# Conceptual questions on proving identity element of a group is unique

1. Aug 28, 2014

### "Don't panic!"

Hi,

I'm hoping to clear up a few uncertainties in my mind about proving that the identity element and inverses of elements in a group are unique.

Suppose we have a group $\left(G, \ast\right)$. From the group axioms, we know that at least one element $b$ exists in $G$, such that $a \ast b = b \ast a = a \quad \forall \; a\in G$. Let $b,c \in G$ be any two elements in $G$ satisfying $a \ast b = b \ast a = a$ and $a \ast c = c \ast a = a \quad \forall \; a \in G$. We have then, that $$b= b \ast c = c$$ Hence, as $b$ and $c$ are arbitrary (other than satisfying the "identity property" stated above), the only way this can be true is if, in fact, there is only one, unique, identity element.

Is this correct?

2. Aug 28, 2014

### haruspex

Yes, though it would be usual to start with the assumption that it is not unique, writing b ≠ c, then show b = c. Hence the identity is unique by reductio ad absurdum.

3. Aug 28, 2014

### "Don't panic!"

Ah ok, so it's a proof by contradiction then?

Is argument I gave for why, if any two elements, satisfying the properties of an identity element, are equivalent, then it must be that the identity element is unique, correct?

4. Aug 28, 2014

### haruspex

Your argument was fine, except that you seemed to be struggling to word the final part of it. I certainly got the impression you weren't convinced it was valid. Recasting it as a proof by contradiction makes it more obviously right.

5. Aug 28, 2014

### "Don't panic!"

Yes, you're right. I thought I understood it, but then I read a text that does a proof by just assuming that $a$ and $b$ are both identity elements and then showing that $a=b$, and from this they conclude that the identity is unique. I was really just trying to justify why this is so in my mind?!
My thoughts on the matter were that if we know that $a$ is an identity element, then if we assume that $b$ is any other element that satisfies the properties of an identity, we find that in fact $a=b$, i.e. $a$ is the unique inverse.

Last edited: Aug 28, 2014
6. Aug 28, 2014

### haruspex

Quite so.

7. Aug 28, 2014

### "Don't panic!"

I guess my confusion has arisen because they most often don't state the proof of existence part first, i.e. there exists an element $a$ which satisfies the properties of an identity. At this point one should then go on to show that any other element that satisfies these properties is equivalent to $a$ and hence uniqueness follows. Sorry for the recapitulation, I'm fairly new to the more formal approach and just want to check that I'm following the correct logical steps. Appreciate all your help!