Subgroup of $\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z}$.

Question

I want to check my solution of this (simple) problem: find all subgroups $H$ of $\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z}$, such that $|H|=36$.

My attempt: $|(\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z})/H|=3$, so $$ (\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z}) \cong \mathbb{Z}/3\mathbb{Z}; $$ using the correspondence theorem I can calculate the subgroups of $\mathbb{Z}/3\mathbb{Z}$, that is only $\{0\}$.

In $\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z}$ not exixts an element of order $36$, so the unique subgroup is $\mathbb{Z}/9\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$.

Do I do some mistakes?

Answer 1

The subgroups of $\mathbb{Z}/n\mathbb{Z}$ are all of the form $m(\mathbb{Z}/n\mathbb{Z})$ where $m \mid n$.

For example, the subgroups of $\mathbb{Z}/6\mathbb{Z}= \{\bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5\}$ are

\begin{align} 1(\mathbb Z/6\mathbb Z) &= \{\bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5\} \\ 2(\mathbb Z/6\mathbb Z) &= \{\bar 0, \bar 2, \bar 4 \} \\ 3(\mathbb Z/6\mathbb Z) &= \{\bar 0, \bar 3 \} \\ 6(\mathbb Z/6\mathbb Z) &= \{\bar 0 \} \\ \end{align}

Note also that $| m(\mathbb{Z}/n\mathbb{Z}) | = \dfrac nm$.

The divisors of $6$ are $m \in \{1,2,3,6\}$ and the divisors of $18$ are $n \in\{1, 2, 3, 6, 9, 18\}$

If you want $|m(\mathbb{Z}/6\mathbb{Z}) \times n(\mathbb{Z}/18\mathbb{Z})| = 36$ then you need to find all $m$ and $n$ such that $\dfrac 6m \cdot \dfrac{18}{n} = 36$, which simplifies to $mn = 3$.

So your subgroups are

$1(\mathbb{Z}/6\mathbb{Z}) \times 3(\mathbb{Z}/18\mathbb{Z})$

$3(\mathbb{Z}/6\mathbb{Z}) \times 1(\mathbb{Z}/18\mathbb{Z})$

This can be "simplified" to

$ \mathbb{Z}/6\mathbb{Z}\times \{\bar 0, \bar 3, \bar 6, \bar 9, \overline{12},\overline{15}\}$

$\{\bar 0, \bar 3 \} \times \mathbb{Z}/18\mathbb{Z}$