Question concerning cyclic modules and modules where the socle of $M$ is irreducible

Question

Let $R$ be a ring and $M$ be an $R$-module of finite length $n$ (given by the number of composition factors). I am trying to prove or disprove:

1. If $\mathrm{Soc}(M)$ (the socle of $M$) is a unique irreducible module, then $M$ is cyclic.

2. The converse statement to 1.

3. If $\mathrm{Soc}(M)$ and $M$ is cyclic and $M$ is of finite length, then every submodule of $M$ is cyclic and of finite length

I have been trying to prove/disprove these things but keep running around in circles. Would I need to put extra conditions on R or M to make the above statements true? From what I found here, maybe I need In addition that all composition factors are pairwise non isomorphic? Thanks!

My attempt at a solution

1. For 1, we could induct on the length $n$. The statement is clearly true for $n=1,n=2$. So suppose that statement is true for $k< n$ and let $M$ be finite length $n$. Then since any two submodules have a nontrivial intersection, we know that $M$ is indecomposable, and so there can only be up to $n-1$ distinct propor submodules of length $n-1$, and (more generally, at most $C(n-1,k)$ distinct submodules of length $n-k$, althought not sure to show this). By the inductive hypothesis, each of these modules of length $n-1$ are cyclic. If there is only one such unique submodule of length $n-1$, then any element of $M$ not in that submodule would generate $M$. Now suppose there is more than one distinct submodule of length $n-1$. Assuming everything up to this point is okay, here is where I am unsure. I think that $M$ should be cyclic and can be generated by a vector that is the sum of generators for each of these up cyclic modules of length $n-1$. On the other hand, maybe that is wrong and you should take the sum of pullbacks of generators for each of each of the composition factors that could correspond to modules of the form $M/K$, where $K$ is a submodule of length $n-1$.

Answer 1

1. Does not hold: A bit involved (at least for my standards) example is as follows: take a local Gorenstein Artinian ring $R$. Then it holds that $\dim_k \operatorname{Soc} R = 1$ (where $k=R/\mathfrak{m}$ is the residue field), and whenever $I$ is a nonzero ideal of $R$, we have $0 \neq \operatorname{Soc} I \subseteq \operatorname{Soc} R$. So one only needs to consider a non-principal ideal $I \subseteq R$ as a counterexample. (I will update this if I think of something more reasonable, i.e. less involved.) In more concrete terms, I think that $$R:=k[X, Y, Z]/(X^2, Y^2, XZ, YZ, Z^2-XY), \; M=I=\mathfrak{m}=(X, Y, Z)$$ fits into this picture.

2. This does not hold, and the counterexample is easier. Take $\mathbb{Z}/6\mathbb{Z}\simeq \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$ as a $\mathbb{Z}$-module. Alternatively, if you don't want your socle to be even cyclic, you can take $M=R=k[X, Y]/(X^2, XY, Y^2)$. Then $\dim_k \operatorname{Soc}R=2$, and so the socle is $2$-generated, even though $M=R$ is clearly cyclic.

3. If I didn't make a mistake in 1., I guess you may use it to disprove this as well: just take $M:=R$, and then the ideal $I$ as described above gives you a counterexample.