Problem 5.2.4 (Which pairs of abelian groups listed are isomorphic ?)

Proof. Two abelian groups are isomorphic if and only if their invariant factors are the same. We give in the second column the invariant factors of the group in the first column.

(a)

$$\begin{array}{cc}\{4,9\}\hfill & (36)\hfill \\ \{6,6\}\hfill & (6,6)\hfill \\ \{8,3\}\hfill & (24)\hfill \\ \{9,4\}\hfill & (36)\hfill \\ \{6,4\}\hfill & (12,2)\hfill \\ \{64\}\hfill & (64)\hfill \end{array}$$

So $Z_4\times Z_9$ and $Z_9\times Z_4$ are the only isomorphic groups in this list.

(b)

$$\begin{array}{cc}\{2^2,2\cdot 3^2\}\hfill & (36,2)\hfill \\ \{2^2\cdot 3,2\cdot 3\}\hfill & (12,6)\hfill \\ \{2^3\cdot 3^2\}\hfill & (72)\hfill \\ \{2^2\cdot 3^2,2\}\hfill & (36,2)\hfill \end{array}$$

So $Z_{2^2}\times Z_{2\cdot 3^2}\simeq Z_{2^2\cdot 3^2}\times Z_2$ : these two groups are both isomorphic to $Z_{36}\times Z_2$.

(c)

$$\begin{array}{cc}\{5^2\cdot 7^2,3^2\cdot 5\cdot 7\}\hfill & (3^2\cdot 5^2\cdot 7^2,5\cdot 7)\hfill \\ \{3^2\cdot 5^2\cdot 7,5\cdot 7^2\}\hfill & (3^2\cdot 5^2\cdot 7^2,5\cdot 7)\hfill \\ \{3\cdot 5^2,7^2,3\cdot 5\cdot 7\}\hfill & (3\cdot 5^2\cdot 7^2,3\cdot 5\cdot 7)\hfill \\ \{5^2\cdot 7,3^2\cdot 5,7^2\}\hfill & (3^2\cdot 5^2\cdot 7^2,5.7)\hfill \end{array}$$

So

$$\begin{array}{llll}\hfill Z_{5^2\cdot 7^2}\times Z_{3^2\cdot 5\cdot 7}& \simeq Z_{3^2\cdot 5^2\cdot 7}\times Z_{5\cdot 7^2}& \hfill & \\ \hfill & \simeq Z_{5^2\cdot 7}\times Z_{3^2\cdot 5}\times Z_{7^2}.& \hfill & \end{array}$$

(d)

$$\begin{array}{cc}\{2^2\cdot 5\cdot 7,2^3\cdot 5^3,2\cdot 5^2\}\hfill & (2^3\cdot 5^3\cdot 7,2^2\cdot 5^2,2\cdot 5)\hfill \\ \{2^3\cdot 5^3\cdot 7,2^3\cdot 5^3\}\hfill & (2^3\cdot 5^3\cdot 7,2^3\cdot 5^3)\hfill \\ \{2^2,2\cdot 7,2^3,5^3,5^3\}\hfill & (2^3\cdot 5^3\cdot 7,2^2\cdot 5^3,2)\hfill \\ \{2\cdot 5^3,2^2\cdot 5^3,2^3\cdot 7\}\hfill & (2^3\cdot 5^3\cdot 7,2^2\cdot 5^3,2)\hfill \end{array}$$

So

$$Z_{2^2}\times Z_{2\cdot 7}\times Z_{2^3}\times Z_{5^3}\times Z_{5^3}{\simeq }_{2\cdot 5^3}\times Z_{2^2\cdot 5^3}\times Z_{2^3\cdot 7}.$$