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

• Samuelb88
In summary, it is possible to prove that a group of order 35 must contain an element of order 5 or an element of order 7, but not both. This can be done by assuming that all elements have order 5 and showing that this leads to a contradiction, and then repeating the process assuming all elements have order 7. Sylow's theorems cannot be used in this proof.
Samuelb88

## Homework Statement

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

*I must prove all my answers.

## 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:
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?

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 $|G|=35$.

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

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

## 1. How can we prove that a group of order 35 contains elements of order 5 and 7?

One way to prove this is by using the Sylow theorems, which state that if a prime number p divides the order of a group, then the group contains a subgroup of order p. In this case, we can use the Sylow theorems to show that the group of order 35 must contain subgroups of order 5 and 7, and therefore, elements of order 5 and 7.

## 2. Is it possible for a group of order 35 to not contain elements of order 5 and 7?

No, it is not possible. By the Sylow theorems, any group of order 35 must contain subgroups of order 5 and 7, and therefore, elements of order 5 and 7. This is because the prime factors of the order of a group must also divide the order of its subgroups.

## 3. Can we prove that a group of order 35 contains elements of order 5 and 7 without using the Sylow theorems?

Yes, there are other methods for proving this. One way is by using the Cauchy's theorem, which states that if a prime number p divides the order of a group, then the group contains an element of order p. In this case, we can use Cauchy's theorem to show that the group of order 35 must contain elements of order 5 and 7.

## 4. Are there any other ways to prove that a group of order 35 contains elements of order 5 and 7?

Yes, there are multiple ways to prove this. Another approach is by using the Lagrange's theorem, which states that the order of any subgroup must divide the order of the group. In this case, we can show that the only possible orders of subgroups in a group of order 35 are 1, 5, 7, and 35, and therefore, the group must contain subgroups of order 5 and 7.

## 5. Can we extend this proof to show that a group of order 35 contains elements of order 5, 7, and other prime numbers?

Yes, the same logic can be applied to show that a group of order 35 must contain subgroups of order 5, 7, and any other prime number that divides 35. This can be done by using the Sylow theorems or other techniques, such as Cauchy's theorem or Lagrange's theorem.

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