# Thread: Solutions for this group

1. ## Solutions for this group

I'm not sure what the question means by e, is that the identity element? or just another element in G?

Also, I am not quite sure on how to go about this question... any help would be greatly appreciated!

2. yes, the notation "e" is fairly standard notation for the identity element (or neutral element) in an abstract group where we don't know exactly what the group operation is.

some textbooks use the notation "1" instead, but this can lead to the notion that the elements of a groups are numbers, which may not be true.

3. Originally Posted by usagi_killer
I'm not sure what the question means by e, is that the identity element? or just another element in G?
Also, I am not quite sure on how to go about this question... any help would be greatly appreciated!
Yes the identity.
{a^1,a^2,...,a^n}, if e is in that set you are done. WHY?
If not, are two of those equal?

4. Originally Posted by Plato
Yes the identity.
{a^1,a^2,...,a^n}, if e is in that set you are done. WHY?
If not, are two of those equal?
show that Plato's second sentence, implies the first.

5. $e$ is the identity element.
If we assume that $a^m\neq e$ for $n\in\left\{1,\cdots,n\right\}$ then $a^{k_1}\neq a^{k_2}$ if $k_1,k_2\in \left\{1,\cdots,n\right\}$ and $k_1\neq k_2$. Hence $e,a,\cdots,a^n are$ $n+1$ elements of $e$.

6. hmmm, thanks guys but i'm still not quite sure how to solve it

7. consider the set of all positive powers of a: {a, a^2, a^3, a^4, a^5,....}.

can this set be infinite? if not, then wouldn't two different powers, say a^k and a^m with k ≠ m, have to be equal?

couldn't we pick k to be the bigger of the two? what would this mean a^(k-m) was equal to?

8. thanks guys, i think ive got it now... my reasoning is as follows:

9. that's pretty much it.