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

I.N.Herstein in Topics of Algebra defines a group as a set having a special element $i$ such that:

$a,i\in A(S) $ which satisfies $i\cdot a = a\cdot i = a$

In this way, this follows commutativity holds true for identity element, then doesnt it run contradictory to the non-necessary condition that $a \cdot b \neq b \cdot a$ ?

In similar vein is the set of integers under subtraction a group? {my reasoning is no it isnt, because $a-0 = a \neq 0-a$}


share|cite|improve this question
That is not what a group is, and I doubt Herstein has made such an error. Read his definition again, in full. – Chris Eagle Jul 26 '12 at 16:03
@ChrisEagle This is one of the properties of a group. I believe the OP skipped the other three since they are not related to his question. – Code-Guru Jul 26 '12 at 16:07
@Code-Guru Indeed perhaps I had been vague, and gave off an impression that it is the only property, but it is one of the four necessary properties. I left those off because they are not pertinent. – Soham Jul 26 '12 at 16:09
Just because a condition is "non-necessary" doesn't mean it can't hold. – Ted Jul 26 '12 at 16:13
up vote 3 down vote accepted

The non-necessary condition is $\forall a,b \in A, a \cdot b \not= b \cdot a$. This doesn't rule out that it may be true for some $a, b \in A$. In fact, it could be true for all elements in the group. Then we call it an Abelian group, which is still a group, nonetheless.

And you are correct, the integers (or rationals or real numbers) with subtraction does not form a group. You gave one reason. You could also check associativity.

share|cite|improve this answer
Yes, thanks I checked all the 4, and indeed it fails. – Soham Jul 26 '12 at 16:07

The commutativity of the identity in a group is part of its definition, even in a non-commutative group. And you are correct about subtraction not being a valid group operator: it's not even associative.

share|cite|improve this answer

Non-commutative denotes the negation of the commutative law $$\rm\:\color{#C00} \lnot\ [\color{#C00}\forall\,x,y\!:\ xy \color{#C00}= yx]\, \equiv\, \color{#C00}\exists\,x,y\!:\ xy\color{#C00}\ne yx\:$$ In words, commutativity fails to be true universally (for all elements), precisely when there exists at least one counterexample. It doesn't mean that commutativity always fails.

share|cite|improve this answer
+1 for awesome use of symbols! – Soham Jul 26 '12 at 16:26
@Soham Because confusion of quantifiers is a common problem when learning mathematics, it sometimes helps to mathematically "reify" the quantifiers, so that one may learn to rigorously manipulate them algebraically (vs. intuitively, in natural language). One needs to learn to symbolically translate logical word problems" just as one does for algebra word problems. And, in fact, logic can be completely algebraicized (cf work of Tarski, Henkin, Halmos, et al). – Bill Dubuque Jul 26 '12 at 16:36
@BillDubuque : What does the first symbol that u have used mean ? – Theorem Jul 26 '12 at 17:25
@Theorem The first symbol $\:\lnot\:$ denotes logical negation. For the others see the Wikipedia articles on $\:\forall = $ Universal Quantification and $\:\exists = $ Existential Quantification. – Bill Dubuque Jul 26 '12 at 17:32

The "unnecessary condition" is that $a\cdot b= b \cdot a $ for all $a,b$. We don't need this fact to prove that the identity commutes. Let's assume there are two: an left identity $i$ and a right identity $j$.

We find that $i\cdot j=i=j$, since we can use the definition of either identity. So they're actually the same. In the immortal words of highlander, "there can be only one". So the same identity element has to work for both sides if it's going to work at all.

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.