in Set Theory & Algebra
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60 votes
60 votes

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

  1. $2$
  2. $3$
  3. $4$
  4. $5$
in Set Theory & Algebra
19.4k views

4 Comments

edited by

I could not resist myself to comment again here because this question is quite related to the great indian mathematician Srinivasa Ramanujan.

S. Ramanujan worked on partitions with G.H. Hardy and obtained the asymptotic solution for partition function :

$P(n) \sim \frac{1}{4n\sqrt{3}}\; e^{\pi \sqrt{2n/3}}$

I have uploaded some video clips from the movie "The Man who knew infinity". Anyone who is interested in mathematics can see these video clips.

1) Partitions

2) Ramanujan with P.C. Mahalonobis(Founder of ISI, father of indian statistics and a FRS too like Ramanujan)

3) Extra

4) Number 1729

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I watched that movie 2years back. But didn't noticed a single point about partitions..

Thanks for sharing.. :)
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6
That time I didn't even knew the importance of partitions.
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1

5 Answers

105 votes
105 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.

Correct Answer: $A$

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4 Comments

For a set to become a partition their intersection must be phi..or null
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Someone Please explain 3rd  point ?

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@Shaikh727 it's a product of number of partitions of powers 

IF order=24 then prime factorization is 

$2^{^{3}}*3^{^{1}}$

now number of partition of 3 is 

{1,1,1},{1, 2},{3}

partition of 1 is {1}

​​​​​​product of number of partitions 3*1=3

 

 

 

 

 

 

 

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6 votes
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.

edited by
2 votes
2 votes

Watch this

Start from 6:34

 

0 votes
0 votes

The answer is Option A (only two partition )

Answer:

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