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The following exercise of from Guide to Abstract Algebra by Carol Whitehead, 1st Edition 1988.

Let $\bullet $ denote a binary operation on a non-empty set $S$. Suppose that $\bullet $ admits a left identity $e$ and a right identity $f$. Prove that $e = f$.

Although I am a beginner on this topic, I am pretty sure the question is wrong.

One way the question could be correct is if to said the • binary operation was commutative. This would mean:

$$ \begin{align} e \bullet x&=x\\ x \bullet f&=x \implies f \bullet x=x \\ \text{} \\ \therefore e&=f \end{align} $$

Question: Is the original question, as reproduced here, valid?

Penelope
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    Use $\bullet$ for $\bullet$. – Shaun Apr 18 '23 at 01:43
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    It is worth pointing out that the argument you give doesn't work. It appears that your arguement is $e \bullet x = x$ and $f \bullet x = x$ for all $x$ implies $e=f$. This is not true. Consider the binary operation $\rightsquigarrow$ defined on any set which simply returns the right element (i.e. $ 1 \rightsquigarrow2 = 2$). Clearly every element is a left identity but if there are more than two elements they are not all equal. (Of course the statement you are trying to prove is true but your argument is flawed) – Fishbane Apr 18 '23 at 10:20

1 Answers1

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No, the textbook's assertion is correct. We have $$ e = e \cdot f = f, $$ the first equality since $f$ is a right-identity, the second as $e$ is a left-identity.

Randall
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