Group Theory Permutation (Hints and )

[MellyVG257]

Homework Statement

1. Let n ≥ 2. Let H = {σ ∈ S_n: ord(σ) = 2}. Decide whether or not H is a subgroup of S_n.
2. Let G be a group of even order. Show that the cardinality of the set of elements of G that have order 2 is odd.

The Attempt at a Solution

1. I have no idea where to start with this. I tried looking at the rule and using the 3 rules for H to be a subgroup to prove it, but I don't know what to do with the order. Really lost with this. Any ideas on how to start it?

2. I tried to solve it using order as being the cardinality of the set. But I think I am missing something. I know what a group with order 2 is [(12)(24), right?] and I know that the sgn for even and odd and 1, -1 respectively. But this is where I became stuck. Any ideas?

 

[anathema]

1. To show that H is a subgroup of S_n, you need to prove that it satisfies the three conditions for being a subgroup: closure, identity, and inverse.

First, let's define what H is. H is the set of all permutations in S_n that have an order of 2. This means that when you apply the permutation twice, you get back to the original element. For example, (12) has an order of 2 because (12)^2 = (1)(2) = identity element.

Now, to show closure, we need to show that if we take two elements from H, their composition (or product) is also in H. Let's take two elements from H, say (12) and (34). Their composition is (12)(34) = (1432), which has an order of 2. So, (1432) is also in H and H is closed under composition.

Next, we need to show that H contains an identity element. The identity element in S_n is the permutation that leaves all elements unchanged, which is (1)(2)(3)...(n). This permutation has an order of 1, which is not 2. So, the identity element is not in H. Therefore, H is not a subgroup of S_n.

2. Let's start by defining the set of elements in G that have an order of 2. We'll call this set E. We know that the elements in E have an order of 2, which means that when we apply them twice, we get back to the identity element.

Now, let's consider the elements in G that do not have an order of 2. These elements have an order of 1, which means that when we apply them once, we get back to the identity element.

Since G has an even order, we know that there must be an even number of elements in G. If we take out all the elements in G that have an order of 2, we are left with an odd number of elements because an even number minus an even number is an odd number. Therefore, the cardinality of E must be odd.

 

[kerimek]

1. To determine whether H is a subgroup of S_n, we need to check if it satisfies the three conditions for a subgroup: closure, identity, and inverses.
For closure, we need to show that the composition of any two elements in H is also in H. Since the operation in S_n is composition, this means we need to show that the composition of two permutations of order 2 is also of order 2. This can be easily proven by considering the possible cycle structures of permutations of order 2.
For identity, we need to show that the identity element (the identity permutation) is in H. This is true since the identity permutation has order 1, which is not equal to 2.
For inverses, we need to show that for any element in H, its inverse is also in H. Since the order of a permutation is equal to the least common multiple of the lengths of its cycles, this means that the inverse of a permutation of order 2 must also have order 2. Therefore, the inverse of any element in H is also in H. Thus, H satisfies all three conditions and is a subgroup of S_n.

2. Let G be a group of even order, say 2k. We want to show that the cardinality of the set of elements in G with order 2 is odd.
Let H be the set of elements in G with order 2. Since the order of an element divides the order of the group, we know that the order of any element in H must divide 2k. This means that the possible orders of elements in H are 1, 2, k, and 2k.
Since G has an even order, we know that there must be an even number of elements in G with order 1 and k, since these are the only orders that divide 2k and are not equal to 2. This means that the total number of elements in G with order 1 or k is even.
Since the total number of elements in G is 2k, and the number of elements with order 1 or k is even, the number of elements with order 2 must be odd. Therefore, the cardinality of H, the set of elements with order 2, is odd.