Derived Subgroups Of Sn And Dn: A Detailed Explanation

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Hey guys! Today, we're diving deep into the fascinating world of group theory, specifically focusing on derived subgroups. We'll be tackling the symmetric group $S_n$ and the dihedral group $D_n$, building upon the knowledge that the derived subgroup of $S_3$ is $A_3$ and that of $D_4$ is cyclic of order 2. Let's unravel the mysteries behind calculating these derived subgroups for general $S_n$ and $D_n$.

What are Derived Subgroups?

Before we jump into the specifics of $S_n$ and $D_n$, let's refresh our understanding of derived subgroups. The derived subgroup, also known as the commutator subgroup, of a group G, denoted as G', [G, G], or $G^{(1)}$, is a subgroup generated by all commutators of G. A commutator is an element of the form aba⁻¹b⁻¹, where a and b are elements of G. Think of it as a measure of how "non-commutative" a group is. If G is abelian (commutative), then every commutator is the identity element, and the derived subgroup is trivial (containing only the identity).

Keywords in Focus: Derived subgroup, commutator, group theory, symmetric group, dihedral group, abelian group, identity element.**

The derived subgroup $G'$ is crucial because it helps us understand the structure of a group. It's the smallest normal subgroup of G such that the quotient group G/ G' is abelian. This means that G' essentially captures all the non-abelian behavior within G. To calculate the derived subgroup, we first identify the commutators in the group and then find the subgroup generated by these commutators. This might sound straightforward, but it can become quite intricate for larger groups.

Importance of Derived Subgroups:

1. Measuring Non-commutativity: The derived subgroup G' quantifies the extent to which a group G deviates from being abelian. If G' is trivial, the group is abelian, and as G' grows, G becomes less commutative.
2. Abelianization: The quotient group G/ G' is abelian, providing an "abelianized" version of G. This quotient group simplifies the analysis and allows us to focus on the abelian aspects of G.
3. Solvability: The concept of derived subgroups is central to defining solvable groups. A group G is solvable if its derived series, formed by repeatedly taking derived subgroups (G, G', G'', ...), eventually reaches the trivial subgroup {e}. Solvable groups have significant applications in Galois theory and the solution of polynomial equations.
4. Structure Analysis: Derived subgroups help decompose complex group structures into simpler, more manageable components. Understanding the derived subgroups often reveals crucial information about the group's composition and properties.
5. Applications in Physics: In physics, particularly in quantum mechanics, commutators of operators play a crucial role. The derived subgroup concept helps in understanding the algebraic structures underlying physical systems.

Let's solidify our understanding with an example. Consider the symmetric group $S_3$, which consists of all permutations of three elements. The derived subgroup of $S_3$ is the alternating group $A_3$, which contains the even permutations (identity and 3-cycles). This means that the "non-commutative part" of $S_3$ is precisely captured by $A_3$, and the quotient group $S_3$/$A_3$ is isomorphic to $\mathbb{Z}_2$, which is abelian.

Now, let's move on to the main focus of our discussion: finding the derived subgroups of $S_n$ and $D_n$ for general n.

Derived Subgroup of the Symmetric Group $S_n$

Let's tackle the derived subgroup of the symmetric group, $S_n$. The symmetric group $S_n$ is the group of all permutations of n elements, and its order is n!. The alternating group, $A_n$, is the subgroup of $S_n$ consisting of all even permutations. Our goal is to determine the commutator subgroup $[S_n, S_n]$.

Keywords in Focus: Symmetric group, alternating group, even permutations, odd permutations, commutator subgroup, generators, cycles, conjugation.**

It turns out that the derived subgroup of $S_n$ is closely related to the alternating group $A_n$. Specifically:

• For n ≥ 2, the derived subgroup of $S_n$ is $A_n$.

Let's break down why this is the case. We know that the alternating group $A_n$ is generated by 3-cycles. A 3-cycle is a permutation that cyclically permutes three elements and leaves the others fixed. For example, in $S_4$, (1 2 3) is a 3-cycle that sends 1 to 2, 2 to 3, and 3 to 1, while leaving 4 unchanged.

To prove that the derived subgroup $[S_n, S_n]$ is equal to $A_n$, we need to show two things:

1. Every commutator in $S_n$ is an element of $A_n$.
2. Every element of $A_n$ can be written as a product of commutators in $S_n$.

First, let's show that every commutator in $S_n$ is an element of $A_n$. Consider a commutator xyx⁻¹y⁻¹, where x and y are elements of $S_n$. Recall that the sign of a permutation is +1 if it's an even permutation and -1 if it's an odd permutation. The sign is a homomorphism from $S_n$ to the multiplicative group {+1, -1}. Therefore,

sign(xyx⁻¹y⁻¹) = sign(x)sign(y)sign(x⁻¹)sign(y⁻¹) = sign(x)sign(y)sign(x)⁻¹sign(y)⁻¹ = (+1)(+1) = 1.

This implies that xyx⁻¹y⁻¹ is an even permutation, and thus it belongs to $A_n$. So, every commutator in $S_n$ is an element of $A_n$, which means that the derived subgroup $[S_n, S_n]$ is a subgroup of $A_n$.

Now, let's show that every element of $A_n$ can be written as a product of commutators in $S_n$. We know that $A_n$ is generated by 3-cycles. Therefore, if we can show that every 3-cycle is a commutator (or a product of commutators), we'll have proven that $A_n$ is contained in the derived subgroup $[S_n, S_n]$.

Consider a 3-cycle (a b c). We can express it as a commutator in the following way:

(a b c) = (a b)(b c)(a b)⁻¹(b c)⁻¹ = (a b)(b c)(a b)(b c)

However, this is incorrect, but it illustrates the idea that 3-cycles can be written using transpositions. A more accurate approach to show that a 3-cycle is in the commutator subgroup is to use the identity:

(a b c) = (a b)(a c)(a b)(a c) = [(a b), (a c)]

This shows that every 3-cycle can be expressed as a commutator of two transpositions. Since $A_n$ is generated by 3-cycles, and every 3-cycle is a commutator, every element of $A_n$ can be written as a product of commutators. Thus, $A_n$ is contained in the derived subgroup $[S_n, S_n]$.

Combining these two results, we conclude that for n ≥ 2, the derived subgroup of $S_n$ is $A_n$. For the special case of $S_1$, which is the trivial group with only the identity element, the derived subgroup is also trivial.

Derived Subgroup of the Dihedral Group $D_n$

Now, let's shift our focus to the dihedral group $D_n$. The dihedral group $D_n$ is the group of symmetries of a regular n-sided polygon, including rotations and reflections. It has 2n elements. We want to find the derived subgroup of $D_n$.

Keywords in Focus: Dihedral group, rotations, reflections, generators, relations, commutators, cyclic subgroups, derived series.**

Recall that $D_n$ can be represented by generators r and s, where r is a rotation by 2π/n and s is a reflection. These generators satisfy the relations:

• rⁿ = e (where e is the identity element)
• s² = e
• srs⁻¹ = r⁻¹

To find the derived subgroup, we need to compute the commutators in $D_n$. A general element in $D_n$ can be written as rⁱsʲ, where 0 ≤ i < n and j = 0 or 1. Let's consider the commutator of two general elements rⁱsʲ and rᵏsˡ:

[rⁱsʲ, rᵏsˡ] = (rⁱsʲ)(rᵏsˡ)(rⁱsʲ)⁻¹(rᵏsˡ)⁻¹

Let's break this down into cases:

1. Commutator of two rotations:

[rⁱ, rᵏ] = rⁱrᵏr⁻ⁱr⁻ᵏ = e (since rotations commute with each other)

2. Commutator of a rotation and a reflection:

[rⁱ, s] = rⁱs r⁻ⁱ s⁻¹ = rⁱs r⁻ⁱ s = rⁱ (sr⁻ⁱ s) = rⁱ rⁱ = r²ⁱ

3. Commutator of two reflections:

[s, rᵏs] = s(rᵏs)s⁻¹(rᵏs)⁻¹ = s rᵏ s s r⁻ᵏ = s rᵏ s r⁻ᵏ = (s rᵏ s) r⁻ᵏ = r⁻ᵏ r⁻ᵏ = r⁻²ᵏ

From these calculations, we see that commutators in $D_n$ are of the form r²ⁱ. This means that the derived subgroup will be generated by elements of the form r². Let's analyze this further:

• If n is odd, then the set {2i mod n | i ∈ \mathbb{Z}} generates all integers modulo n. In this case, the derived subgroup is the cyclic group generated by r, which is <r>.
• If n is even, say n = 2m, then the set {2i mod n | i ∈ \mathbb{Z}} generates the subgroup of even integers modulo n. In this case, the derived subgroup is the cyclic group generated by r², which is <r²>.

Therefore, the derived subgroup of $D_n$ can be summarized as follows:

• If n is odd, [Dₙ, Dₙ] = <r>, which is isomorphic to $\mathbb{Z}_n$.
• If n is even, [Dₙ, Dₙ] = <r²>, which is isomorphic to $\mathbb{Z}_{n/2}$.

Let's look at some examples to make this clearer:

• For D₄ (symmetries of a square), the derived subgroup is <r²>, which is a cyclic group of order 2. This matches our initial information.
• For D₅ (symmetries of a pentagon), the derived subgroup is <r>, which is a cyclic group of order 5.
• For D₆ (symmetries of a hexagon), the derived subgroup is <r²>, which is a cyclic group of order 3.

Summary and Key Takeaways

Let's recap what we've learned about the derived subgroups of $S_n$ and $D_n$:

• For the symmetric group $S_n$ (n ≥ 2), the derived subgroup is the alternating group $A_n$. This means that the commutators in $S_n$ generate all the even permutations.
• For the dihedral group $D_n$, the derived subgroup depends on whether n is odd or even:
  • If n is odd, the derived subgroup is the cyclic group of rotations, isomorphic to $\mathbb{Z}_n$.
  • If n is even, the derived subgroup is a cyclic subgroup generated by twice the rotation, isomorphic to $\mathbb{Z}_{n/2}$.

Understanding derived subgroups helps us to dissect the structure of groups and understand their non-commutative behavior. The concepts and methods discussed here can be extended to other groups as well.

I hope this comprehensive guide has shed some light on the derived subgroups of $S_n$ and $D_n$. Keep exploring the fascinating world of group theory, and you'll uncover many more exciting results!