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Prove that the identity element in a group is unique.

Suppose $e_1$ and $e_2$ are two identity elements of a group $G$, then for $g\in G$ we have $e_1g = ge_1 = e_1$ and $e_2g = ge_2 = e_2$ then $e_2e_1g = ge_1e_2 = e_2$. The proof in the book stops here, saying $e_1e_2 =e_2$.

This is the proof in the book, and I did the same in my attempt. However, the book follows the above by simply stating, "hence $e_1 = e_2$". How is this?

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    $\begingroup$ Is that really the proof in the book? It seems a little garbled, especially the unnecessary introduction of an extra element $g$ in the penultimate equation. $\endgroup$
    – cardinal
    Commented Aug 28, 2011 at 16:08
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    $\begingroup$ Can you tell us what book it is, so I can add it to my personal do-not-recommend-list? $\endgroup$
    – Arturo Magidin
    Commented Aug 28, 2011 at 19:12
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    $\begingroup$ I hope that's not transcribed correctly. If $e$ is an identity, $eg = g$, not $e$. $\endgroup$
    – jwodder
    Commented Aug 28, 2011 at 19:37

2 Answers 2

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I would have written $e_2 = e_1 e_2 = e_1$ myself

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This is more work than necessary. All you need to say is that $e_1 = e_1 e_2 = e_2$.

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