18. Let a be a generator of a cyclic group G. If |G| = n and \langlea ^ 10\rangle = \langlea ^ 5\rangle prove that n is odd.

We are tasked with proving that \( n \), the order of the cyclic group \( G \), is odd, given that \( a \) is a generator of \( G \) (i.e., \( G = \langle a \rangle \)) and \( \langle a^{10} \rangle = \langle a^5 \rangle \).

### Step 1: Analyze the given condition

Since \( G = \langle a \rangle \) and the order of \( G \) is \( n \), the order of any element \( a^k \) in \( G \) must divide \( n \). Now, we are given the following equality of cyclic subgroups:

\[
\langle a^{10} \rangle = \langle a^5 \rangle
\]

This means that both subgroups are the same, i.e., they generate the same set of elements.

### Step 2: Structure of cyclic subgroups

For any integer \( k \), the subgroup \( \langle a^k \rangle \) consists of powers of \( a^k \). The order of \( a^k \) in \( G \) is \( \frac{n}{\gcd(n,k)} \). Specifically:
- The order of \( a^{10} \) is \( \frac{n}{\gcd(n,10)} \).
- The order of \( a^5 \) is \( \frac{n}{\gcd(n,5)} \).

Since we are given that \( \langle a^{10} \rangle = \langle a^5 \rangle \), the orders of \( a^{10} \) and \( a^5 \) must be the same. This implies:

\[
\frac{n}{\gcd(n,10)} = \frac{n}{\gcd(n,5)}
\]

### Step 3: Simplify the condition

The greatest common divisor (gcd) function behaves as follows:
- \( \gcd(n,5) \) divides 5.
- \( \gcd(n,10) \) divides 10.

Since \( \gcd(n,10) \geq \gcd(n,5) \), the equality \( \frac{n}{\gcd(n,10)} = \frac{n}{\gcd(n,5)} \) implies that \( \gcd(n,10) = \gcd(n,5) \).

This is only possible if 2 does not divide \( n \), because if \( n \) were divisible by 2, we would have \( \gcd(n,10) = 2 \cdot \gcd(n,5) \), which would contradict the equality.

### Step 4: Conclude that \( n \) is odd

Since \( \gcd(n,10) = \gcd(n,5) \) implies that \( n \) is not divisible by 2, we conclude that \( n \) must be odd.

Thus, we have proven that \( n \) is odd.

---

Would you like further details on any step, or have any questions?

Here are 5 related questions for deeper understanding:
1. What is the order of an element in a cyclic group and how does it relate to subgroups?
2. How can we determine the order of a cyclic subgroup given a generator?
3. How does the greatest common divisor (gcd) affect the order of elements in cyclic groups?
4. What happens if \( n \) is even in this problem, and how would the solution change?
5. Can a similar approach be applied to different values of powers of \( a \), like \( a^k \), for different \( k \)?

**Tip:** In cyclic groups, the order of any element always divides the order of the group. This property is crucial when working with subgroups and their generators.

Let a be a generator of a cyclic group G. If |G| = n and ⟨a^10⟩ = ⟨a^5⟩ prove that n is odd.

This page provides a detailed proof that the order of a cyclic group G, where |G| = n and ⟨a^10⟩ = ⟨a^5⟩, must be odd. Explore key concepts in group theory, such as the order of elements, subgroups, and gcd properties. Suitable for undergraduate mathematics students.

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