Because first two properties of group are satisfied but I have read in local author book it is not semi-group.

  • 2
    $\begingroup$ What book is that? Either the author's wrong, or you're misreading it, or they're understanding "semigroup" as what is generally called "monoid" and they don't consider $0$ to be a natural number (monoids have to have identity elements). Btw I didn't downvote, but this needs more context to be answerable. $\endgroup$ – Noah Schweber Oct 4 at 19:37
  • $\begingroup$ Okay Sir you are right author is wrong.Thank you so much $\endgroup$ – Navid Gujjar Oct 5 at 20:06

Normally you would need to specify with respect to which operation you are checking if the given set is a semigroup. For example, the set of natural numbers is a semigroup with respect to addition and so is with respect to multiplication, but, for example:

  • You cannot define subtraction in an ordinary way on $\mathbb N$ - this is sometimes called out as "$\mathbb N$ not being a semigroup with respect to subtraction" (although this is incorrect - it is just that subtraction is not defined on $\mathbb N$)
  • A lot of other binary operations can be defined on $\mathbb N$ but are not associative. Think of exponentiation: $x^{(y^z)}$ is generally not the same as $(x^y)^z$.

So my guess is that maybe the author of the book is thinking of one of the two cases above, or of something similar.

  • $\begingroup$ Notice that $\mathbb N$ with subtraction is also not a semigroup for another reason, which is that, even where defined, it isn't associative. So for example, $(\mathbb Z,-)$, with $-$ seen as binary, is not a semigroup, as $$a-(b-c)=a-b+c\neq a-b-c=(a-b)-c,$$ where the inequality is verified if $c\neq 0$. $\endgroup$ – amrsa Oct 5 at 16:38
  • $\begingroup$ Okay sir! I have clear my point.so much thank you giving me information $\endgroup$ – Navid Gujjar Oct 5 at 20:10

Definition: A semigroup is a set $S$ together with a binary operation $*$ that is, a function $* :S\times S\rightarrow S$ that satisfies the associative property:

For all $a,b,c\in S$, the equation $(a * b) * c=a* (b * c)$ holds.

The natural number $\mathbb{N}$ with usual addition and multiplication is semigroup.

If we define binary operation on $\mathbb{N}$ as follows \begin{align*} *: \mathbb{N}\times \mathbb{N} \to \mathbb{N}\\ (n, m) \mapsto n^{m} \end{align*} then this binary operation is not associative, because \begin{align*} 2*(3*4)=2^{3^{4}}\neq (2^{3})^{4}=(2*3)*4. \end{align*} So $(\mathbb{N}, *)$ is not Semigroup.


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