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.

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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.