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Chapter 6: Problem 9

Show that if \(H\) is a subgroup of an abelian group \(G,\) then a set \(K \subseteq H\) is a subgroup of \(G\) if and only if \(K\) is a subgroup of \(H\).

Short Answer

Expert verified
Question: Show that if \(H\) is a subgroup of an abelian group \(G\) and \(K\subseteq H\), then \(K\) is a subgroup of \(G\) if and only if \(K\) is a subgroup of \(H\). Answer: To show that \(K\) is a subgroup of \(G\) if and only if \(K\) is a subgroup of \(H\), we verified three properties: identity, closure, and inverses. We used the given condition that \(K \subseteq H\) and the fact that \(G\) is an abelian group, which implies that \(H\) is also an abelian group. By checking each property, we confirmed that \(K\) is a subgroup of \(G\) if and only if it is a subgroup of \(H\).

Step by step solution

01

Identity element

We need to show that the identity element of \(G\) is also in \(K\). This is true because, as \(K \subseteq H\) and \(H\) is a subgroup of \(G\), the identity element of \(G\) is also the identity element of \(H\). Therefore, the identity element of \(G\) and \(H\) is also in \(K\).
02

Closure

We need to show that if \(a, b \in K\), then \(a\ast b \in K\). Since \(K \subseteq H\), we have that \(a, b \in H\). Since \(H\) is a subgroup and thus closed under the operation, we get \(a \ast b \in H\). As \(K\) is a subgroup of \(H\), it must also be closed. Thus, \(a \ast b \in K\).
03

Inverses

We need to show that if \(a \in K\), then the inverse of \(a\) is also in \(K\). Since \(K \subseteq H\), we have that \(a \in H\). The inverse of \(a\), denoted \(a^{-1}\), must belong to \(H\) as \(H\) is a subgroup of \(G\). Since \(K\) is a subgroup of \(H\), the inverses must also exist in \(K\). Therefore, if \(a \in K\), then its inverse \(a^{-1}\) exists in \(K\) as well. By verifying these three properties (identity, closure, and inverses), we have shown that if \(H\) is a subgroup of an abelian group \(G\) and \(K\subseteq H\), then \(K\) is a subgroup of \(G\) if and only if \(K\) is a subgroup of \(H\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Group Theory
Group theory is a fascinating area of mathematics fundamentally concerned with the study of symmetry, structure, and transformations. A group is a set combined with an operation that satisfies four key properties:
  • There must be an identity element that when combined with any element of the group, leaves it unchanged.
  • The operation must be associative, meaning that the way elements are grouped during the operation does not change the outcome.
  • Every element must have an inverse element that when combined with the original, results in the identity element.
  • The group must be closed under the given operation, implying any operations performed on elements of the set yield a result that is also in the set.

In the context of the exercise, we're interested in subgroups, which are subsets of a group that themselves form a group with the same operation. This brings us to analyze specific properties that a subset must satisfy to be considered a subgroup.
Identity Element
In group theory, the identity element is a cornerstone concept. It is the element within a group that, when you perform the group's operation between any element and the identity, you get the original element back. Think of it as the number '1' in multiplication; no matter what you multiply 1 with, you'll always end up with the other number.

In the example of the subgroup, it's essential to verify that the identity element of the larger group, \(G\), is also present in the potential subgroup \(K\). This is because to fulfill the requirements of being a subgroup, \(K\) must mirror the properties of the larger group, including having this neutral element. Without the identity element, \(K\) could not possibly be a subgroup as it would lack a fundamental characteristic of group structure.
Closure Property
The closure property is quite intuitive. It is the principle that when you take any two elements from a group and combine them using the group's operation, the result must also be within the group.

For subgroups, this implies a slightly more complex situation. When you have a subgroup \(H\) of a group \(G\), and a set \(K\) is within \(H\), for \(K\) to be a subgroup of \(G\), any combination of elements from \(K\) must remain within \(K\). Thus, if you pick any two elements from \(K\) and apply the group operation, the result doesn't just have to be in \(H\) (since \(K\subseteq H\)), but must also be an element of \(K\). Therefore, verifying closure is crucial when determining whether a subset is a subgroup.
Inverse Elements
In group theory, the inverse of an element is like a puzzle piece that, when combined with the original piece, creates a complete picture of the identity element. An inverse element effectively cancels out another element. So in groups, for every element, there is another that combined with it (in any order, if the group is Abelian), returns the identity element.

In the case of subgroups, this means if you take any element in a potential subgroup \(K\), not only must an inverse exist in the larger group \(G\), but this inverse must also be present within \(K\). This feature ensures that every aspect of the group's structure is preserved within the subgroup, which is essential for the subgroup to inherit the group's properties and qualify as a bona fide subgroup itself.

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Most popular questions from this chapter

Let \(G\) be an abelian group and \(H\) a subgroup with \([G: H]=2 .\) Show that if \(a, b \in G \backslash H,\) then \(a+b \in H\).

Suppose that \(G, G_{1},\) and \(G_{2}\) are abelian groups, and that \(\rho\) : \(G_{1} \times G_{2} \rightarrow G\) is a group isomorphism. Let \(H_{1}:=\rho\left(G_{1} \times\left\\{0_{G_{2}}\right\\}\right)\) and \(H_{2}:=\) \(\rho\left(\left\\{0_{G_{1}}\right\\} \times G_{2}\right) .\) Show that \(G\) is the internal direct product of \(H_{1}\) and \(H_{2}\).

Let \(G\) be an abelian group. (a) Show that if \(H\) is a subgroup of \(G, h \in H,\) and \(g \in G \backslash H,\) then \(h+g \in G \backslash H\) (b) Suppose that \(H\) is a non-empty subset of \(G\) such that for all \(h, g \in G:\) (i) \(h \in H\) implies \(-h \in H,\) and (ii) \(h \in H\) and \(g \in G \backslash H\) implies \(h+g \in G \backslash H\). Show that \(H\) is a subgroup of \(G\)

Let \(G\) be an abelian group, and let \(\sim\) be an equivalence relation on \(G\). Further, suppose that for all \(a, a^{\prime}, b \in G,\) if \(a \sim a^{\prime},\) then \(a+b \sim a^{\prime}+b .\) Let \(H:=\left\\{a \in G: a \sim 0_{G}\right\\} .\) Show that \(H\) is a subgroup of \(G,\) and that for all \(a, b \in G\) we have \(a \sim b\) if and only if \(a \equiv b(\bmod H)\).

Show that if \(G=G_{1} \times G_{2}\) for abelian groups \(G_{1}\) and \(G_{2},\) and \(H_{1}\) is a subgroup of \(G_{1}\) and \(H_{2}\) is a subgroup of \(G_{2},\) then we have a group isomorphism \(G /\left(H_{1} \times H_{2}\right) \cong G_{1} / H_{1} \times G_{2} / H_{2}\).
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