Question

Find all Abelian groups of cardinality 36, and all Abelian groups of cardinality 54.

Answer 1

For the abelian groups of order 36 :

$36=2^2 3^2$

We will try to decompose this group into a product and using the Chinese remainder theorem to regroup all the possible decompositions.

So there are forcefully elements of order 2 and 3 respectively. Now there are different cases :

1. There are elements of order 4 and 9. In this case their product gives un an element of order 36 and thus $G\simeq \mathbb{Z}/36\mathbb{Z}$
2. There is an element of order 4, but no element of order 9. Therefore there is an element of order 18 (say $g$) and an element of order 2 which is not in $<g>$ (due to cardinality reasons). Therefore we can decompose $G$ in a direct product : $G\simeq \mathbb{Z}/18\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ .
3. There is an element of order 9, but no element of order 4. This case is completely symmetrical and we have a product $G \simeq \mathbb{Z}/12\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$ .
4. If there is no element of order 4 or 9, we still have an element of order 6 (by multiplying the elements of order 2 and 3), say $g$ , and there is another element of order 6 which is not in $<g>$ Therefore our product is $G\simeq (\mathbb{Z}/6\mathbb{Z})^2$

For abelian groups of order $54=2\cdot3^3$ we have essentially the same algorithm:

There are forcefully the elements of order 2 and 3. Now there are cases:

1. There is an element of order 27. In this case the product of elements of order 2 and 27 will give an element of order 54 and thus $G\simeq\mathbb{Z}/54\mathbb{Z}$ .
2. There is no element of order 27, but there is an element of order 9. In this case there is an element of order 18, say $g$ and there is an element of order 3 that is not in $<g>$ . Thus we can decompose $G\simeq \mathbb{Z}/2\mathbb{Z}\times (\mathbb{Z}/9\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}) \simeq \mathbb{Z}/18\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ .
3. There are no elements of order 9. In this case there is an element of order 6, we decompose $G$ into a product $G\simeq \mathbb{Z}/6\mathbb{Z} \times H$ for $H$ a group of cardinal 9 without any element of order 9, so this is $(\mathbb{Z}/3\mathbb{Z})^2$ . Therefore we have $G\simeq \mathbb{Z}/6\mathbb{Z}\times (\mathbb{Z}/3\mathbb{Z})^2$ .