How many different non-isomorphic Abelian groups of order $4$ are there?
Some Reference -->
https://math.stackexchange.com/questions/786525/abelian-groups-of-order-n
https://en.wikipedia.org/wiki/Partition_(number_theory)
Hi Guys,
Many people have given correct answer for this question by using direct formula, but if you can mention some simple proof for these direct formula then it will be great help because it will clear remaining how and why ? :)
Additional Info: The number of partitions of a number $n$ can be found out using the following diagram:
\begin{array}{c c c c c} \color{red}{1}\\ 1 &\color{red}{2}\\ 2 &3 &\color{red}{5}\\ 5 &7 &10 &\color{red}{15} \cdots \end{array}
The number of ways a set of $n$ elements can be partitioned into nonempty subsets is called a Bell number and can be obtained by constructing a triangle as above. The first row starts with a $1$. For the first element of next row you copy the last element of previous row. And find the sum with number above and right it in the next column, till $n^{th}$ row having $n$ elements.
Refer the link below for question related to Bell Number along with Great Explanation..
https://gateoverflow.in/2259/gate1997-6-3
krish__ According to your formula, number 4 has 15 partitions. Since 4th Bell number is 15. Bur number 4 has 5 partitions (1+1+1+1, 1+2+1,2+2,3+1,0+4)
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
how Partition of 2 are {1,1}, {2,0}
2^{1} * 2^{1} =4 and 2^{2} * 2^{0}=4 @asu
Credits : @Kaluti , I have commented the same thing here because I felt the best answer chosen here lacked clarity ,so just for making things clear for other users.
Wont we consider 5 in prime factorization of 600 as you stated as under:
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.
Please Clearify my doubt.
Thanks in advance.
Here the number of partitions of K must be done into positive integers($Z^+)$.
Don't be trapped by assuming that the number of partitions of K will be given by bell triangle.
Bell triangle provides the number of partitions of a set n(with distinct elements)
But here you have to find the number of ways in which K can be written as a sum of positive integers
THERE IS A DIFFERENCE.
Consider K=3.
Number of ways in which you can write 3 as the sum of 3 positive integers
3
2+1
1+1+1
3 ways
But Bell triangle for 3 is
1
1 2
2 3 5
Which gives the answer as 5 ways to partition a set of 3 distinct elements into non-empty partitions.
Here, in this problem, you need the former method.
Answer: A
Number of different non-isomorphic abelian groups of order n are $\prod_{i=1}^{w(n)} p(a_i)$, where $p(a_i)$ is the number of partitions of the i^{th} prime number. Here the order is only divisible by 2 and number of partitions of 2 are 2: {0,2} and {1,1}.
Hence, the answer is 2.
The answer is Option A (only two partition )
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.
you can refer this