Problem 18
Question
Let \(G\) be a cyclic \(p\) -group with subgroups \(H\) and \(K\). Prove that either \(H\) is contained in \(K\) or \(K\) is contained in \(H\).
Step-by-Step Solution
Verified Answer
Question: In a cyclic \(p\)-group \(G\), given two subgroups \(H\) and \(K\), either \(H\) is a subset of \(K\) or \(K\) is a subset of \(H\). Explain why this is true.
Answer: In a cyclic \(p\)-group \(G\), every subgroup is also cyclic and has an order that is a power of \(p\). When examining the intersection of subgroups \(H\) and \(K\), we find that its order is also a power of \(p\). Depending on which order is smaller, either \(H\) becomes a subset of \(K\) or \(K\) becomes a subset of \(H\), as their intersection must also be a power of \(p\). This results from the property that the order of an intersection of subgroups divides the orders of the original subgroups.
1Step 1: Identify the properties of a cyclic group
A cyclic group \(G\) is a group that is generated by a single element, denoted as \(g\). This means that every element in \(G\) can be written as \(g^n\) for some integer \(n\). The order of \(G\), denoted as \(|G|\), is the smallest positive integer \(n\) such that \(g^n = e\), where \(e\) is the identity element in \(G\).
In this exercise, we are given that \(G\) is a cyclic \(p\)-group, meaning that its order \(|G|\) is a power of a prime number \(p\).
2Step 2: Identify the properties of subgroups of a cyclic group
Every subgroup \(H\) of a cyclic group \(G\) is itself a cyclic group. This means it will be generated by an element \(h\), with every element in \(H\) written as \(h^m\) for some integer \(m\).
A key property of subgroups of cyclic groups is that the order of the subgroup \(|H|\) divides the order of the group \(|G|\). In other words, if \(|G|\) is a power of a prime number \(p\), the order of any subgroup \(H\) will also be a power of \(p\).
3Step 3: Use properties of orders of subgroups of a cyclic group
As mentioned in Step 2, the order of a subgroup \(H\) of a cyclic group \(G\) must divide the order of \(G\). Let \(|H| = p^a\) and \(|K| = p^b\), where \(a\) and \(b\) are non-negative integers. Since \(|G| = p^n\) for some positive integer \(n\), we have \(p^a\) divides \(p^n\) and \(p^b\) divides \(p^n\).
Note that \(|H \cap K|\) divides both \(|H|\) and \(|K|\) as \(H\cap K\) is a subgroup of both \(H\) and \(K\). Since \(|H|\) and \(|K|\) are both powers of \(p\), we can conclude that \(|H\cap K|\) must also be a power of \(p\).
4Step 4: Analyze the intersection of subgroups H and K
Let \(|H \cap K| = p^c\), where \(c \leq \min(a, b)\). Now, we will consider two cases:
1) If \(c = a\), then \(|H \cap K| = |H|\). In this case, we can conclude that \(H = H \cap K \subseteq K\).
2) If \(c = b\), then \(|H \cap K| = |K|\). In this case, we can conclude that \(K = H \cap K \subseteq H\).
In either case, we see that either \(H \subseteq K\) or \(K \subseteq H\), which was the goal of the exercise.
Key Concepts
Subgroupsp-groupIntersection of SubgroupsOrder of a Group
Subgroups
In group theory, a subgroup is a smaller group within a given group that itself satisfies the group properties. For a group \( G \) to have a subgroup \( H \), \( H \) must:
- Be closed under the group operation, meaning if any two elements \( a \) and \( b \) are in \( H \), then their product \( ab \) is also in \( H \).
- Contain the identity element of \( G \), which acts as a neutral element.
- Include the inverse of each of its elements, ensuring that if \( a \) is in \( H \), then its inverse \( a^{-1} \) is also in \( H \).
p-group
A \( p \)-group is a type of group in which the order (size) of the group is a power of a prime number \( p \). For example, if the order of the group is \( p^n \) where \( n \) is a non-negative integer, it is called a \( p \)-group. Cyclic \( p \)-groups are particularly simple because they are generated by a single element.
Some interesting properties of \( p \)-groups include:
Some interesting properties of \( p \)-groups include:
- They are always solvable, meaning they can be broken down into abelian groups through a series of normal subgroups.
- Their center, which is the set of elements that commute with all other elements in the group, is non-trivial.
- Every subgroup of a \( p \)-group is also a \( p \)-group.
Intersection of Subgroups
The intersection of two subgroups \( H \) and \( K \) of a given group \( G \) is a new subgroup consisting of elements that are in both \( H \) and \( K \). Symbolically, it is denoted as \( H \cap K \).
To qualify as a subgroup itself, the intersection must:
To qualify as a subgroup itself, the intersection must:
- Be non-empty, which is always true since the identity element must belong to any subgroup.
- Be closed under the group operation.
- Contain the inverse of every element within the intersection.
Order of a Group
The order of a group refers to the total number of elements it contains. For a finite group \( G \), the order is simply \(|G|\). Understanding the order is fundamental in group theory as it offers insights into the group's structure and properties.
For instance, in a cyclic group which has an order that is a power of \( p \) (making it a \( p \)-group), we find useful characteristics:
For instance, in a cyclic group which has an order that is a power of \( p \) (making it a \( p \)-group), we find useful characteristics:
- The order of any cyclic group is tied to the least power such that the group's generator raised to that power equals the identity element.
- Any subgroup's order divides the order of the group itself, according to Lagrange's Theorem.
- The order of a subgroup within a \( p \)-group must be a power of \( p \) as well.
Other exercises in this chapter
Problem 15
Let \(G\) be a solvable group and \(N\) a normal subgroup of \(G\). Prove that \(G / N\) is solvable.
View solution Problem 17
Suppose that \(G\) has a composition series. If \(N\) is a normal subgroup of \(G,\) show that \(N\) and \(G / N\) also have composition series.
View solution Problem 19
Suppose that \(G\) is a solvable group with order \(n \geq 2\). Show that \(G\) contains a normal nontrivial abelian subgroup.
View solution Problem 21
Suppose that \(G\) is a solvable group with order \(n \geq 2 .\) Show that \(G\) contains a normal nontrivial abelian factor group.
View solution