Question
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  1. Suppose that is a finite group with an element of order 5 and an element of
    order 7 . Why must ?

Ask by Stephens Vaughan. in the United States
Mar 22,2025

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Tutor-Verified Answer

Answer

To ensure that the group has elements of order 5 and 7, its order must be at least 35.

Solution

  1. Since is an element of order 5, the subgroup generated by is cyclic of order 5. By Lagrange’s theorem, the order of any subgroup divides the order of the group. Thus,
  1. Similarly, since is an element of order 7, the subgroup generated by is cyclic of order 7. Therefore,
  1. Since 5 and 7 are relatively prime, if a number is divisible by both 5 and 7, then it must be divisible by their product:
  1. Hence, the smallest possible order for that satisfies these divisibility conditions is 35, i.e.,

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Extra Insights

To understand why , we can use a nice little piece of group theory known as the LCM (Least Common Multiple) principle. The orders of the elements and tell us a lot! The order of an element divides the order of the group, which means the order of must be a multiple of both 5 and 7. Since 5 and 7 are coprime (they share no common factors), their least common multiple is simply their product: . Thus, for to accommodate both elements, its order must be at least 35!
Additionally, since groups tend to hold all sorts of fascinating secrets, knowing an element’s order can unveil many possibilities. When you find such elements in any finite group, it hints at larger structural dynamics at play. If had an order smaller than 35, it wouldn’t have the capacity to house both an element of order 5 and one of order 7. Hence, is a necessary conclusion, nearly like a delightful mathematical puzzle solved!

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