Hw9 Problem 3

### Problem 3

Find all abelian groups, up to isomorphism of the following orders.
(a) 81
(b) 2200

Solution

(a) Since groups are to be of the finite order 81, no factors $\mathbb{Z}$ will appear in the direct product. And $81=3^4$. Then using Theorem 11.12, we get
$\mathbb{Z}_3\times \mathbb{Z}_3\times \mathbb{Z}_3\times \mathbb{Z}_3$
$\mathbb{Z}_3\times \mathbb{Z}_3\times \mathbb{Z}_9$
$\mathbb{Z}_9\times \mathbb{Z}_9$
$\mathbb{Z}_3\times \mathbb{Z}_{27}$
$\mathbb{Z}_{81}$
Thus there are five different abelian groups of order 81.

(b) Since groups are to be of finite order 2200, no factors $\mathbb{Z}$ will appear in the direct product. And $2200=2^3*5^2*11$ Then using Theorem 11.12, we get
$\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_5\times \mathbb{Z}_5\times \mathbb{Z}_{11}$
$\mathbb{Z}_2\times \mathbb{Z}_4\times \mathbb{Z}_5\times \mathbb{Z}_5\times \mathbb{Z}_{11}$
$\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_{25}\times \mathbb{Z}_{11}$
$\mathbb{Z}_2\times \mathbb{Z}_4\times \mathbb{Z}_{25}\times \mathbb{Z}_{11}$
$\mathbb{Z}_8\times \mathbb{Z}_5\times \mathbb{Z}_5\times \mathbb{Z}_{11}$
$\mathbb{Z}_8\times \mathbb{Z}_{25}\times \mathbb{Z}_{11}$
Thus there are six different abelian groups of order 2200.