Question

t (ℤ, +) be a group and show that
a) < 4 > is a subgroup of ℤ.
b) < 4 > is an abelian group

Answer 1

To show that \( \langle 4 \rangle \) is a subgroup of \( \mathbb{Z} \) (the group of integers under addition), we will follow the subgroup criteria commonly used in group theory, and then we will demonstrate that it is an abelian group.

### Part a: Show that \( \langle 4 \rangle \) is a subgroup of \( \mathbb{Z} \)

The notation \( \langle 4 \rangle \) represents the set of all integer multiples of \( 4 \):
\[
\langle 4 \rangle = \{ 4n \mid n \in \mathbb{Z} \} = \{ \ldots, -8, -4, 0, 4, 8, \ldots \}
\]

To show that \( \langle 4 \rangle \) is a subgroup of \( \mathbb{Z} \), we need to check the following criteria:

1. **Non-empty**: The subgroup must contain the identity element of \( \mathbb{Z} \).
- The identity element for \( \mathbb{Z} \) under addition is \( 0 \).
- Clearly, \( 0 = 4 \cdot 0 \) is in \( \langle 4 \rangle \).

2. **Closure**: The sum of any two elements in \( \langle 4 \rangle \) must also be in \( \langle 4 \rangle \).
- Let \( a = 4m \) and \( b = 4n \), where \( m, n \in \mathbb{Z} \).
- Then, \( a + b = 4m + 4n = 4(m + n) \).
- Since \( m+n \) is an integer, \( a + b \in \langle 4 \rangle \).

3. **Inverses**: Every element in \( \langle 4 \rangle \) must have an inverse also in \( \langle 4 \rangle \).
- Let \( a = 4m \) for some \( m \in \mathbb{Z} \).
- The inverse of \( a \) is \( -a = -4m = 4(-m) \).
- Since \(-m\) is also an integer, \( -a \in \langle 4 \rangle \).

Since all three criteria are satisfied, we conclude that \( \langle 4 \rangle \) is a subgroup of \( \mathbb{Z} \).

### Part b: Show that \( \langle 4 \rangle \) is an abelian group

An abelian group is one where the group operation (in this case, addition) is commutative. To show that \( \langle 4 \rangle \) is abelian, we need to show that for any two elements \( a, b \in \langle 4 \rangle \), the equation \( a + b = b + a \) holds.

Let \( a = 4m \) and \( b = 4n \) for \( m, n \in \mathbb{Z} \).

Then,
\[
a + b = 4m + 4n = 4(m + n)
\]
and
\[
b + a = 4n + 4m = 4(n + m).
\]

Since addition of integers is commutative (i.e., \( m + n = n + m \)), we have:
\[
4(m + n) = 4(n + m).
\]
Thus, \( a + b = b + a \).

Therefore, \( \langle 4 \rangle \) is an abelian group.

### Conclusion

We have shown that \( \langle 4 \rangle \) is a subgroup of \( \mathbb{Z} \) and that it is an abelian group.