Let $S$ be a submodule of an $R$-module $M$. A submodule $C⊆M$ is said to be a complement to $S$ (in $M$) if $C$ is maximal with respect to the property that $C∩S=0$. (This does exist by Zorn Lemma.) I search for a complement in $M$ (a submodule which is a complement to a submodule of $M$) that is not a direct summand in $M$. Over $R=\mathbb Z$, I considered the module $M=C⊕S$ where $C=\langle c\rangle$ has order $8$ and $S=\langle s\rangle$ has order $2$. My question is:
Could $C_1=\langle c_1\rangle$ of order 4 generated by $c_1=2c+s$ be a candidate for a complement of $M$ which is not a direct summand there of?
I would appreciate for any help!
$C'$ is not a direct summand in $M$: if there is $C''$ such that $C'+C''=M$ and $C'\cap C''=0$, then $|C''|=4$, so $C''$ is either cyclic or Klein.
Since the elements of order four are $(2,0)$, $(2,1)$, $(6,0)$, respectively $(6,1)$ it's clear that $C''=\langle(2,0)\rangle=\langle(6,0)\rangle$. But $|C'+C''|=8$, so $C'+C''\ne M$.
When $C''$ is of Klein type, then it contains three elements of order two, that is, $(0,1)$, $(4,0)$ and $(4,1)$. But $(4,0)\in C'$, a contradiction.
$C'$ is a complement in $M$: if $C'\subsetneq C''$ such that $C''\cap S=0$, then $C''\oplus S=M$ (note that $|C''|=8$). So $C''\simeq\mathbb Z/8\mathbb Z$ and thus $(2,1)\in 2C''$ (note that the order of $(2,1)$ is four). It follows that $(2,1)\in2M=2C$, a contradiction.