Show that $G$ is isomorphic to $\mathbb{Z}_{12}\times \mathbb{Z}_{18}$

Question

The group $\mathbb{Z}_{12}\times \mathbb{Z}_{18}$ has 32 elements of order 12 and 54 elements of order 18. Let $G$ be finite commutative group of order $12\times 18$ such that $G$ has 32 elements of order 12 and 54 elements of order 18. To justify whether $G\simeq \mathbb{Z}_{12} \times \mathbb{Z}_{18}$.

Here is what I did.

It is known that, if $G'$ be a commutative group and $|a|=m, |b|=n$ then $G'$ must have at least one element of order lcm$(m,n)$. This means $\mathbb{Z}_{12}\times \mathbb{Z}_{18}$ must have at least one element of order lcm$(12,18)=36$, which eventually happens to be the highest possible order.

Now $\mathbb{Z}_{12}\times \mathbb{Z}_{18}=\langle (1,0), (0,1)\rangle$. Since $G$ has at least one element of order $12$ and at least one element of order 18, so $G$ much have at least one element of order 36. We assume $|a|=12, |b|=18$ in $G$.

Let $f:\mathbb{Z}_{12}\times \mathbb{Z}_{18}\rightarrow G$ be defined by $f(1,0)=a$ and $f(0,1)=b$. Then $f(\alpha, \beta)=a^{\alpha} b^{\beta}$ for all $(\alpha, \beta)\in \mathbb{Z}_{12}\times \mathbb{Z}_{18}$. Although this is not an isomorphism. and here I got stuck.

Please help how to establish it.

Answer 1

This is how I would solve this problem. Since $|G|=2^3\cdot3^3$, the maximal $2$-subgroup of $G$ is isomorphic to one of three groups: $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$, $\mathbb{Z}_2\times\mathbb{Z}_4$ or $\mathbb{Z}_8$.

Similarly, the maximal 3-subgroup of $G$ is isomorphic to either $\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_3$ either $\mathbb{Z}_3\times\mathbb{Z}_9$ or $\mathbb{Z}_{27}$.

Count the number of elements of order $2$, $4$, $3$, $9$ in each case.

$$ \begin{array}{c|ccc} Order/Group & \mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2 & \mathbb{Z}_2\times\mathbb{Z}_4 & \mathbb{Z}_8 \\ \hline 2 & 7 & 3 & 1\\ 4 & 0 & 4 & 2\\ \end{array} $$ $$ \begin{array}{c|ccc} Order/Group & \mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_3 & \mathbb{Z}_3\times\mathbb{Z}_9 & \mathbb{Z}_{27}\\ \hline 3 & 26& 8 & 3\\ 9 & 0 & 18& 5\\ \end{array} $$

The rest should be clear in my opinion.