Mathematics Asked by Ramesh Karl on December 22, 2020

My attempts for part (a):

As 4 and 7 are 2-cycles, and 1, 2, 3, 5, 6 is another 5-cycles,

therefore the product of disjoint cyclic permutations

= (1 2 3 5 6)(4 7)= (1 2)(1 3)(1 5)(1 6)(4 7)

My attempt for part (b):

As the permutation σ consists of 5 disjoint cyclic permutations, and 5 is an odd number, therefore the permutation σ is an odd permutation.

My attempt for part (c):

As the permutation σ can be written as (1 2 3 5 6)(4 7).

(1 2 3 5 6) will become the identity permutation after 5 times of σ and (4 7) will become the identity permutation every 2 times of σ.

Combining the above results, the permutation σ will become the identity permutation for every 5 x 2 = 10 times of σ. So the order of σ is 10.

For (d), somehow I had managed to construct something, but not sure if it’s correct or not, it goes like this: (177)(127)(177)(167)(177)(137)(147)(157)

My question is:

I am not sure if my attempts on (a), (b), (c), (d) are correct or not.

Thank you.

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