# Uniqueness of the neutral element $e$ in a group

Let $$(G, e, ·)$$ be a group. Show the following:

(a) If $$b$$$$G$$ satisfies that $$∀a$$$$G$$ : $$a · b = a = b · a$$ then $$b = e$$. (uniqueness of the neutral element)

(b) If there are $$a, b$$$$G$$ that satisfy $$a · b = 1 = b · a$$, then $$b = a^{−1}$$ and $$a = b^{−1}$$ . (uniqueness of the inverse element)

My attempt:

a) For this part, I have two attempts:

First one:

$$a · b = b · a$$ (given)

$$a · b = b' · a$$

$$a · b -b' · a = 0$$

$$a · (b - b') = 0$$

$$b - b' = 0$$ which implies that $$b=b'=e$$, thus the uniqueness of $$e$$.

Second one:

$$a · b = a = b · a$$ (given)

$$a · b = a = b' · a$$

$$a · b - b' · a = a-b' · a$$

$$a · (b-b')= a · (1- b')$$

$$b - b' = 1-b'$$ which implies that $$b=1$$ (solution to given), thus the uniqueness of $$e$$.

Which attempt is better?

b) $$a · b = b · a = 1$$ (given)

$$a · b = b' · a = 1$$

$$a · b - b' · a = 1 - b' · a$$

$$a · (b - b') = 1 - b' · a$$

$$a · (b - b') + b' · a = 1$$

$$a · (b - b' + b') = 1$$

$$a · b = 1$$

$$a=\frac1b = b^{-1}$$, thusthe uniqueness of the inverse element.

And I did the same steps to find that $$b=a^{-1}$$. Are my attempts correct?

One of the great lessons of Group Theory is that of unlearning what you have learned. When you are working with a group, the ONLY operation that exists is the operation of the group. In your case, you are stating that operation to be the 'dot' operation.

In each of your attempts, you have used a second operation (which I think you are considering to be subtraction, since you're using the m-dash to indicate it). This operation does not exist, and its presence completely invalidates your proof.

So, I'm sorry to say that none of them work.

As a piece of advice, additionally, you will find that others can read your proofs much more easily if you declare the variable and how it is to be defined before you actually include it in any expressions.

So, if you want to use a as a generic element of the group and both b and b' as identity elements, then you probably want to start your proof with lines like this:

1. Let a, b, and b' be elements of G such that b is an identity of G and b' is an identity of G.

Then, people can follow when you start treating b and b' as identities.

All attempts are wrong. How can you subtract elements in the group and what is $$0$$? In the group you have a unique binary operation which is usually denoted as multiplication (though it can be something else, we just denote it like this), and there is an identity $$e$$ with respect to that operation. You can't add some other operations to this. So there is no meaning to $$ab-b'a=0$$.

The exercise itself is very simple. Suppose $$e,e'$$ are two identity elements. This implies that $$ee'=e$$ and also $$ee'=e'$$, because an identity works from both sides. And hence $$e=e'$$.

As for inverses, suppose $$b,c$$ are two inverses of $$a$$, which means $$ab=ba=ac=ca=e$$. Then $$cab=c(ab)=ce=c$$ and also $$cab=(ca)b=eb=b$$. Hence $$b=c$$.

Let $$e, e'$$ be identities in $$(G, e, \cdot)$$. Then $$e\stackrel{(1)}{=}e\cdot e'\stackrel{(2)}{=}e',$$ where $$(1)$$ holds because $$e'$$ is an identity and $$(2)$$ holds because $$e$$ is an identity; hence $$e=e'$$.

Let $$g\in G$$. Then if $$g'$$ and $$\hat{g}$$ are each inverses of $$g$$ with respect to the $$\cdot$$ operation, we have \begin{align} g'&=g'\cdot e\\ &=g'\cdot (g\cdot \hat{g}) \\ &=(g'\cdot g)\cdot \hat{g} \\ &=e\cdot \hat{g} \\ &=\hat{g}, \end{align} so $$g'=\hat{g}$$.