Here is the problem: Consider the group (S3, o) under composition, which consist
ID: 2986755 • Letter: H
Question
Here is the problem:
Consider the group (S3, o) under composition, which consists of List the elements of each cyclic subgroup explicitly in (1)-(2). Then complete (3)-(5). [f5] = {f?....f?}; avoid repetition. [fa] ={f?....f?}; avoid repetition. Is [f5] n [f6] a subgroup of S3? Is [f5] U [f0] a subgroup of S3? Justify your claim made in (4). Hint. Which element is the identity of S3? What are the orders of f5 and f6? Now, can you determine [f5] and [f6]? There is no need to justify your answer to (3). Consider (Z12, +), which is a quotient group of (Z, +) modulo its normal subgroup N = 12Z = {12m. | m Z}. Thus Z12 = Z/N = {0 + N, 1 + N, ..., 11 + N}. Find o(3 + N) and [8 + N). Show your work. Determine [3 + N] and [8 + N] by listing their distinct elements explicitly. Hint. Note that the operation of Z12 is +. Which element is the identity of Z12? For an element k + N Z12, you may/should write (k + N) + ... + (k + N) as n(k + N).Explanation / Answer
Problem III.1
As given in hint we first recognize f1 as identity.
Now we shall raise f5 and f6 to powers to find the cyclic group generated by them in parts 1 and 2.
(1)
f5 : 1->3, 2->1, 3->2
f5^2 = f5of5: 1->2, 2->3, 3->1 = f4
f5^3 = f5of5^2 = f5of4 : 1->1, 2->2, 3->3 = f1(identity)
Thus <f5> = {f5,f4,f1}
(The above calculations are just compositions of functions for example to find f5of4, we proceed as follows f5of4(1) = f5(f4(1)) = f5(2) = 1 etc)
(2)
f6 : 1->3, 2->2, 3->1
f6^2 = f6of6 : 1->1, 2->2, 3->3 = f1(identity)
Thus <f6> = {f6,f1}
(3)
Yes,
intersection = {f1}, trivial subgroup
In fact intersection of any two subgroups is always a subgroup, caution that the same cannot be said about union as can be seen in next part
(4)
No
(5)
union = {f1,f4,f5,f6}
f4of6 : 1->1, 2->3, 3->2 = f2 not in union, hence union is not closed under group operation and hence not a subgroup.
Problem III.2
(1)
3+N is distinct from 0+N since 3 is not in N
(3+N)+(3+N) = 6+N not 0+N (since 6 is not in N)
(3+N)+(3+N)+(3+N) = 9+N not 0+N (since 9 is not in N)
(3+N)+(3+N)+(3+N)+(3+N) = 12+N = 0+N (since 12 is in N)
Thus 4 is the smallest positive integers such that (3+N)+...+(3+N) n times = 0+N
Hence o(3+N) = 4
8+N not 0+N (since 8 is not in N)
(8+N)+(8+N) = 16+N = 4+N (since 16 = 12+4 and 12 is in N) not 0+N (since 16 is not in N as it is not multiple of 12)
(8+N)+(8+N)+(8+N) = 24+N = 0+N (since 24 is in N)
Thus o(8+N) = 3
(2)
From the above calculations we get
<3+N> (subgroup generated by 3+N) as {3+N,6+N,9+N,0+N}
<8+N> = {8+N,4+N,0+N}