Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $$ \left( \begin{array}{cc} 1 & 37 \\ 0 & 0 \end{array}\right) $$

share|cite|improve this answer

Just say $1 = 11' = 1'$. Then you're done.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.