is isomorphic to D_5 because if you label the vertices of a pentagon with 1 at the top and the rest in order, (1 2 3 4 5) describes the action of a rotation and (2 5)(3 4) describes reflecting across the vertical axis.
For the order 2 elements of A_5, you could also pick any of 5 objects to be the one that doesn't appear, and then any of the 3 not-least objects of the remaining objects to be in the same cycle with the least object, giving 5x3 without any double counting.
In the first exercise you say what the answers are but not why those are the answers. For A_6 you correctly give witnesses for orders 1, 3, 5, 2, and 4, namely the identity and the cycle types (a b c), (a b c d e), (a b)(c d), and (a b)(c d e f), but it should be mentioned why (a b)(c d e f) is even. The reason why 1,2,3,4,5 are the _only_ orders, which needs to be shown, is because the only other possible cycle type in A_6 is (a b c)(d e f), (this is just a matter of listing out all the possible cycle types in S_6 and determining which are even) which has order lcm(3, 3) = 3 which we already have. Similarly, for A_7 you mention examples for orders 1, 3, 5, 7, 2, and 4 but not for order 6 - the only example is the cycle type (a b c)(e f)(g h). The reason why 1,2,3,4,5,6,7 are the only orders is because A_7 consists of the identity and the cycle types (a b c d e f g), (a b c d e), (a b c), (a b)(c d), (a b c)(d e f), (a b c d)(e f), and (a b c)(d e)(f g), which it is easy to check do not give any other orders. Note that it is not obvious that the orders of elements in A_n are ≤ n - indeed this is false, as shown by the last part of the exercise.
So I found an alternate proof of exercise 3. Could someone check if it holds up? case 1: all elements are even case 2: not that Lemma: you can "go" from any odd cycle to any other odd cycle via an even cycle (notice: even × even -> even, odd × even -> odd, odd × odd -> even) a, b are odd => a⁻¹ is odd => a⁻¹b = h which is even => b = ah let G≤H denote all elements of H that are even. Via the subgrouptest: x = τ1 τ2 τ3 ... τn, where n is even => x ∈ G y = µ1 µ2 µ3 ... µm, where m is even => y ∈ G xy⁻¹ = τ1 τ2 τ3 ... τn µm... µ3 µ2 µ1 which is of length m+n => even => xy⁻¹ ∈ G => G is a subgroup Let h ∈ H be an odd cycle => h ∉ G Then by lemma, any odd cycle is an element of the coset hG => if an element is even, it's an element of G. if it's odd, it's an element of hG. => all elements are either in G or hG => G ∪ hG = H and by theorems about cosets of subgroups, we know that |G| = |hG| => [H:G] = 2 => exactly half of all cycles are even, the others are odd
That would be an alternative proof, so congratulations on that. Although I think that parts of the reasoning could be simplified. Just a couple of suggestions: 1. To prove that G is a subgroup notice that G is the intersection of H with An so it's a subgroup. 2. Once you have proved that |G| = |hG|, you have proved that there are exactly the same amount of even and odd permutations, so there is no need to continue, you have reached your goal. Finally, just so that your proof is completed, take notice that you proved your lemma proving that b = ah so b ∈ Gh, and in your proof you use hG, this is just a technicality.
is isomorphic to D_5 because if you label the vertices of a pentagon with 1 at the top and the rest in order, (1 2 3 4 5) describes the action of a rotation and (2 5)(3 4) describes reflecting across the vertical axis.
For the order 2 elements of A_5, you could also pick any of 5 objects to be the one that doesn't appear, and then any of the 3 not-least objects of the remaining objects to be in the same cycle with the least object, giving 5x3 without any double counting.
In the first exercise you say what the answers are but not why those are the answers. For A_6 you correctly give witnesses for orders 1, 3, 5, 2, and 4, namely the identity and the cycle types (a b c), (a b c d e), (a b)(c d), and (a b)(c d e f), but it should be mentioned why (a b)(c d e f) is even. The reason why 1,2,3,4,5 are the _only_ orders, which needs to be shown, is because the only other possible cycle type in A_6 is (a b c)(d e f), (this is just a matter of listing out all the possible cycle types in S_6 and determining which are even) which has order lcm(3, 3) = 3 which we already have.
Similarly, for A_7 you mention examples for orders 1, 3, 5, 7, 2, and 4 but not for order 6 - the only example is the cycle type (a b c)(e f)(g h). The reason why 1,2,3,4,5,6,7 are the only orders is because A_7 consists of the identity and the cycle types (a b c d e f g), (a b c d e), (a b c), (a b)(c d), (a b c)(d e f), (a b c d)(e f), and (a b c)(d e)(f g), which it is easy to check do not give any other orders. Note that it is not obvious that the orders of elements in A_n are ≤ n - indeed this is false, as shown by the last part of the exercise.
So I found an alternate proof of exercise 3. Could someone check if it holds up?
case 1: all elements are even
case 2: not that
Lemma: you can "go" from any odd cycle to any other odd cycle via an even cycle
(notice: even × even -> even, odd × even -> odd, odd × odd -> even)
a, b are odd => a⁻¹ is odd => a⁻¹b = h which is even => b = ah
let G≤H denote all elements of H that are even.
Via the subgrouptest:
x = τ1 τ2 τ3 ... τn, where n is even => x ∈ G
y = µ1 µ2 µ3 ... µm, where m is even => y ∈ G
xy⁻¹ = τ1 τ2 τ3 ... τn µm... µ3 µ2 µ1 which is of length m+n => even =>
xy⁻¹ ∈ G
=> G is a subgroup
Let h ∈ H be an odd cycle => h ∉ G
Then by lemma, any odd cycle is an element of the coset hG
=> if an element is even, it's an element of G. if it's odd, it's an element of hG.
=> all elements are either in G or hG => G ∪ hG = H
and by theorems about cosets of subgroups, we know that |G| = |hG|
=> [H:G] = 2 => exactly half of all cycles are even, the others are odd
That would be an alternative proof, so congratulations on that. Although I think that parts of the reasoning could be simplified.
Just a couple of suggestions:
1. To prove that G is a subgroup notice that G is the intersection of H with An so it's a subgroup.
2. Once you have proved that |G| = |hG|, you have proved that there are exactly the same amount of even and odd permutations, so there is no need to continue, you have reached your goal.
Finally, just so that your proof is completed, take notice that you proved your lemma proving that b = ah so b ∈ Gh, and in your proof you use hG, this is just a technicality.
Right, that makes sense. Thank you!
9:40 How does one find that solution???
Hello,
Can abstract algebra help me analyze future events??
No
No, but Markov Chains, probability theory, and Stochastics may help
Can't you use chalk on a board? That's much easier to watch!