Problem 23
Question
Prove that the number of distinct conjugates of a subgroup \(H\) of a finite group \(G\) is \([G: N(H)]\)
Step-by-Step Solution
Verified Answer
Question: Prove that the number of distinct conjugates of a subgroup \(H\) of a finite group \(G\) is given by the index of the normalizer of \(H\) in \(G\), denoted by \([G: N(H)]\).
Answer: To prove this, we consider the conjugation action of \(G\) on its subgroups, identify the normalizer and the stabilizer of \(H\) in \(G\), and apply the Orbit-Stabilizer theorem. The result shows that the number of distinct conjugates of \(H\) is equal to \([G: N(H)]\), the index of the normalizer of \(H\) in \(G\).
1Step 1: Identify the normalizer of H in G#
The normalizer of the subgroup \(H\) in the group \(G\), denoted as \(N(H)\), is defined as the set of all elements \(g \in G\) such that \(gHg^{-1} = H\). In other words, \(N(H) = \{g \in G \mid gHg^{-1} = H\}\).
2Step 2: Identify the stabilizer of H under the conjugation action#
Consider the conjugation action of \(G\) on its subgroups. For an element \(g \in G\) and subgroup \(H\), the conjugation action is defined as \(\sigma_g(H)=gHg^{-1}\). The stabilizer of the subgroup \(H\) under this action, denoted \(Stab(H)\), is the set of all elements \(g \in G\) that fixes \(H\) under this action, i.e., \(Stab(H) = \{g \in G \mid \sigma_g(H) = H\}\). Notice that this is precisely the definition of the normalizer of \(H\) in \(G\). Therefore, \(Stab(H) = N(H)\).
3Step 3: Apply the Orbit-Stabilizer theorem to the conjugation action#
The Orbit-Stabilizer theorem states that for a group acting on a set, the order of the group (\(|G|\)) is equal to the size of the orbit of an element (in this case, the number of distinct conjugates of \(H\)) multiplied by the size of the stabilizer of that element (which we have established as \(N(H)\)). Mathematically, this can be written as:
$$
|G|=|\text{distinct conjugates of H}||Stab(H)| = |\text{distinct conjugates of H}||N(H)|.
$$
4Step 4: Calculate the number of distinct conjugates of H#
To find the number of distinct conjugates of \(H\) (the orbit size), we can rearrange the Orbit-Stabilizer theorem equation as follows:
$$
\text{number of distinct conjugates of H} = \frac{|G|}{|N(H)|}.
$$
By definition, the number on the right-hand side of the equation is the index of the normalizer of \(H\) in \(G\), i.e., \([G: N(H)]\). Therefore, the number of distinct conjugates of \(H\) is given by:
$$
\text{number of distinct conjugates of H} = [G: N(H)].
$$
This completes the proof.
Key Concepts
Conjugate SubgroupsNormalizerOrbit-Stabilizer TheoremFinite Groups
Conjugate Subgroups
In group theory, the concept of conjugate subgroups is crucial to understanding the dynamics within a group. For any subgroup \(H\) of a group \(G\), a conjugate of \(H\) by an element \(g\) in \(G\) is the subgroup \(gHg^{-1}\). This operation involves taking each element \(h\) in \(H\), transforming it by \(g\) and its inverse, \(g^{-1}\).Some important points about conjugate subgroups:
- The conjugate group \(gHg^{-1}\) is essentially a 'reshuffling' of \(H\) in a different guise.
- The set of all conjugates of a subgroup forms what is known as the orbit of \(H\) under the conjugation action of \(G\).
- Conjugate subgroups can provide insights into the symmetry and structure of the group, as they reflect the internal alignments and equivalences among subgroup forms.
Normalizer
The normalizer of a subgroup \(H\) in a group \(G\), denoted \(N(H)\), plays a pivotal role in understanding the ambient structure surrounding \(H\). Simply put, \(N(H)\) is the largest subgroup of \(G\) in which \(H\) is normal. It includes all elements \(g\) in \(G\) for which the group \(gHg^{-1}\) equals \(H\).Key insights about normalizers:
- If \(g \in N(H)\), then \(gHg^{-1} = H\), meaning that conjugating \(H\) by \(g\) doesn't change its structure.
- The normalizer can be considered the 'stabilizer' of \(H\) under the conjugation action, maintaining its form during the transformation.
- The size of the normalizer \(|N(H)|\) helps determine the number of distinct conjugates of \(H\) within \(G\).
Orbit-Stabilizer Theorem
The Orbit-Stabilizer Theorem is a fundamental result in group theory that connects the action of a group on a set with key measurements of size. It states:\[ |G| = |\text{orbit of an element}| \times |\text{stabilizer of that element}| \]When applying this theorem to subgroup conjugation, the orbit size is the number of distinct conjugates of \(H\), and the stabilizer is the normalizer \(N(H)\).Essential points about the theorem:
- The theorem provides a beautiful equation linking the group's size, orbit's size, and stabilizer's size.
- In the context of conjugates, it asserts that the number of distinct forms \(H\) can take (via conjugation) is \(\frac{|G|}{|N(H)|}\).
- It ensures group symmetry and helps calculate quantities that reveal depth about group actions.
Finite Groups
In the realm of mathematics, finite groups are those composed of a finite number of elements. This finiteness brings about properties and characteristics that facilitate deep mathematical exploration. Here’s why they matter:
- Finite groups have a precise size, often denoted \(|G|\), which allows for comprehensive calculations and proofs supplied by exact numbers.
- They form the basis for understanding more complex structures like rings and fields in abstract algebra.
- Their study leads directly to key results in mathematics, such as Lagrange's Theorem, which connects subgroups and their sizes directly with the entire group's size.
Other exercises in this chapter
Problem 21
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Prove that a Sylow 2-subgroup of \(S_{5}\) is isomorphic to \(D_{4}\).
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(a) Suppose \(p\) is prime and \(p\) does not divide \(m\). Show that $$ p \nmid\left(\begin{array}{c} p^{k} m \\ p^{k} \end{array}\right) . $$ (b) Let \(\mathc
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