I was interesed in finding subgroups of $G= \mathbb{Z}/{4\mathbb{Z}} \times \mathbb{Z}/{2\mathbb{Z}}$.

To count the cylic ones I know from experience that I can use the formula $\frac{x \in G : \,\text{ord}(x)=n}{\phi(n)}$. (Where $\text{ord}(x)=\text{mcm}[\text{ord}(a),\text{ord}(b)]$, with $a \in \mathbb{Z}/{4\mathbb{Z}}, b \in \mathbb{Z}/{2\mathbb{Z}}$).

So I think (if I'm not mistaken) there should be $6$ cyclic subgroups ($2$ of order $4$, $3$ of order $2$ and the trivial one).

I'd like to know how to precisely find the others, the non-cyclic ones, because I think at least the Klein four-group is in there but I'd like in general, given other groups, to find them all.

Any tips or help would be appreciated, thanks anyway.


Suppose this non-cyclic subgroup is generated by elements (a,0), (b,1) (in other cases it would be cyclic, also any non-cyclic subgroup of a two-generated abelian group will be two-generated) and $a$ is nonzero. If $a$ generates $\mathbb{Z}_4$, then $\langle(a, 0), (b, 1)\rangle = \mathbb{Z}_4 \times \mathbb{Z}_2$. If $b$ generates $\mathbb{Z}_4$ and $a$ does not, then $\langle(a, 0), (b, 1)\rangle$ is cyclic. All other situations can be listed further: $\langle(2,0), (0, 1)\rangle$ and $\langle(2,0), (0, 1)\rangle$ both generate the same subgroup isomorphic to Klein four-group (which appears to be the only non-cyclic proper subgroup)

  • $\begingroup$ How much this can be generalized ? $\endgroup$ – jacopoburelli Jul 14 '18 at 17:52
  • $\begingroup$ @jacopoburelli, this proof seems to be valid for all groups of the type $\mathbb{Z}_{p^2} \times \mathbb{Z}_{p}$ where $p$ is prime. More generally a similar method can be used for $\mathbb{Z}_{p^k} \times \mathbb{Z}_{p}$ where $p$ is prime and $k$ is arbitrary, but the things become more complicated, as for larger $k$ there will be larger number of different cases to analyse... $\endgroup$ – Yanior Weg Jul 14 '18 at 18:00

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