Proving Group of Order 35 Contains Elements of Order 5 & 7

[Samuelb88]

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

So at this point, I want to suppose that x doesn't have order 35. But since G 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 x \in G is of order 5. Then each x generates an additional three unique elements: x, \, \, x^2, \, \, x^3. But this would imply that the order of G would be either 33 or 37 since elements of order 5 yield sets of four unique elements \{ x , \, \, x^2, \, \, x^3, \, \, x^4\}, a contradiction to the fact that |G| = 35. Thus G contains at least one element of order 7.

 

[HallsofIvy]

You have said, correctly, that G must have a member of order 5, 7, or 35. You have noted, correctly, that if it has a member of order 35, it must be cyclic and so have members of order 5 and 7.

Now, it G does not have a member or order 35, then it must have a member of order 5 or 7. Yes, assume, first, that there is a member of order 5. Can you then prove it must also have a member of order 7? Lastly assume there is a member of order 7. Can you then prove it must also have a member of order 5?

 

[Samuelb88]

Yes, I believe I can. Is the part where I assumed all members of G have an order 5 then showed that G must contain element of order 7 correct? If so, I can play the same game again and suppose all members of G have an order of 7 and show that this would imply the order of G must be either 31 or 37; again a contradiction to the fact that |G|=35.

 

[Char. Limit]

Are you allowed to use Sylow's theorems? If so, they might be useful here.

 

[Samuelb88]

No, I am not allowed to use Sylow's theorems.