# A Group with no Composition Series

## A Group with no Composition Series

### Spookiness in Group Theory

The Jordan-Holder theorem is a group-theoretic analogue of the uniqueness of prime factorizations. A “factorization” of a (finite) group is called a composition series.

Definition. Let $G$ be a group. A composition series of $G$ is a chain of normal subgroups

$1= N_0 \trianglelefteq N_1 \trianglelefteq ... \trianglelefteq N_k = G$

where $N_{i+1}/N_i$ is simple.

There is a theorem stating that composition series of finite groups are unique.

Theorem (Jordan-Holder). Let $G$ be a finite group. Consider two composition series

$1= N_0 \trianglelefteq N_1 \trianglelefteq ... \trianglelefteq N_k = G$

and

$1= M_0 \trianglelefteq M_1 \trianglelefteq ... \trianglelefteq M_l = G$

of $G.$ Then, $k=l$ and the list $\{N_1/N_0,...,N_k/N_k/N_{k-1}\}$ is a permutation of $\{M_1/M_0,...,M_k/M_{k-1}.\}$

So, a composition series tells us a lot about a group. Therefore, we like them. Lucky for us then, there is a theorem stating that all finite groups admit a composition series. But, this is not true for infinite groups.

This bring us to our counterexample: $\mathbb{Z}$ has no composition series. To see this, think about what the first inclusion $N \trianglelefteq \mathbb{Z}$ must be (hint: the quotient must be a finite simple group). Those of us reading this post any time after 2004 have the privilege of being able to say that there is a list of all finite simple groups; only $\mathbb{Z}/p\mathbb{Z}$ will do here.

This yields $p\mathbb{Z} \trianglelefteq \mathbb{Z}.$ Contining forces an infinite regress

$...\mathbb{Z}/p^3\mathbb{Z} \trianglelefteq \mathbb{Z}/p^2\mathbb{Z} \trianglelefteq \mathbb{Z}/p\mathbb{Z} \trianglelefteq \mathbb{Z}$