# 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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