# Show that if $*$ is a binary operation on the set $S$ which has an identity, then the identity is unique.

This is how I went about the problem, but I'm wanting a second opinion.

Prove by contradiction. Suppose the identity is not unique. Let $a,b\in S$ such that $a$ and $b$ are identities in $*$. Then $a*b=a$ and $a*b=b$ so

$a=a*b=b \implies a=b$

but $a\neq b$ because the identity is not unique.

Therefore, the identity must be unique.

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Just to be sure: is the identity two-sided? Also, use your argument without the contradiction :) it saves you one line of writing. – Clayton Oct 20 '13 at 19:10
The question does not say whether it is or isn't. Should I do it both ways to be safe? – TheMobiusLoops Oct 20 '13 at 19:11
The problem is that if it is only a one-sided identity, your technique doesn't work (and I don't think the statement would be true, but I can't think of a counterexample off-hand). – Clayton Oct 20 '13 at 19:14
I assume the OP has been given the definition of an identity as being a two-sided identity, as otherwise the statement is false. – Dan Rust Oct 20 '13 at 19:16
If your definition of identity is as an element $e$ such that for all elements $a$ we have $a*e=a=e*a,$ then you're dealing with two-sided identities. A left-identity is an element $e$ such that for all elements $a$ we have $e*a=a.$ A right-identity is an element $e$ such that for all elements $a$ we have $a*e=a$. Also, I recommend that you take @Clayton's advice. If you ever manage to prove what you're trying to prove while searching for a contradiction, then you're probably better off proceeding directly. – Cameron Buie Oct 20 '13 at 19:30

Let $S$ be the set of 2 by 2 matrices with bottom row entries $0$, so $$\left( \begin{array}{cc} a & b \\ 0 & 0 \end{array}\right),$$ with ordinary matrix multiplication. $$\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)$$ is a left identity.
But so is $$\left( \begin{array}{cc} 1 & 37 \\ 0 & 0 \end{array}\right)$$
Just say $1 = 11' = 1'$. Then you're done.