Problem 38 - Algebra: Pure and Applied

Question

In Exercises 35 through 38 determine whether the indicated map \(\phi\) is an isomorphism. Justify your answer. $$ \phi: \mathbb{Z} \rightarrow \mathbb{Z}, \text { where } \phi(n)=3 n $$

Step-by-Step Solution

Answer

The map \( \phi(n) = 3n \) is not an isomorphism as it is not surjective.

Step 1: Define the Isomorphism

An isomorphism between two groups is a bijective homomorphism. This means the map must satisfy three properties: it must be (1) a homomorphism, (2) injective (one-to-one), and (3) surjective (onto).

Step 2: Check Homomorphism Property

To check if \(u000Pphi\) is a homomorphism, verify \(phi(n + m) = phi(n) + phi(m)\) for all \(n, m \) in \(mathbb{Z} \). Calculate \[\phi(n+m) = 3(n+m) = 3n + 3m = phi(n) + phi(m)\] Thus, \(phi\) is a homomorphism.

Step 3: Check Injective Property

A map is injective if \( phi(n) = phi(m) \) implies \(n = m \) for all \( n, m \) in \( mathbb{Z}\). Assume \( phi(n) = phi(m) \), which gives \(3n = 3m\).Dividing both sides by 3, we get \(n = m\), thus \( n = m \), proving \( phi \) is injective.

Step 4: Check Surjective Property

To be surjective, for every \( y \) in \( mathbb{Z} \), there must exist an \( x \) in \( mathbb{Z} \) such that \( phi(x) = y\). For any \( y \) in \( mathbb{Z}\), let \( x = \frac{y}{3}\). Since \( x \) must be an integer, \( y \) needs to be a multiple of 3. As not every integer is divisible by 3, \( phi \) is not surjective.

Step 5: Final Conclusion

Since \( phi \) is not surjective, it fails to satisfy the condition for being an isomorphism (even though it is a homomorphism and injective). Therefore, \( phi \) is not an isomorphism.

Key Concepts

Homomorphism

A homomorphism is a crucial concept in group theory. It is a function between two groups that preserves the group structure. This means if you have a group operation, like addition or multiplication, the homomorphism respects these operations.

For example, if you have a homomorphism function, say \( \phi \), from a group \( G \) to a group \( H \), then for any two elements \( a \) and \( b \) in \( G \), \( \phi(a + b) = \phi(a) + \phi(b) \).

In our specific case of the map \( \phi(n) = 3n \) from \( \mathbb{Z} \) to \( \mathbb{Z} \), this property holds because:

• \( \phi(n + m) = 3(n + m) = 3n + 3m = \phi(n) + \phi(m) \)

Hence, \( \phi \) is indeed a homomorphism. Understanding this helps us see how the map interacts with elements from the two groups, preserving their relationship.

Injective Mapping

An injective map, often referred to as "one-to-one", is a function in which different elements from the domain map to different elements in the codomain. No two different elements are sent to the same place, which means each output is unique to a single input.

To determine if \( \phi:\mathbb{Z}\to\mathbb{Z}\) defined by \( \phi(n) = 3n \) is injective, we need to verify that if \( \phi(n) = \phi(m) \), then this implies \( n = m \).
Consider:

• Start with the assumption \( \phi(n) = \phi(m) \), which leads to \( 3n = 3m \).
• By dividing both sides by 3, we find \( n = m \).

Therefore, \( \phi \) is injective since different integers \( n \) and \( m \) will not map to the same integer.

Surjective Mapping

A surjective map, also called "onto", is a function where every element in the codomain corresponds to an element in the domain. In other words, every possible output is covered by the function.

To check for surjectivity in \( \phi(n) = 3n \) from \( \mathbb{Z} \) to \( \mathbb{Z} \), make sure that for every integer \( y \), there is an integer \( x \) such that \( \phi(x) = y \).
Try to solve for \( x \) in terms of \( y \):

• Let \( \phi(x) = y \), then \( 3x = y \).
• This implies \( x = \frac{y}{3} \).

For \( x \) to be an integer, \( y \) must be a multiple of 3. Because not all integers \( y \) are multiples of 3, \( \phi \) is not surjective. This means \( \phi \) cannot map every number in the codomain \( \mathbb{Z} \) back to an integer in the domain.

Isomorphism Criteria

To determine if a function is an isomorphism, it must simultaneously satisfy the conditions of being a homomorphism, injective, and surjective. An isomorphism indicates a perfect structure-preserving correspondence between two groups.

The function \( \phi(n) = 3n \) is a homomorphism and is injective, as we've shown. However, it is not surjective, since there are elements in the codomain \( \mathbb{Z} \) that don't have a pre-image in \( \mathbb{Z} \) under \( \phi \).
Thus, due to the lack of surjectivity, \( \phi \) fails the criteria of being an isomorphism despite fulfilling two of the three necessary conditions. A reminder that an isomorphism requires all three properties:

• Homomorphism: Preserves group structure.
• Injective: Unique mapping for each input.
• Surjective: Covers every element in the codomain.

Therefore, \( \phi \) does not establish an isomorphism between \( \mathbb{Z} \) and \( \mathbb{Z} \).