By Fact (1), we have that there exist finite non-isomorphic complete groups $G$ and $H$ such that $G$ and $H$ are composition factor-equivalent. We prove that these $G$ and $H$ 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.

Assumption from which we derive the contradiction: Suppose there exists a group $A$ with a normal subgroup $B$ isomorphic to $G$ and a normal subgroup $C$ isomorphic to $H$, such that $A/B \cong A/C$.
1 There exist subgroups (possibly equal, possibly distinct) $D,E \le A$ such that $A$ is an internal direct product of $B$ and $D$ (and hence is isomorphic to the external direct product $B \times D$) and is also an internal direct product of $C$ and $E$ (and hence is isomorphic to the external direct product $C \cong E$). Fact (2)
2 $D \cong A/B$ and $E \cong A/C$. Step (1) Follows from basic properties of internal direct products.
3 $D \cong E$. $A/B \cong A/C$ Step (2) Step-given direct.
4 $A$ is isomorphic both to $B \times D$ and $C \times D$. Thus, $B \times D \cong C \times D$. Steps (1), (3) Use that $D \cong E$ to replace $E$ by $D$ in the expression $C \times E$.
5 $B \cong C$. Fact (3) $A$ is finite. Step (4) direct.
6 $G \cong H$, the desired contradiction. We use that $B \cong G, C \cong H$. Step (5) direct.