# Showing that a cyclic group of prime power order has only 1 composition series

I am trying to show that a cyclic group of prime power order has only 1 composition series. Is the following correct?

Let $G=C_{p^n}$. Then as cyclic groups are abelian we have that there is a subgroup of every order dividing $p^n$, which is $1,p,p^2,...$, then these are all normal (as $G$ is abelian) and are cyclic and so are unique up to isomorphism.

So if we start of with any subnormal series $\displaystyle{\{1\}\triangleleft C_{p^i} \triangleleft C_{p^n}}$ then we can always refine it so that it is:

$\{1\}\triangleleft C_p\triangleleft C_{p^2}\triangleleft .....\triangleleft C_{p^m}$

-
That's the idea. Induction on the exponent is another possibility, but it is to an extent a matter of taste how you organize the proof. The key is (as you observed) that the cardinality of a subgroup determines it uniquely. – Jyrki Lahtonen Dec 22 '12 at 12:30
Yeah thats how it was done in my answers but I was wondering if this way was good too thanks :) – Joe cabel Dec 22 '12 at 12:32
@JyrkiLahtonen: Sorry for asking, but can we think of the number of composition series- for example- for $\mathbb Z_4\times\mathbb Z_9$ regarding to this problem? Thanks. – Babak S. Dec 22 '12 at 17:46

## 1 Answer

That looks correct. The facts you need are:

1. A (finite) cyclic group has at most one subgroup of a given order (or, to put it differently, any cyclic group has at most one subgroup of a given index).
2. A quotient of a cyclic group is cyclic.
3. A subgroup of a cyclic group is cyclic.
4. A nontrivial cyclic group is simple iff it is of prime order.
-