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# Understand the problem

The permutation group $S_{10}$ has an element of order 30.
##### Source of the problem
TIFR 2018 Part A Problem 23
Group Theory
Easy
##### Suggested Book
Dummit and Foote

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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