# Solutions for this group

• Apr 17th 2011, 10:07 AM
usagi_killer
Solutions for this group
http://img855.imageshack.us/img855/7...sforagroup.jpg

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!
• Apr 17th 2011, 10:16 AM
Deveno
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.
• Apr 17th 2011, 10:17 AM
Plato
Quote:

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?
• Apr 17th 2011, 10:19 AM
Deveno
Quote:

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.
• Apr 17th 2011, 10:19 AM
girdav
$e$ 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 $e$.
• Apr 17th 2011, 10:30 PM
usagi_killer
hmmm, thanks guys but i'm still not quite sure how to solve it :(
• Apr 17th 2011, 10:44 PM
Deveno
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?
• Apr 19th 2011, 09:02 AM
usagi_killer
thanks guys, i think ive got it now... my reasoning is as follows:

http://img689.imageshack.us/img689/4...ionfinally.jpg
• Apr 19th 2011, 09:26 AM
Deveno
that's pretty much it.