How to determine whether $\mathbb Z_{p^m}$ is decomposable into (normal) subgroups

Question

QUESTION:

Consider the group $G = \mathbb Z_{p^m}$, where $p$ is prime and the integer $m \ge 2$. How can I determine whether or not $G$ is an internal direct sum of its (normal) subgroups?

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At least I assume the definition is equivalent to the following: there exists positive integer $k$ s.t. there exist proper, nontrivial, normal subgroups $N_1, \dotsc, N_k$ of $G$ where

1. $G$ is the join of $N_i$'s (i.e. $G$ is equal to the subgroup generated by the union of $N_i$'s) and
2. the subgroups satisfy the independence property that for each $j=1,\dotsc,k$, we have that $N_j \cap \bigvee_{i \ne j} N_i = \{0_G\}$, the singleton containing the identity $0_G$ of $G$.

• Idea 1: I think Keith Conrad (Eg 2.3) ('A cyclic group of prime-power order is indecomposable.') says precisely there does not exist a $k$, I believe. However, we're not yet allowed to use Cauchy's theorem (which Keith Conrad does)! See below for Keith Conrad (Eg 2.3)

  Example 2.3. A cyclic group of prime-power order is indecomposable. Let $A$ be cyclic of order $p^k$ where $k \geq 1$. If $A = B \oplus C$ where $B$ and $C$ are nontrivial subgroups of $A$ then $B$ and $C$ have $p$-power order greater than $1$ and thus $B$ and $C$ each contain a subgroup of order $p$ by Cauchy's theorem. That implies $A$ has more than one subgroup of order $p$, but in a cyclic group there is at most one subgroup per size. Thus $A$ is indecomposable.

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Update:

• Idea 1 (expanded): Actually, I don't have to use the full Cauchy's theorem (see wiki or Keith Conrad. in each link, there's a proof wherein the proof has like a case for Abelian).

1. There's a version of Cauchy's theorem for Abelian groups.

2. Better yet, I should recall that Cauchy's theorem is merely a partial converse to Lagrange's theorem. What's another partial converse to Lagrange's theorem? converse of Lagrange's theorem for finite cyclic groups! see 'proof of the converse of Lagrange’s theorem for finite cyclic groups'.

How we continue in Keith Conrad Example 2.3 is like we have that both $B$ and $C$ have respective subgroups of order $p$ $H_B$ and $H_C$. We have in turn that $H_B$ and $H_C$ are subgroups of order $p$ of the original $G$. We have that this violates either

1. (assuming they are distinct subgroups indeed) converse of Lagrange's theorem for finite cyclic groups again! Well at least the stronger version that says there is exactly 1 subgroup. Here it's just at least 1 subgroup.

2. (assuming they are equal) well $B$ and $C$ intersect trivially so if $H_B=H_C$, then $H_B=H_C=\{0_G\}$. But the trivial subgroup doesn't have order $p$ unless the group is trivial, which is of course not the case.

And then in my case here...I just choose we can think of $N_i$'s as any decomposition of $m$ by sums like $m=a+b$ or $m=a_1+a_2+...+a_k$ and the extend by the argument above.

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Please help for the non-(prime power) case: Whether or not a (cyclic and thus Abelian) group $\mathbb Z_{n \ne p^m}$ (not prime power) is decomposable into (normal) subgroups

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References.

1 What is internal direct sum or internal direct product in Dummit and Foote?

Answer 1

Apart from the unnecessary rant in the comments, I think now the question is quite simple to answer in simple terms.

I am using multiplicative notation because it is how one usually starts when studying finite groups, so the group $\mathbb Z_n$ is the group whose elements are $\{1, a, a^2, \ldots , a^{n-1}\}$ You only need to use the following "theorem":

If $n$ is a natural number, the group $G = \mathbb Z_n$ has one subgroup for each divisor $d$ of $n$, call it $G_d$, and it can be described explicitly as follows: $$ G_d = \{g \in G \mid g^d = 1\} $$ This can be proven only using Lagrange's theorem and a little bit of manipulations, but it is really easy and you should try to prove it by yourself.

Now, concerning the groups $\mathbb Z_{p^n}$. Note that $p^n$ has exactly $n+1$ divisors: $\{1, p, \ldots , p^n\}$. If $g \in G_{p^k}$, $g \in G_{p^{k+1}}$ (this is by the description of these groups, and therefore we have an ascending chain $$ \{1\} = G_1 \subset G_p \subset \ldots \subset G_{p^{n-1}} \subset G_{p^n} = G $$ This proves that the group is idecomposable: If you wanted to write G as the product of some $N_k$, then no $N_k$ can be $G$ (condition (2) would not be guaranteed), but then all of them are contained in $G_{p^{n-1}}$ and therefore, so is their join, and so condition (1) is not guaranteed.