Consider the group $\Bbb Z_8\oplus\Bbb Z_4\oplus\Bbb Z_2$. How many nontrivial cyclic groups does it contain? I know the answer is $27$, so my question is how to get it effectively. I was trying to solve it by solving the system of modular equations: find all n for which there are such a, b, c in $\Bbb Z_8, \Bbb Z_4, \Bbb Z_2$ respectively that satisfy:
$\\ an (mod 8)=0,\\ bn (mod 4)=0,\\ cn (mod 2)=0,\\$
but I'm not sure if it is really effective. Also it would be great to see if this task is solvable by using Young tableaux.
The answer is $28$. Let $1,1,1$ be the additive generators of the three direct factors. Then every cyclic subgroup is generated by a triple $(m, n, k)$ where $m\in\{0,...,7\}, n\in\{0,1,2,3\}, k\in\{0,1\}$. If $gcd(m,8)=1$, then some odd multiple of $(m,n, k)$ is of the form $(1,n',k')$ and is generating the same cyclic subgroup of order $8$. So we can assume that in this case $m=1$. The choices for $n,k$ are arbitrary: $4\cdot 2=8$ cyclic subgroups of order $8$.
If $gcd(m,8)=2$, then some odd multiple of $(m,n,k)$ is of the form $(2,n',k')$ and is generating the same cyclic subgroup of order $4$. So we can assume that in this case $m=2$. The choices for $n,k$ are again arbitrary, $8$ cyclic subgroups of that form.
If $m=4$, then we can assume that $n=1,2$ or $0$,arbitrary choices of $k$ give different cyclic subgroups, $6$ cyclic subgroups of this form.
Finally if $m=0$, we have to look at $n$. If $n=1,3$ we can assume $n=1$, we have $2$ such cyclic subgroups generated by $(0,1,1),(0,1,0)$. If $n=2$ we again have two cyclic subgroups, $k=0,1$, generated by $(0,2,0), (0,2,1)$ and if $n=0$, we have two cyclic subgroups, one of which is trivial. So altogether there are $8*2+6*2=28$ cyclic subgroups including the trivial one.