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How does this proof work?

Theorem.$\quad$Let $G$ be a group. Then $G$ has a unique identity.

Proof.$\quad$Assume that $e$ and $f$ are two identities in $G$. Since $e$ is an identity, $ef=f$; and since $f$ is an identity, $ef=e$. Thus $e=ef=f$.

I think need to get my understanding of variables sorted out, because when I read the first line of the proof I picture $e$ as an object different from $f$ and it's confusing to then read the conclusion that $e$ and $f$ are equal. Also, how does this show that $G$ has a unique identity?

2 Answers2

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Having a unique identity means exactly that such scenarios when there are really two different identities cannot occur, and that's what is proven here: assuming letters $e$ and $f$ both denote identities in the same group $G$, we arrive that $e=f$ must hold, due to the definition of an identity.

Note also that in this proof actually we only used that $e$ is a left identity and $f$ is a right identity.
That said, we also proved that whenever a binary operation has both a left and a right identity, then they are equal, and in this case any left identity and any right identity must be the same.

Berci
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    So let me clarify. We allow $e$ and $f$ to take on any values (not necessarily different) in $G$ that are identities. Actually, because of how groups are defined, it turns out that $e$ and $f$ take on the same value since there is exactly one identity in $G$, and we show that in the proof. So, basically the proof shows that any objects representing an identity are equal? –  May 30 '17 at 21:35
  • Exactly. $,!,!$ – Berci Jun 01 '17 at 22:21
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This is called a proof by contradiction. Your goal is to show that you have a unique identity element, and the way you go about proving this is by assuming you don't have a unique identity, i.e you have at least two.

Suppose you have two distinct identity elements $e$ and $f$. Then as you showed above, $e=f$ which contradicts your assumption that $e$ and $f$ are distinct and so your group has a unique identity.

Rocket Man
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    The proof in the OP's question isn't technically proof by contradiction, since it isn't strictly assumed that $e \neq f$. Though your proof is a proof by contradiction. – Mosquite May 30 '17 at 21:20