Question

List all abelian groups (up to isomorphism) of the given order:1160

Short answer

The abelian groups of order 600 are

$\{{\mathbb{Z}}_{2^3}\oplus {\mathbb{Z}}_5\oplus {\mathbb{Z}}_{29},{\mathbb{Z}}_2\oplus {\mathbb{Z}}_{2^2}\oplus {\mathbb{Z}}_5\oplus {\mathbb{Z}}_{29},{\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\oplus {\mathbb{Z}}_5\oplus {\mathbb{Z}}_{29}\}$

Step-by-step solution

Step-by-Step Solution Step 1: Finite Abelian Group

Every finite abelian group G is the direct sum of cyclic groups, each of prime power orders.

Abelian groups of order 1160 

The number 1160 can be written as a product of prime powers in 3 different ways:

$\{(2^3\cdot 5\cdot 29),(2\cdot 2^2\cdot 5\cdot 29),(2\cdot 2\cdot 2\cdot 5\cdot 29)\}$

Therefore, the abelian group can be written as follows:

$\begin{array}{l}{\mathrm{\mathbb{Z}}}_{2^3}\oplus {\mathrm{\mathbb{Z}}}_5\oplus {\mathrm{\mathbb{Z}}}_{29}\\ {\mathrm{\mathbb{Z}}}_2\oplus {\mathrm{\mathbb{Z}}}_{2^2}\oplus {\mathrm{\mathbb{Z}}}_5\oplus {\mathrm{\mathbb{Z}}}_{29}\\ {\mathrm{\mathbb{Z}}}_2\oplus {\mathrm{\mathbb{Z}}}_2\oplus {\mathrm{\mathbb{Z}}}_2\oplus {\mathrm{\mathbb{Z}}}_5\oplus {\mathrm{\mathbb{Z}}}_{29}\end{array}$

Hence, the abelian groups of order 600 are:

$\{{\mathbb{Z}}_{2^3}\oplus {\mathbb{Z}}_5\oplus {\mathbb{Z}}_{29},{\mathbb{Z}}_2\oplus {\mathbb{Z}}_{2^2}\oplus {\mathbb{Z}}_5\oplus {\mathbb{Z}}_{29},{\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\oplus {\mathbb{Z}}_5\oplus {\mathbb{Z}}_{29}\}$