1. finite group question

Prove of disprove: there exists a finite group with an element of infinite order.

I am not convinced this holds, because a finite group will have the form $\displaystyle \{e,a_{1},a_{2},...,a_{n}\}$, where $\displaystyle e$ is the identity element. Since groups are closed under their operation, I would think each element would eventually have to be the identity for some $\displaystyle n\in{\mathbb{N}}$; namely, $\displaystyle a_{i}^{n}=e$ for some $\displaystyle n\in{\mathbb{N}}$.

Not sure though, we just introduced the concept of order yesterday. I looked ahead in the book and read a little about cyclic groups, but I don't think any knoweldge of that is needed for this.

2. Originally Posted by Danneedshelp
Prove of disprove: there exists a finite group with an element of infinite order.

I am not convinced this holds, because a finite group will have the form $\displaystyle \{e,a_{1},a_{2},...,a_{n}\}$, where $\displaystyle e$ is the identity element. Since groups are closed under their operation, I would think each element would eventually have to be the identity for some $\displaystyle n\in{\mathbb{N}}$; namely, $\displaystyle a_{i}^{n}=e$ for some $\displaystyle n\in{\mathbb{N}}$.

Not sure though, we just introduced the concept of order yesterday. I looked ahead in the book and read a little about cyclic groups, but I don't think any knoweldge of that is needed for this.

If a group $\displaystyle G$ has an element $\displaystyle g$ of infinite order, then by closedness of the operation we have that

the set $\displaystyle \{g,\,g^2,\,g^3\,\ldots,\,g^n,\,\ldots\}$ belongs to $\displaystyle G$ for all $\displaystyle n\in\mathbb{N}$ , and since the order of $\displaystyle g$ is

infinite this set is infinite, so the group cannot be finite.

Interestingly, the other way around is true: there are infinite groups all of which elements have finite order.

Tonio