Understand the problem

The permutation group \(S_{10}\) has an element of order 30.
Source of the problem
TIFR 2018 Part A Problem 23
Topic
Group Theory
Difficulty Level
Easy
Suggested Book
Dummit and Foote

Start with hints

Do you really need a hint? Try it first!

Consider S={1,2,…,10}.\(S_{10}\) be the permutation group on S.
What will you do if one asked for a subgroup of order 3!=6?
  • We all would have taken the subgroup of \(S_3\) embedded in \(S_{10}\) right? Call the subgroup H taking identity transformation on {4,5,6,..10} and embedding in {1,2,3}.
What do you do if one asked for a subgroup of order 5?
  • We will try to form a cyclic subgroup of order 5. We need to find a generator. Can you see it?
  • Keeping {1,2,.,5} same and taking 5 elements {6,7,..,10} and observe the permutation \(i\mapsto i+1\) for i=6,7,8,9 and \(10\mapsto 7\).Take the subgroup generated by this element.Observe this is a cyclic subgroup of order 5.Call this subgroup K.
  • Now any idea how to combine this?
  • Observe that HK is a set of 30 elements. Does it seem HK is a subgroup?
  • Lemma: H and K are two subgroups of G. HK is a subgroup of G iff KH=HK.
  • Using the lemma prove that HK is really a subgroup of \(S_{10}\) of order 30.
  • Observe that the selection of disjoint elements of H and K is the main reason behind this!
  • Hence the answer is True.

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