Determine whether or not the group (S3, ∘) is isomorphic to the group (T, ∘)
Determine whether or not the group (S3, ∘) is isomorphic to the group (T, ∘), where S3 is the set of permutations of three objects, and T is the set of symmetries of an equilateral triangle.
Group (S3, ∘) is A permutation of a set of objects is a rearrangement of the objects in the set. For example, objects in the set S = {1, 2, 3} can be arranged (or listed) in various orders: {1, 2, 3} (i.e., the original arrangement, called the
trivial permutation), {1, 3, 2}, {2, 1,3}, {2, 3, 1}, {3, 1, 2} and {3, 2, 1}. So, we would say that setS has six permutations.
Each permutation can be considered a function using the original arrangement as the domain and the re-arrangement as the range. Using a table of values representation
of a function, the above six permutations can be written as follows:
P1: 1 1
2 2
3 3
P2: 1 1
2 3
3 2
P3: 1 2
2 1
3 3
P4: 1 2
2 3
3 1
P5: 1 3
2 1
3 2
P6: 1 3
2 2
3 1
We could also use function notation to express the result for each permutation. For
example, for permutation
P3 we could write:
P3(1) = 2,
P3(2) = 1,
P3(3) = 3. For the set S = {1, 2, 3}, we’ll let S3 (since S has three objects) represent the set of permutations of the objects of set S, i.e., S3 = {P1, P2, P3, P4, P5, P6}. To define an operation between objects (i.e., permutations) in S3, we can use the idea of the composition of functions. Two permutations can be combined by using the result of one permutation as the domain of the next permutation. For example, using the symbol ∘ as the operation between two permutations, P4 ∘ P5 would mean, "apply the permutation P5 first to the domain {1, 2, 3}; then apply the permutation P4 onto the range values from P5. So, P5 applied to {1, 2, 3} results in {3, 1, 2}; then applying P4 to {3, 1, 2} results in {1, 2, 3} since the P4 permutation pairs 3 with 1, 1 with 2, and 2 with 3. Thus the combined permutations applied to {1, 2, 3} result in {1, 2, 3}, which is equivalent to the permutation P1. So, we can write P4 ∘ P5 = P1.
| ∘ | P1 | P2 | P3 | P4 | P5 | P6 |
| P1 | P1 | P2 | P3 | P4 | P5 | P6 |
| P2 | P2 | P1 | P4 | P3 | P6 | P5 |
| P3 | P3 | P5 | P1 | P6 | P2 | P4 |
| P4 | P4 | P6 | P2 | P5 | P1 | P3 |
| P5 | P5 | P3 | P6 | P1 | P4 | P2 |
| P6 | P6 | P4 | P5 | P2 | P3 | P1 |
Group (T, ∘) is
| ∘ | I | R1 | R2 | F1 | F2 | F3 |
| I | I | R1 | R2 | F1 | F2 | F3 |
| R1 | R1 | I | R1 | F2 | F3 | F1 |
| R2 | R2 | R2 | I | F3 | F1 | F2 |
| F1 | F1 | F3 | F2 | I | R2 | R1 |
| F2 | F2 | F1 | F3 | R1 | I | R2 |
| F3 | F3 | F2 | F1 | R2 | R1 | I |
Determine whether or not the group (S3, ∘) is isomorphic to the group (T, ∘)
Two groups are isomorphic if there exists a bijective function f: G -> H such that for all a, b in G, f(a ∘ b) = f(a) ∘ f(b).
S3 is the symmetric group of order 3, which contains 3! = 6 elements, and it is the set of all permutations of three objects. The elements of S3 are {P1, P2, P3, P4, P5, P6}.
T is the group of symmetries of an equilateral triangle, which contains six elements: {I, R1, R2, F1, F2, F3}.
In both groups, there are six elements, and their group operations are associative. We can find a bijective function between these two groups that preserve the group operation:
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