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Samuelb88

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## Homework Statement

Does a group of order 35 contain an element of order 5? of order 7?

*I must prove all my answers.

## Homework Equations

## The Attempt at a Solution

I'm not really sure how to prove this and could really use some hints. Here's what I've got so far:

**Pf.**Let [itex]G[/itex] be a group of order 35, and let [itex]x \in G[/itex] such that [itex]x \neq e[/itex], where [itex]e[/itex] is the identity element of [itex]G[/itex]. It follows from Lagrange's Thoerem that [itex]x[/itex] will be of order 5, 7, or 35. If [itex]x[/itex] is of order 35, then [itex]G[/itex] is cyclic and thus has elements of order 5 and 7.

So at this point, I want to suppose that [itex]x[/itex] doesn't have order 35. But since [itex]G[/itex] has 35 elements, the only other possibilities for the orders of the other nontrivial elements are either 5 or 7. Is this reasoning okay? Being optimistic, I will proceed and claim that each element [itex]x \in G[/itex] is of order 5. Then each [itex]x[/itex] generates an additional three unique elements: [itex]x, \, \, x^2, \, \, x^3[/itex]. But this would imply that the order of [itex]G[/itex] would be either 33 or 37 since elements of order 5 yield sets of four unique elements [itex]\{ x , \, \, x^2, \, \, x^3, \, \, x^4\}[/itex], a contradiction to the fact that [itex]|G| = 35[/itex]. Thus [itex]G[/itex] contains at least one element of order 7.

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