Textbook proof for uniqueness of inverse in a group

Question

A Textbook of Abstract Algebra by Pinter gives the following proof of the property that an element in a group can have only one inverse (consider $*$ to be the operation and $e$ to be the identity element):

Suppose an element $a$ has two inverses, $a_1$ and $a_2$.

Then, $a_1*(a*a_2) = a_1*e = a_1$, and

($a_1*a)*a_2 = e*a_2 = a_2$

The book then says by associativity, the two-left hand sides are equal, and hence so are $a_1$ and $a_2$.

I have an objection to this proof. The identity element is defined as $a*e=a$, and since commutativity is not necessary in a group, the argument that $e*a_2 = a_2$ doesn't look watertight to me.

Any comments?

Ah, thank you! That made me recheck the definitions and this indeed is the case. :) — ankush981, Apr 04 '14 at 16:46

score 3 · Accepted Answer · edited Sep 15 '19 at 15:53

3

Irrespective of whether the group is commutative or not, $aa^{−1}=a^{−1}a=e$ and that's why the relation which you mentioned commutes.

$$aa^{−1}=e \implies a=e\cdot a $$ $$\text{ and }$$ $$ a^{−1}a=e \implies a=a\cdot e$$

edited Sep 15 '19 at 15:53

kelalaka

1,637

answered Apr 04 '14 at 16:52

MathMan

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I don't understand the last two relations, though. How does $aa^{-1}=e$ imply that $a = ea$?, etc. – ankush981 Apr 04 '14 at 17:02
1

Start with $aa^{-1} = e$, multiply both sides on the right by $a$, and then associate to get $ae = e*a$. – user7530 Apr 04 '14 at 17:40
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Btw, you can use \implies to get $\implies$. – Ben West Apr 04 '14 at 18:17
@BenWest Oh, cool! Thanks a lot for the tip. :) I always used \Rightarrow – ankush981 Apr 05 '14 at 03:25

score 1 · Answer 2 · edited Apr 04 '14 at 18:15

1

If you have already proven that the identity is unique in a group, then you can use $$e*a=a*e=a$$ for all $a\in G$.

edited Apr 04 '14 at 18:15

egreg

238,574

answered Apr 04 '14 at 17:36

Maryam

19

Textbook proof for uniqueness of inverse in a group

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