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$\begingroup$

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 $a-0 = a \neq 0-a$}

Soham

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    That is not what a group is, and I doubt Herstein has made such an error. Read his definition again, in full.2012-07-26
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    @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.2012-07-26
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    @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.2012-07-26
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    Just because a condition is "non-necessary" doesn't mean it *can't* hold.2012-07-26

4 Answers 4

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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.

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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.

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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.

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    @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).2012-07-26
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    @BillDubuque : What does the first symbol that u have used mean ?2012-07-26
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    @Theorem The first symbol $\:\lnot\:$ denotes logical negation. For the others see the Wikipedia articles on $\:\forall = $ [Universal Quantification](http://en.wikipedia.org/wiki/Universal_quantification) and $\:\exists = $ [Existential Quantification.](http://en.wikipedia.org/wiki/Existential_quantification)2012-07-26
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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.

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    Yes, thanks I checked all the 4, and indeed it fails.2012-07-26