# Quick question about identity element

• Sep 27th 2011, 05:44 PM
tangibleLime
Quick question about identity element
Say I have a group $\displaystyle G$ and an unknown binary operation $\displaystyle *$, and $\displaystyle a*a=e, \forall a \in G$. ($\displaystyle e$ being the identity element)

Is it true that the only value $\displaystyle a$ can be is the identity element, $\displaystyle e$?
• Sep 27th 2011, 05:52 PM
FernandoRevilla
Re: Quick question about identity element
Quote:

Originally Posted by tangibleLime
Say I have a group $\displaystyle G$ and an unknown binary operation $\displaystyle *$, and $\displaystyle a*a=e, \forall a \in G$. ($\displaystyle e$ being the identity element). Is it true that the only value $\displaystyle a$ can be is the identity element, $\displaystyle e$?

Not true. Choose for example $\displaystyle (\mathbb{Z}_2,+)$ , we have $\displaystyle 0+0=1+1=0$ .
• Sep 27th 2011, 05:55 PM
tangibleLime
Re: Quick question about identity element
Ah, didn't consider that. Thanks for the response!
• Sep 27th 2011, 08:01 PM
Deveno
Re: Quick question about identity element
what you can say, if a*a = e,and a is NOT the identity, is that a is of order 2, and that a is its own inverse (showing that elements of order 2 and of order 1(the identity is the only element of order 1) share this property).

a good example of an element of order 2 is the following group, which is a subset of the actions you can perform on a light switch:

{flip the switch, do nothing} and the group operation is: first do one thing, then the other.

"flip the switch" is an element of order 2: flip the switch, then flip the switch = do nothing.

(to be honest, this is in all actuality just Z2 in disguise, and is one of the basic ways in which logical circuits can "do math").
• Sep 28th 2011, 02:20 AM
Swlabr
Re: Quick question about identity element
What is true, however, is that if $\displaystyle a\ast a=a$ then $\displaystyle a$ is the identity. This is true because,

$\displaystyle a=a^{-1}\ast a\ast a=a\ast a^{-1}=e$