Let G be a group, let a be an element in G of order 60. Give a generator of <a^(14)> distinct from a^(14)

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.

Let G be a group, let a be an element in G of order 60. Give a generator of <a^(14)> distinct from a^(14).

Explore how to find a generator of the cyclic subgroup <a^(14)> in a group where the element a has an order of 60. This problem utilizes key group theory concepts such as the order of elements, least common multiple, and relatively prime integers. Suitable for abstract algebra students at the undergraduate level.

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