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Consider the following mappings between a permutation group and a cyclic group.
(a) Denote by {A}_{n} the subset of the permutation group {S}_{n} that contains all the even permutations. Show that {A}_{n} is a subgroup of {S}_{n}.
(b) List the elements of {S}_{3} in cycle notation and identify the subgroup {A}_{3}.
(c) For each element X of {S}_{3}, let p(X) = 1 if X belongs to {A}_{3} and p(X) = −1 if it does not. Denote by {C}_{2} the multiplicative cyclic group of order 2. Determine the images of each of the elements of {S}_{3} for the following four mappings:
{Φ}_{1} : {S}_{3} → {C}_{2} X → p(X),
{Φ}_{2} : {S}_{3} → {C}_{2} X →−p(X),
{Φ}_{3} : {S}_{3} → {A}_{3} X → {X}^{2},
{Φ}_{4} : {S}_{3} → {S}_{3} X → {X}^{3}.
(d) For each mapping, determine whether the kernel K is a subgroup of {S}_{3} and if so, whether the mapping is a homomorphism.

Step-by-step

(a) With {A}_{n} as the subset of {S}_{n} that contains all the even permutations, we need to demonstrate that it has the four properties that characterise a group:
(i) If X and Y belong to {A}_{n} so does XY , as the product of two even permutations is even.
This establishes closure.
(ii) From the definition of an even permutation, the identity I belongs to {A}_{n}.
(iii) If X belongs to {A}_{n} so does {X}^{−1}, as the number of pair interchanges needed to change from X to I is the same as the number needed to go in the opposite direction. This establishes the existence, within the subset, of an inverse for each member of the subset.
(iv) Associativity follows from that of the group {S}_{n}.
Thus {A}_{n} does possess the four properties and is a subgroup of {S}_{n}.
(b) (1), (123) and (132) belong to {A}_{3}. The permutations (12), (13) and (23) do not belong, as each involves only one pair interchange.
(c) With the given definition of p(X),
p(X) = 1 for X = (1), (123), (132),
p(X) = −1 for X = (12), (13), (23).
{C}_{2} consists of the two elements +1 and −1.
For {Φ}_{1} : {S}_{3} → {C}_{2} X → p(X), elements in {A}_{3} have image +1; those not in {A}_{3} have image −1.
For {Φ}_{2} : {S}_{3} → {C}_{2} X →−p(X), elements in {A}_{3} have image −1; those not in {A}_{3} have image +1.
For {Φ}_{3} : {S}_{3} → {A}_{3} X → {X}^{2}
(1) → (1)(1) = (1),
(123) → (123)(123) = (132),
(132) → (132)(132) = (123),
(12) → (12)(12) = (1), similarly, (13) and (23).
For {Φ}_{4} : {S}_{3} → {S}_{3} X → {X}^{3}
(1) → (1)(1) = (1),
(123) → (123)(123)(123) = (132)(123) = (1),
(132) → (132)(132)(132) = (123)(132) = (1),
(12) → (12)(12)(12) = (1)(12) = (12), similarly, (13) and (23).
(d) For {Φ}_{1}, the kernel is the set of elements belonging to {A}_{3} and, as already shown, this is a subgroup of {S}_{3}.
We note that the product of two even or two odd permutations is an even
permutation, whilst the product of an odd and an even permutation is an odd
permutation. We also note that +1×+1 and −1×−1 are both equal to +1, whilst
+1×−1 = −1. Since {Φ}_{1} maps even permutations onto +1 and odd permutations
onto −1, the preceding observations imply that {Φ}_{1} is a homomorphism.
For {Φ}_{2}, the kernel is the set of elements not belonging to {A}_{3}. Since this set does not contain the identity (1), it cannot be a subgroup of {S}_{3}.
For {Φ}_{3} the kernel is the set {(1), (12), (13), (23)}. Since, for example, (12)(13) = (132), the set is not closed and so cannot form a group. It cannot, therefore, be a subgroup of {S}_{3}.
For {Φ}_{4} the kernel is the set {(1), (123), (132)}, i.e. the subgroup {A}_{3}. However, for example, [ (12)(13) ]^{\prime} = (132)^{\prime} = (1), whilst (12)^{\prime}(13)^{\prime} = (12)(13) = (132); these two results are not equal, showing that the mapping cannot be a homomorphism.

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