Problem 26
Question
Let \(H\) and \(K\) be subgroups of a group \(G\). Show that \(H K\) is a subgroup of \(G\) if and only if \(H K=K H\).
Step-by-Step Solution
Verified Answer
\( HK \) is a subgroup of \( G \) if and only if \( HK = KH \).
1Step 1: Understanding the Problem
We need to show that the product of two subgroups \( H \) and \( K \), denoted as \( HK \), is also a subgroup of \( G \) if and only if \( HK = KH \). This means we need to show two implications: (1) If \( HK \) is a subgroup, then \( HK = KH \). (2) If \( HK = KH \), then \( HK \) is a subgroup.
2Step 2: Proving First Implication
Assume \( HK \) is a subgroup of \( G \). For any \( h \in H \) and \( k \in K \), \( hk \in HK \). Since \( HK \) is a subgroup, it is closed under taking inverses. Hence, \((hk)^{-1} = k^{-1}h^{-1} \in HK \). This implies there exist \( h' \in H \) and \( k' \in K \) such that \( k^{-1}h^{-1} = h'k' \). Rearranging gives \( hk = k'(h')^{-1} \), showing that each element of \( HK \) can be expressed as any element of \( KH \), hence \( HK \subseteq KH \). By symmetry, \( KH \subseteq HK \), thus \( HK = KH \).
3Step 3: Proving Second Implication
Assume \( HK = KH \). We need to show that \( HK \) is a subgroup. To demonstrate subgroup properties, we need to show closure and inverses are contained in \( HK \). For closure, take any \( hk, h'k' \in HK \). Since \( KH = HK \), reorder and combine: \( hk \cdot h'k' = h(hk'h')k' \in HK \). For inverses, \( (hk)^{-1} = k^{-1}h^{-1} \). Since \( KH = HK \), \( k^{-1}h^{-1} \) can be expressed as an element of \( KH \), and thus in \( HK \), proving that \( HK \) is a subgroup.
4Step 4: Conclusion
We have shown both implications: If \( HK \) is a subgroup then \( HK = KH \), and if \( HK = KH \) then \( HK \) is a subgroup. Therefore, \( HK \) is a subgroup of \( G \) if and only if \( HK = KH \).
Key Concepts
Group OperationCosetsClosure Property
Group Operation
In group theory, a **group operation** is a fundamental concept that defines how elements combine within a group. A group is a set equipped with an operation that takes any two elements of the group and produces another element of the group.
- **Associativity**: The operation must satisfy the associative law, meaning that for any elements \( a, b, \) and \( c \) in the group, \( (ab)c = a(bc) \).
- **Identity Element**: There exists an identity element \( e \) in the group such that for any element \( a \), \( ae = ea = a \).
- **Inverse Element**: For every element \( a \) in the group, there exists an inverse element \( a^{-1} \) such that \( aa^{-1} = a^{-1}a = e \).
Cosets
Cosets are collections derived from a subgroup within a larger group. They are used to partition groups into disjoint subsets. Given a subgroup \( H \) of a group \( G \), any element of \( G \) can be paired with \( H \) to form what is called a **coset**.
- **Left Coset**: For an element \( g \) in \( G \), the left coset of \( H \) in \( G \) is \( gH = \{ gh : h \in H \} \).
- **Right Coset**: Similarly, the right coset of \( H \) with respect to \( g \) is \( Hg = \{ hg : h \in H \} \).
Closure Property
The **closure property** is a vital component of group theory that ensures for any elements \( a \) and \( b \) in a group, the result of the group operation, \( ab \), is also in the group. This property is a required condition for any subset to be considered a group.When dealing with sets like \( HK \) (the product of subgroups \( H \) and \( K \) of a group \( G \)), closure ensures that every possible combination of elements from \( H \) and \( K \) stays within the group \( G \).
- **Element Pairing**: If \( h \in H \) and \( k \in K \), then the product \( hk \) must be in \( HK \) if it is a group.
- **Subgroup Combination**: The operation must behave such that any combination \( hk \) and further products \( (hk)(h'k') \) remain in \( HK \), verifying the closure law for \( HK \).
Other exercises in this chapter
Problem 25
Find the remainder of \(9^{1573}\) when divided by 11 .
View solution Problem 25
In Exercises 23 through 30 construct an example of a nontrivial homomorphism between the two indicated groups, if this is possible, or explain why this is not p
View solution Problem 26
Show that in \(S_{4}\) the subgroup generated by \\{(12),(1234)\\} (in the sense of the preceding Exercise 25 ) is the whole group: \(\langle(12),(1234)\rangle=
View solution Problem 26
Compute \(\phi\left(p^{2}\right),\) where \(p\) is prime.
View solution