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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?

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  • $\begingroup$ Ah, thank you! That made me recheck the definitions and this indeed is the case. :) $\endgroup$ – dotslash Apr 4 '14 at 16:46
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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=>a=e∗a $ and $ a^{−1}a=e=>a=a∗e$

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  • $\begingroup$ I don't understand the last two relations, though. How does $a*a^{-1}=e$ imply that $a = e*a$?, etc. $\endgroup$ – dotslash Apr 4 '14 at 17:02
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    $\begingroup$ Start with $a*a^{-1} = e$, multiply both sides on the right by $a$, and then associate to get $a*e = e*a$. $\endgroup$ – user7530 Apr 4 '14 at 17:40
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    $\begingroup$ Btw, you can use \implies to get $\implies$. $\endgroup$ – Ben West Apr 4 '14 at 18:17
  • $\begingroup$ @BenWest Oh, cool! Thanks a lot for the tip. :) I always used \Rightarrow $\endgroup$ – dotslash Apr 5 '14 at 3:25
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If you have already proven that the identity is unique in a group, then you can use $$e*a=a*e=a$$ for all $a\in G$.

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