# Group of order 35

1. Jul 10, 2011

### Samuelb88

1. The problem statement, all variables and given/known data
Does a group of order 35 contain an element of order 5? of order 7?

*I must prove all my answers.

2. Relevant equations

3. 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.

Last edited: Jul 10, 2011
2. Jul 10, 2011

### HallsofIvy

Staff Emeritus
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?

3. Jul 10, 2011

### 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 $|G|=35$.

4. Jul 10, 2011

### Char. Limit

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

5. Jul 10, 2011

### Samuelb88

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