Let $G$ be a group, and let $a$ be an element of $G$ with order 60. This means that $a^{60} = e$, where $e$ is the identity element of $G$, and no smaller positive integer than 60 satisfies $a^n = e$.

Now, consider the subgroup generated by $a^{14}$, denoted $\langle a^{14} \rangle$. To find the order of this element, observe that the order of $a^{14}$ is the smallest positive integer $k$ such that:

$(a^{14})^k = e.$

This implies $a^{14k} = e$, so $14k$ must be a multiple of 60. The smallest such $k$ is the least common multiple of 14 and 60, divided by 14, which is:

$k = \frac{\text{lcm}(14, 60)}{14}.$

To compute the least common multiple, we first find the greatest common divisor (gcd) of 14 and 60:

$\gcd(14, 60) = 2,$

so the least common multiple is:

$\text{lcm}(14, 60) = \frac{14 \times 60}{\gcd(14, 60)} = \frac{14 \times 60}{2} = 420.$

Thus,

$k = \frac{420}{14} = 30.$

Therefore, the order of $a^{14}$ is 30, meaning $\langle a^{14} \rangle$ is a cyclic subgroup of order 30. The generators of this subgroup are of the form $(a^{14})^m = a^{14m}$, where $m$ is relatively prime to 30 (i.e., $\gcd(m, 30) = 1$).

The integers $m$ that satisfy $\gcd(m, 30) = 1$ are:

$m = 1, 7, 11, 13, 17, 19, 23, 29.$

Thus, the generators of $\langle a^{14} \rangle$ are $a^{14}, a^{14 \cdot 7}, a^{14 \cdot 11}, a^{14 \cdot 13}, a^{14 \cdot 17}, a^{14 \cdot 19}, a^{14 \cdot 23}, a^{14 \cdot 29}$.

Since you want a generator distinct from $a^{14}$, a possible choice is $a^{14 \cdot 7} = a^{98}$.

Final Answer:

A generator of $\langle a^{14} \rangle$ distinct from $a^{14}$ is $a^{98}$.

Would you like more details or further explanation on any part of the solution?

Here are some related questions:

1. How do you determine the order of an element in a group?
2. What is the significance of the least common multiple in this context?
3. Why are integers relatively prime to 30 important for generating subgroups?
4. What happens if $a$ had an order different from 60?
5. Can we use a similar method for finding generators of any cyclic subgroup?

Tip: When working with cyclic subgroups, the generators are the elements that are powers of the main generator, where the exponents are relatively prime to the order of the subgroup.