Finite not implies composition factor-permutable

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., finite group) need not satisfy the second group property (i.e., composition factor-permutable group)
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Statement

We can have a finite group $G$ such that not every ordering of its composition factors arises from a composition series.

Proof

Example of the symmetric group

Further information: symmetric group:S3

Let $G$ be the symmetric group on the three-element set $\{ 1,2,3\}$. $G$ has only one proper nontrivial normal subgroup: the group $H$ generated by a 3-cycle. $H$ is also the alternating group on three letters. We have:

$H = \{ (1,2,3), (1,3,2), () \}$.

Thus, the only possible composition series for $G$ is:

$\{ () \} \le H \le G$.

The composition factors for this (from left to right) are the cyclic groups of order three and two respectively. Since this is the only composition series, it is in particular impossible to have a composition series where the first composition factor is cyclic of order two and the second composition factor is cyclic of order three.