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I had a question in one of my group theory exercises, it goes as follows.

Suppose that in the definition of a group $G$, the condition that $\exists$ an element $e$ with the property
$$ae = ea = a \,\,\,\forall\, a \in G$$
is replaced by $ae = a \,\,\,\forall\, a \in G$. Then show that $ea = a \,\,\,\forall\, a \in G.$

The solution that was provided for this problem used cancellation property $$ab = ac \implies b= c$$

Solution:
We're given that $$ae = a$$
Pre-multiplying $a^{-1}$ on both sides we get, $$ a^{-1}(ae) = a^{-1}a$$ $$ (a^{-1}a)e = a^{-1}a$$ Post multiplying both sides by a $$ (a^{-1}a)ea = a^{-1}aa$$ $$ \therefore eea = ea$$ By the cancellation property we get $$ ea = a$$

But this cancellation property is proved using the very condition i.e. $ea = a \,\,\,\forall\, a \in G$ that we're trying to prove here.

So am I correct in assuming that the cancellation property as stated above cannot be used? Also is there an alternate approach to solve this question?

grey
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    Are you also assuming the existence of a right inverse? If so, this is a duplicate. – lulu Aug 31 '23 at 19:48
  • As lulu says: You will, of course, also need the part of the definition about "inverse". Since you have not told us what that is, how can we answer? – GEdgar Aug 31 '23 at 19:51
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    Welcome to Mathematics Stack Exchange. If you want us to verify whether the solution provided is correct, I think you would need to provide us with more detail. What does the solution actually say? In the post that lulu links to, I posted an answer which is taken from Lang's Algebra. I think it is the quickest proof. – Joe Aug 31 '23 at 19:53
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    The question didn't mention anything specific about inverses. So I guess we could assume that the axiom for inverses is not altered. Also, I will update the question with the solution that was provided. – grey Aug 31 '23 at 20:07
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    Fix $a$, and let $b$ be such that $ab=ba=e$. Then $a = ae = a(ba) = (ab)a = ea$. If you only know that $G$ has a right identity and right inverses, then see here, – Arturo Magidin Aug 31 '23 at 20:10
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    If the proof of left cancellation used that $ex=x$ for all $x$ (quite likely) then the argument is indeed circular. It is also unnecessarily convoluted. If you assume (as they clearly do) two-sided inverses, then simply start with $ea$, replace $e$ with $aa^{-1}$, and associate. – Arturo Magidin Aug 31 '23 at 20:21

1 Answers1

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Assuming only $ae = a$ for any $a\in G$

Then

$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a $$

Note that for this argument you need both sided inverses

Carlyle
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