Why is $T(t+s)=T(t)T(s)$ called semigroup property?

Question

Let $\{T(t):t>0\}$ be a family of operators. We say it is semi group if $$ T(t+s)=T(t)T(s) \text{ for all }t,s >0$$ On the other hand any set with an associative binary operation is called semi group. I am unable to figured it out that why $T(t+s)=T(t)T(s)$ is called semi-group property.

Answer 1

I guess it's because the map $T:\mathbb R_+\to B(H)$ is a semigroup homomorphism (maps from one semigroup to another one, preserving the semigroup operation). In the same way that a representation $U:\mathbb R\to B(H)$ is a group homomorphism.

Answer 2

As you said, any set with an associative binary operation is a semigroup. In particular, any subset $S$ of $\mathcal{L}(X)$ closed under composition of operators is a semigroup (where the operation is the composition).

Among these subsets, "the most interesting" (words of E. Hille in [1]) are those which are indexed by a semigroup $(A,\circ)$, say $S=\{T(a)\mid a\in A\}$, and satisfies $$T(a\circ b)=T(a)T(b).$$ This equality gives a relation between the operation of the semigroup of operators $S$ and the operation of the index semigroup $A$. This is why it is called "semigroup property". (The most common case is $A=\mathbb R_{+}$).

So, roughly speaking, a semigroup (in the operator sense, also called one-parameter semigroup) is a subsemigroup (in the abstract sense) of $\mathcal{L}(X)$ indexed by another semigroup with the property that the semigroup operation is preserved. The sense of "operation is preserved" is given by the equality which is called "semigroup property".

[1] E. Hille. What is a semi-group? in Studies in real and complex analysis edited by I. I. Hirschman, Jr. (1965).