Pregunta
- Suppose that
is a finite group with an element
of order 5 and an element
of
order 7 . Why must
?
order 7 . Why must
Ask by Stephens Vaughan. in the United States
Mar 22,2025
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To ensure that the group
has elements of order 5 and 7, its order must be at least 35.
Solución
- 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,
- Similarly, since
is an element of order 7, the subgroup generated by is cyclic of order 7. Therefore,
- 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:
- 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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