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How many different non-isomorphic Abelian groups of order $4$ are there?

  1. $2$
  2. $3$
  3. $4$
  4. $5$
asked in Set Theory & Algebra by Veteran (59.7k points) | 4.1k views
+1

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 ? :)

+6

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.

0

Refer the link below for question related to Bell Number along with Great Explanation..

https://gateoverflow.in/2259/gate1997-6-3

4 Answers

+40 votes
Best answer

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


  1. First, find the prime factorization of $n.$ For example, $4$ has prime factorization as $2*2.$ Also, $600$ can be factorized as $2^3∗3^1∗5^2$
  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.
  3. 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 $3∗1∗2=6.$
  4. 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.
answered by Veteran (55.6k points)
edited by
0
what is meant by partitions of k ??
0

how  Partition of 2 are {1,1}, {2,0}

+1

21 * 21 =4 and 22 * 20=4 @asu

+13

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.

  1. 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)
  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 doesn't matter, so for example, partitions {2+1+1} {1+1+2} are 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.
  3. Suppose, in question, order given is 600, , so different powers are 3,1,2. Number of partitions for 3,1,2  are 3,1,2  respectively, so result would have been 3∗1∗2=6.
  4. 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. 
0
@VS ji,
is there any direct formula for finding number of distinct partitions ?
0
very clear explanation. Thanx  a  lot
0
@VS

In this method which step guarantees non isomorphism ?
0
No. of the partition with 3 elements is 5. Please Correct it.
0
@hemant,if we take distinct element like {a,b,c} then total 5 partition.

but here all elements are same {1,1,1} then number of partitions are

{{1,1,1}}

{{1,1},{1}}

{{1},{1},{1}} total 3 partitions
0

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.

+4

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.

+6 votes

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 ith 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.

answered by Boss (34k points)
edited by
0 votes

The answer is Option A (only two partition )

answered by Active (1.2k points)
–2 votes
Option b .
answered by Active (3.3k points)
0
can you explain pls
+2
1. Group with order <=5 always abelian.
if we find no of groups of order 4 that ll be same as no of abelian group of order 4..
in every group of order d  , so  element order 1 and 4 common in every group..
d only thing dat matters is no of groups with or without order 2.
2 non isomorphic group of order 4..
+11
  1. 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)
  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 doesn't matter, so for example, partitions {2+1+1} {1+1+2} are 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.
  3. Suppose, in question, order given is 600, , so different powers are 3,1,2. Number of partitions for 3,1,2  are 3,1,2  respectively, so result would have been 3∗1∗2=6.
0
what is meant by isomorphic and non-isomorphic abelian group,did't get any good reference.

@Kaluti any help?
+5

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 

Answer:

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