Problem 22 - Algebra: Pure and Applied

Question

Let \(G=\left\\{n+m i \mid m, n \in \mathbb{Z}, i^{2}=-1\right\\}\). Show that \(G\) is a subgroup of \(C\) under addition.

Step-by-Step Solution

Answer

\(G\) is a subgroup of \(C\) under addition.

Step 1: Definition of a Subgroup

To show that a subset \(G\) is a subgroup of a group \(C\) under addition, we need to verify three conditions: closure under addition, existence of an identity element, and existence of inverses.

Step 2: Verify Closure Under Addition

We need to show that if we take any two elements \(a = n_1 + m_1 i\) and \(b = n_2 + m_2 i\) from \(G\), their sum \(a + b\) is also in \(G\).Calculate: \[a + b = (n_1 + m_1 i) + (n_2 + m_2 i) = (n_1 + n_2) + (m_1 + m_2) i\]Since \(n_1, n_2, m_1, m_2\) are integers, \(n_1 + n_2\) and \(m_1 + m_2\) are also integers. This means \(a + b\) belongs to \(G\), showing closure under addition.

Step 3: Verify the Identity Element

The identity element for addition in the group of complex numbers is \(0 + 0i\). Since \(0\) and \(0\) are integers, \(0 + 0i\) belongs to \(G\), serving as the identity element.

Step 4: Verify Existence of Inverses

For any element \(a = n + mi\) in \(G\), its additive inverse is \(-a = -n - mi\). Since \(-n\) and \(-m\) are both integers, \(-a\) belongs to \(G\), showing that every element in \(G\) has an inverse in \(G\).

Key Concepts

Understanding Closure Under Addition

To understand why closure under addition is important for subgroups, let's consider what it means when we add two elements in a set. For a subset to be closed under addition, whenever you take two elements from this set, their sum must also be in the same set. This condition ensures that the operation of addition keeps you within the boundaries of the set, not venturing outside.

In our case, with the set \(G\) given as \(\{n + mi \mid n, m \in \mathbb{Z}, i^2 = -1\}\), we take any two elements, say \(a = n_1 + m_1 i\) and \(b = n_2 + m_2 i\). The sum \(a+b\) becomes \((n_1 + n_2) + (m_1 + m_2)i\).

• Since the sum of two integers is always another integer, \(n_1 + n_2\) and \(m_1 + m_2\) maintain their integrality.
• This means that \(a + b\) is expressed in the same form as elements of \(G\), ensuring it stays within the set.

This is why \(G\) is closed under addition—no matter how you add two elements from \(G\), the result still belongs to \(G\).

Identity Element in a Set

The identity element is a true centerpiece in group theory. For any operation in a group, an identity element is that special number which, when combined with any element in the set, leaves that element unchanged. In the context of addition, this is simply the number zero (0).

The identity element in the additive group of complex numbers is \(0 + 0i\). This is because adding zero to any complex number yields the original number back—effectively leaving it unchanged.

To determine if our subset \(G\) has an identity element, we have to check if this "zero" belongs in the same set form as the rest of the elements in \(G\). Since \(0\) and \(0\) are integers, \(0 + 0i\) is indeed part of \(G\).

This verification confirms the presence of the identity element in \(G\), which satisfies one of the core aspects of it being a subgroup under addition.

Recognizing Additive Inverses

Additive inverses are a vital aspect of group structure. For any element \(a\) in a group, there must exist another element \(-a\) such that their sum equals the identity element of the group, which is zero for additive groups.

In our set \(G\), consider an arbitrary element \(a = n + mi\). The additive inverse of this element is \(-a = -n - mi\).

• Sine \(-n\) and \(-m\) are simply the negative integers of \(n\) and \(m\), they remain integers because the opposite of an integer is still an integer.

Thus, \(-a\) retains its structure as an element of \(G\), confirming that for every element \(a\), there exists an element \(-a\) such that \(a + (-a) = 0\). This assurance of having inverses solidifies \(G\) as a subgroup under the operation of addition.