GATE CSE 2007 | Question: 21

Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Comments

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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That time I didn't even knew the importance of partitions.

Answer 1

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

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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: $A$

Comments

what is meant by partitions of k ??

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how  Partition of 2 are {1,1}, {2,0}

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2¹ * 2¹ =4 and 2² * 2⁰=4 @asu

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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. 

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@VS ji,
is there any direct formula for finding number of distinct partitions ?

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very clear explanation. Thanx  a  lot

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@VS

In this method which step guarantees non isomorphism ?

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No. of the partition with 3 elements is 5. Please Correct it.

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@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

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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.

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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.

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I am able to get how the answer is being calculated, but can someone explain the reason behind this procedure ie. why no of partitions of k is the same as no of Abelian groups of order prime^k?

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no of column of ith row in Bell-triangle is no. of partition of i ???

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@Shubhm 

Number of partitions of a number n and Number of partitions of a set (having n elements) both are different.

$n^{th}$ Bell no. represents the number of partitions of a set with n elements, same as no. of equivalence relations on a set containing n elements.

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@Verma Ashish thanks!

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For a set to become a partition their intersection must be phi..or null

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Someone Please explain 3^rd  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

 

 

 

 

 

 

 

Answer 2

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.

Answer 3

Watch this

Start from 6:34

 

Answer 4

The answer is Option A (only two partition )