Problem 2 - A Book of Abstract Algebra

Question

Let \(G\) be the group \(\left\\{10^{n}: n \varepsilon \mathbb{Z}\right\\}\) with respect to multiplication. Prove that \(G \cong \mathbb{Z}\) (Remember that the operation of \(\mathbb{Z}\) is addition.)

Step-by-Step Solution

Answer

The group G is isomorphic to \(\mathbb{Z}\) through the function \(\varphi(10^n) = n\), which is bijective and respects the group operations.

Step 1: Understand the group structure of G

Let's start by analyzing the group \(G = \{10^n : n \in \mathbb{Z}\}\). This group consists of powers of 10 for all integers \(n\). The operation in \(G\) is multiplication.

Step 2: Define a possible isomorphism

To show that \(G \cong \mathbb{Z}\), we need a bijection \(\varphi: G \to \mathbb{Z}\) that respects the group operations. Define \(\varphi(10^n) = n\). This maps elements of \(G\) to integers.

Step 3: Show the function is bijective

A function is bijective if it's both injective (one-to-one) and surjective (onto). - **Injective**: Assume \(\varphi(10^a) = \varphi(10^b)\) implies \(a = b\), since \(\varphi(10^a) = a\) and \(\varphi(10^b) = b\). - **Surjective**: For every integer \(m\), there is an element \(10^m\) in \(G\) such that \(\varphi(10^m) = m\). Thus, \(\varphi\) reaches all integers.

Step 4: Prove operation preservation

We need to show that the homomorphism \(\varphi\) respects the operation of the groups, i.e., multiplication in \(G\) and addition in \(\mathbb{Z}\). \[ \varphi(10^a \cdot 10^b) = \varphi(10^{a+b}) = a+b = \varphi(10^a) + \varphi(10^b) \] Thus, the addition of integers corresponds to the multiplication of powers of 10, preserving the group operation.

Step 5: Conclusion

Since \(\varphi: G \to \mathbb{Z}\) is bijective and operation-preserving, \(G\) is isomorphic to \(\mathbb{Z}\). Therefore, \(G \cong \mathbb{Z}\).

Key Concepts

Group Homomorphism

A group homomorphism is a function between two groups that respects the structure of the groups. In simpler terms, it means the function takes the operation in one group and translates it into the operation of another group without losing any information in the process.
To understand this, imagine you have a group of integers with addition, \ \{ Z_n\ \} . If we have a map or function that takes an element from this group to another group, say of powers of 10 with multiplication, it should preserve the structure for them to be considered a group homomorphism. This means:

• If you add two numbers in the integer group, the result after applying the function should be the multiplication of their corresponding elements in the second group.

Thus, in our example, we defined a function \(\varphi(n) = 10^n\) for the powers of 10 group. It maintains the group structure because:

• The function \varphi(10^a \cdot
• We said it's a map that maintains the group structures from `Z` to powers of 10, so it considers the multiplication aspects in the powers of 10.

This shows that the function respects or maintains the group structure and operation, establishing it as a homomorphism.

Bijection

A bijection is a type of function that establishes a perfect pairing between the elements of two sets, ensuring each element of the first set corresponds to exactly one element of the second set and vice versa.
In our context, it's essential for demonstrating that two groups are isomorphic. For our function \(\varphi: G \to \mathbb{Z}\), we're saying that for every element in the group of powers of 10 (\(G\)), there's a unique corresponding integer (\(\mathbb{Z}\)) and each integer has exactly one corresponding power of 10.
This can be broken down into two main properties:

• Injective: No two elements in \(G\) map to the same integer in \(\mathbb{Z}\).
• Surjective: Every integer in \(\mathbb{Z}\) is mapped to by something in \(G\).

By satisfying these two conditions, \(\varphi\) can be said to be a bijection. It is a crucial step to show that the two groups not only have similar structures but also each element has a perfect match with another.

Injective Function

An injective function, also known as a one-to-one function, ensures that each element of the domain maps to a unique element in the co-domain. This means no two different inputs (from the domain) map to the same output (in the co-domain).
In our exercise, our function \(\varphi: G \to \mathbb{Z}\) needs to be injective to prove that \(G\) and \(\mathbb{Z}\) are isomorphic. How do we prove this? By ensuring that whenever\[\varphi(10^a) = \varphi(10^b),\] it follows that \(a = b\).
Here's how it plays out:

• If two powers of 10, \(10^a\) and \(10^b\), map to the same integer, \(a\) and \(b\) must be equal. Otherwise, our function would mistakenly imply different powers of 10 are represented by the same integer, which can't happen since each integer in our source group should map uniquely to one in our target group.

This injectivity verifies that each distinct element in \(G\) maintains its distinct identity when moved over to \(\mathbb{Z}\).

Surjective Function

A surjective function, or onto function, covers all elements in the codomain. Every element in the codomain has at least one pre-image in the domain. This characteristic ensures that the function's range covers the entire codomain.
In the context of proving isomorphism between \(G\) and \(\mathbb{Z}\), we need \(\varphi: G \to \mathbb{Z}\) to be surjective. What does this look like? It means:

• For every integer \(m\) in \(\mathbb{Z}\), there is a corresponding \(10^m\) in \(G\).
• Thus, every integer has a matching power of 10, and nothing is skipped or left unmatched.

If \(\varphi\) is surjective, it guarantees that the integer group is entirely wrapped up in the powers of 10 representation, with nothing missing. This is crucial to demonstrate the equivalence of structures required in an isomorphic relationship.