The number of Abelian groups of order $P^{k}$ ($P$ is prime) is the number of partitions of $k.$
Here, order is $4$ i.e. $2^{2}$.
Partition of $2$ are $\{1,1\}, \{2,0\}.$
Total $2$ partition so no. of different abelian groups are $2.$
http://oeis.org/wiki/Number_of_groups_of_order_n
- First, find the prime factorization of $n.$ For example, $4$ has prime factorization as $2*2.$ Also, $600$ can be factorized as $2^3∗3^1∗5^2$
- Now, find the number of partitions of all powers, and then multiply them. Number of Partitions of a number $k$ is the number of ways $k$ can be partitioned. For example, number of partitions of $3$ is $3,$ because $3$ can can be partitioned in $3$ different ways: $\{1+1+1\}, \{1+2\}, \{3\}.$ Similarly, $4$ can be partitioned in $5$ different ways: $\{1+1+1+1\},\{2+1+1\},\{2+2\},\{3+1\},\{4\}.$ Note that order of elements in a partition does not matter, for example, partitions $\{2+1+1\}$ and $\{1+1+2\}$ are the same.
So for this question, we will find number of partitions of $2,$ which is $2 : \{1+1\},\{2\}.$ There is no other power, so answer is $2$ only.
- Suppose, in question, order given is $600.$. Then, different powers are $3,1,2$. Number of partitions for $3,1,2$ are $3,1,2$ respectively and result would have been $3∗1∗2=6.$
- Group theorists view two isomorphic groups as follows: For every element g of a group G, there exists an element h of H group such that h 'behaves in the same way' as g(operates with other elements of the group in the same way as g). For instance, if g generates G, then so does h. This implies in particular that G and H are in bijective correspondence. Thus, the definition of an isomorphism is quite natural.
Correct Answer: $A$