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Chapter 16: Problem 10

In \(S_{5}\) find an element of order \(n\), for all \(2 \leq n \leq 5\). Also determine the (cyclic) subgroup of \(S_{5}\) that each of these elements generates.

Short Answer

Expert verified
The elements of order 2, 3, 4, and 5 in \(S_{5}\) are (12), (123), (1234), (12345) respectively. The cyclic subgroups they generate are {e, (12)}, {e, (123), (132)}, {e, (1234), (1243), (1423)}, and {e, (12345), (13452), (14253), (15432)}, respectively.

Step by step solution

01

Find an element of order 2

In order to find an element of order 2 in \(S_{5}\), we can choose two adjacent numbers and swap them. Therefore, we can choose (12) as an element of order 2 because if you apply it twice, you get the identity permutation.
02

Find an element of order 3

To find an element of order 3 in \(S_{5}\), we can choose a cycle of length 3. Here, we choose (123) as element of order 3, because applying it three times gives us back the identity permutation.
03

Find an element of order 4

To find an element of order 4 in \(S_{5}\), we can choose a cycle of length 4. So, let's choose (1234) as an element of order 4, as applying it four times returns to the identity permutation.
04

Find an element of order 5

In the same way to find an element of order 5, we choose a cycle of length 5. We can choose (12345) as an element of order 5 because applying it five times gives us the identity permutation.
05

Determine the generated cyclic subgroups

The cyclic subgroup generated by an element is simply all the powers of that element. Hence the cyclic subgroup generated by (12) is {e, (12)}, the cyclic subgroup of (123) is {e, (123), (132)}, the cyclic subgroup of (1234) is {e, (1234), (1243), (1423)} and that of (12345) is {e, (12345), (13452), (14253), (15432)}.

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