Group Theory Hints: Proving a Subgroup of Order 2n+1 for |G| = 4n+2

[wglmb]

Hi everyone
This is the same question as was asked about in this topic, but I can't post in that one (presumably because it's archived?)

Homework Statement

Suppose G is a group with |G| = 4n+2. Show that there is a subgroup H < G such that |H| = 2n + 1. Use Cauchy's theorem, Cayley's theorem and the fact that any subgroup of [itex]S_n[/itex] has either all of its elements or precisely half of its elements being even.

Homework Equations

Cayley: Every group G is a subgroup of [itex]S_|_G_|[/itex].

Cauchy: If G is a finite group & P [tex]\in[/tex] N is prime with P | |G|, then G contains an element of order P.

The Attempt at a Solution

Cayley [tex]\Rightarrow G \leq[/tex] [itex]S_4_n_+_2[/itex] [tex]\Rightarrow[/tex] all or half of its elements are even (*)

Cauchy [tex]\Rightarrow[/tex] G contains an element of order 2 (since 2 | 4n+2)
[tex]\Rightarrow[/tex] G contains an element that's a product of disjoint transpositions.

Suppose G contains an element that's a product of an odd number of disjoint transpositions. (**)
[tex]\Rightarrow[/tex] Half the elements in G are even, by (*), i.e. 2n+1 of them are even.
These 2n+1 elements form a group (since they're even, their products are even, their inverses are even, and 1 is even), which is a subgroup of G with order 2n+1.

... So what I need to do now is show that G contains an element that's a product of an odd number of disjoint transpositions, i.e. prove that (**) is true.
And that's where I'm stuck... I have no idea what to do.. Can I have a nudge in the right direction?
Thanks

 

[anathema]

for your post and for bringing this topic back up for discussion. To address your question, let's first review the given information:

- We know that |G| = 4n+2 and we want to show that there exists a subgroup H < G such that |H| = 2n+1.
- We are given two theorems: Cayley's theorem and Cauchy's theorem.
- We also have the fact that any subgroup of S_n has either all of its elements or precisely half of its elements being even.

Now, let's think about how we can use these tools to prove our desired result. First, we can use Cayley's theorem to say that G is a subgroup of S_(4n+2). This means that G is a permutation group, and therefore, each element in G can be written as a product of disjoint cycles.

Next, we can use Cauchy's theorem to say that G contains an element of order 2. This means that there exists an element in G that can be written as a product of two disjoint transpositions.

Now, let's think about the possible ways this element can be written as a product of disjoint cycles. Since G is a subgroup of S_(4n+2), any element in G can be written as a product of disjoint cycles with lengths that divide 4n+2. This means that our element with order 2 can be written as a product of two disjoint cycles with lengths that divide 4n+2.

But since we want to show that there exists a subgroup H < G with order 2n+1, we need to find an element in G that can be written as a product of an odd number of disjoint cycles. This is where the given fact comes in - since G is a subgroup of S_n, we know that either all of its elements or precisely half of its elements are even. This means that there must exist an element in G that can be written as a product of an odd number of disjoint cycles.

Using this element, we can form a subgroup H with order 2n+1, as desired. Therefore, we have shown that there exists a subgroup H < G such that |H| = 2n+1.

I hope this helps guide you in the right direction. Let me know if you have any further questions or need clarification on any of the steps. Keep up the good work!

 

[nlightner]

Hello,

Thank you for your question. It seems that you are on the right track with using Cayley's and Cauchy's theorems to prove the existence of a subgroup of order 2n+1 in a group of order 4n+2. To prove that G contains an element that is a product of an odd number of disjoint transpositions, you can use the fact that any element in S_n can be written as a product of disjoint cycles and that the number of transpositions in a cycle decomposition determines whether the element is even or odd.

First, consider the group G as a subgroup of S_{4n+2}. By Cayley's theorem, we know that G is isomorphic to a subgroup of S_{4n+2}. Therefore, any element in G can be written as a product of disjoint cycles in S_{4n+2}. Now, let's assume that all elements in G can be written as a product of an even number of disjoint transpositions. This means that all elements in G are even elements in S_{4n+2}, which contradicts (*) in your attempt at a solution. Therefore, there must exist at least one element in G that is a product of an odd number of disjoint transpositions.

To prove this, you can use the fact that any element in S_n can be written as a product of disjoint cycles, and the number of transpositions in the cycle decomposition determines whether the element is even or odd. Since G is a subgroup of S_{4n+2}, this means that any element in G can also be written as a product of disjoint cycles, and the number of transpositions in the cycle decomposition also determines whether the element is even or odd. Therefore, if all elements in G can be written as a product of an even number of disjoint transpositions, this would imply that all elements in G are even elements in S_{4n+2}, which again contradicts (*) in your attempt at a solution.

In conclusion, by assuming that all elements in G can be written as a product of an even number of disjoint transpositions, we arrive at a contradiction. Therefore, there must exist at least one element in G that is a product of an odd number of disjoint transpositions, which means that G contains at least one odd element. This odd element, along with the identity element, forms a subgroup of order 2n+1 in G.

I hope this helps