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Ajay posted an Question

December 18, 2020 • 16:06 pm

30 points

IIT JAM
Mathematics (MA)

Please give detailed solution

Please give detailed solution as to how to figure out such kind of problems

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A
Anonymous User

See attachment
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S
Srinath

See, there is no straight up formula for finding the number of elements of a given order, in ANY group. You have to check whether the group is cyclic/abelian/finite.. (These are mostly asked in exam). There will be key points, regarding finite groups.
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Yes but can you please show in detail how you arrived at the solution in this question?
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Like how do you know there will be 25 transpositions?
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Do ask if you have doubts.
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Sir I have problem only in the step where you calculate the combinations. Can you please explain how you arrive at 10 and 15 ?
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Consider (a b). Now a has 5 possibilities, right? So, b will have 4. ( Think why!) Then there is a total of 5*4 combinations of (a b). But in this counting process, you would take some (a1 , b1) and (b1, a1) to be distinct elements. (Think why!) Therefore total is 5*4/2 (cancelling the repetitions). So first one is 10. Second one is very similar. Except that now you have to consider repetitions of the form (a b)(c d) and (c d)(a b), which explains the 'third' 2 in the denominator. You follow the idea?
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S
Srinath

Also, the identity element is it's own inverse , so the answer is 26.
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S
Srinath

It is just a small trick. You must remember that in a group elements of order 2 are their own inverse. So basically, you just have to find out the number of elements of order 2 in S5, which is 25.
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How to find number of elements of a particular order in any group?
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please show in this example
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