Show that $\mathbb{Z}$ has no composition series.

Question

Please, read the whole post before trying to answer.

Remark: here $\subset$ means "strict inclusion".

I need to prove that the group $\mathbb{Z}$ has no composition series. That is, no normal series of subgroups $d_i\mathbb{Z}$ such that

$$\{ 0 \} = d_0 \mathbb{Z} \subset ... \subset d_n \mathbb{Z} = \mathbb{Z},$$

for some $n \in \mathbb{N}$. Since $\mathbb{Z}$ is abelian, all its subgroups are normal, so any series of subgroups is normal. Let us have a series of a finite length

$$\{ 0 \} = d_0 \mathbb{Z} \subset ... \subset d_n \mathbb{Z} = \mathbb{Z},$$

of some subgroups of $\mathbb{Z}$ (such that inclusions are well-defined).

We need to prove that $\frac{d_{i+1}\mathbb{Z}}{d_i\mathbb{Z}}$ is not simple for any subgroups of $\mathbb{Z}$ such that $d_i \mathbb{Z} \subset d_{i+1} \mathbb{Z}$.

First question. Is it the right way to reformulate the task?

Next: $\frac{d_{i+1}\mathbb{Z}}{d_i\mathbb{Z}}$ is nor simple if there is a normal subgroup $\frac{k\mathbb{Z}}{d_i\mathbb{Z}}$ of $\frac{d_{i+1}\mathbb{Z}}{d_i\mathbb{Z}}$ such that $d_i \mathbb{Z} \subset k \mathbb{Z} \subset d_{i+1} \mathbb{Z}$.

Second question. is is true that $K/H$ is normal in $P/H$ iff $K$ is normal in $P$?

I suppose yes, but would like to hear the confirmation.

If the statement above is true, than it suffices to show that $$\forall \ \ n,m \in \mathbb{N}: n \mathbb{Z} \subset m \mathbb{Z} \ \ \exists k \in \mathbb{N}: n \mathbb{Z} \subset k \mathbb{Z} \subset m \mathbb{Z}.$$

Third question. How can we show it?

Assuming the answer to the previous questions is "yes".

Answer 1

Hint: every proper nonzero subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$

About your points.

A quotient $a\mathbb{Z}/b\mathbb{Z}$ can be simple, precisely when $a$ divides $b$ and $b/a$ is prime.

It is true that, assuming $K$ is a subgroup of $P$ and $H$ is normal in $P$, then $K/H$ is normal in $P/H$ if and only if $K$ is normal in $P$. This is unimportant here, because we're in an abelian group.

Answer 2

Your first reformulation of the problem is not correct; it suffices to show that one of the quotients is not simple, not all of them.

It is true that $K/H$ is normal in $P/H$ if and only if $K$ is normal in $P$. But you don't need this fact because all groups concerned are abelian, so all subgroups in all quotients are normal.

The last statement you wish to prove is false; take $m=1$ and $n=2$. Here's another approach:

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Suppose there exists a composition series $$\{0\}=d_0\Bbb{Z}\subset\ldots\subset d_n\Bbb{Z}=\Bbb{Z}.$$ The fact that $d_{i+1}\Bbb{Z}/d_i\Bbb{Z}$ is simple means it is of prime order, so $d_i=p_id_{i+1}$ for some prime $p_i$. Then $\prod_{i=0}^{n-1}p_i=d_0=0$, a contradiction.