Computing the order of a group

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SUMMARY

The discussion focuses on computing the orders of elements in a group, specifically for an element 'a' with an order of |a| = 15. The calculations involve determining the orders of a^3, a^6, a^9, and a^12. The participant identifies that a^3 generates a cyclic subgroup containing elements up to a^42, but questions the rationale behind stopping at a^42, suggesting that it may represent the 15th power of a^3. The key takeaway is the importance of understanding the cyclic nature of group elements and their orders.

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Homework Statement



Let a be an element of a group an let |a| = 15. Compute the orders of the following elements of G

a) a^3, a^6, a^9, a^12

Homework Equations





The Attempt at a Solution



for the first part of part a, would a^3 be <a^3>=<e,a^3,a^6,a^9,a^12,a^15,a^18,a^21,a^24,a^27,a^30, a^33,a^36,a^39,a^42>
 
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|a|=15 means that 15 is the lowest power of a that is equal to the identity.

So, for example,
a^{42}=a^{15}\times a^{15} \times a^{12}=e \times e \times a^{12}=a^{12}
so you've got too many things in your list.
 
What I want to know is why the OP stopped after a^42 in particular. I mean going beyond a^15 is clearly wrong, but why stop at a^42? Is that the 15th power of a^3? I think so, from quickly scanning the list.
 

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