Exploring One-Sided vs. Two-Sided Identity Elements in Groups

[HyperActive]

Hi
I'm taking a math course at university that covers introductory group theory. The textbook's definition of the identity element of a group defines it as two sided; that is, they say that a group ##G## must have an element ##e## such that for all ##a \in G##, ##e \cdot a = a = a \cdot e## .

Is it possible to define a group's identity element as one-sided, and then prove two-sidedness as a theorem? Or is it an intrinsic property of group identities?

I started with a left-identity ##e \cdot a = a## and tried to prove that ##e \cdot a = a = a \cdot e## and kept hitting walls, so I though I'd better check if it's doable.

Thanks.

 

[Stephen Tashi]

If you change the assumption that there exists a two-sided identity, how are you going to modify the assumption that each element has an inverse? That assumption mentions the two-sided identity.

 

[HyperActive]

Thanks for the reply Stephen Tashi. :)
In my book, the assumption that each element has an inverse is stated as follows: for all ##a \in G## , there exists a ##b \in G## such that ##a \cdot b = e##. That doesn't seem to rely on the two-sidedness of the identity element (and neither it seems to me does the left-handed version of the statement, for all ##a \in G## , there exists a ##b \in G## such that ##b \cdot a = e##). So other than noting that in these definitions ##e## refers to a left identity, I'm not seeing how I would need to modify them.

Could you please clarify this?

 

[rubi]

You can weaken the group axioms to require only the existence of a left inverse and a left neutral element:
(i) ## e g = g##
(ii) ## g^{-1} g = e##

Then every left inverse is a right inverse:
##g g^{-1} = e g g^{-1} = (g^{-1})^{-1} g^{-1} g g^{-1} = (g^{-1})^{-1} g^{-1} = e##
And every left neutral element is a right neutral element:
## g e = g g^{-1} g = e g = e##.
(The proof uses the fact that ##g^{-1}## is a right inverse, which I proved before.)

You can of course replace "left" by "right" and get similar results. You can't mix them, though: Having left inverses and a right neutral element doesn't suffice.

 

[Stephen Tashi]

: for all ##a \in G## , there exists a ##b \in G## such that ##a \cdot b = e##. That doesn't seem to rely on the two-sidedness of the identity element

The point is that the meaning of the statement [itex] a \cdot b = e [/itex] depends on what you mean by the notation "[itex] e [/itex]".

 

[HyperActive]

Thank you both very much!