# Associativity of a binary operator

I am reading the book "A first course in abstract algebra", and i am reading the chapter called subgroups. It says that, for general situations we do not use "*" to denote the binary operation on a group, but we simply use addition symbol "+" or multiplication symbol ".". Then it says, we denote the product a.a.a.a.a...a for n factors by an. But i am confused here: is that always associative? For example, for 3 factors, a.a.a is a3 according to what book says, but are we sure that a.(a.a)=(a.a).a=a3 is always true? Thank you.

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Do you know what "group" means? – Chris Eagle Mar 14 '13 at 20:53
Yes, i do. Why? – bigO Mar 14 '13 at 20:54
Then you know the answer to this question. – Chris Eagle Mar 14 '13 at 20:54
This is why i am asking this, i must be missing something – bigO Mar 14 '13 at 20:55
Note that not all binary operations need be associative. For example, take the integers under subtraction: $-4=(1-2)-3\neq 1-(2-3)=0$. – user1729 Mar 15 '13 at 14:54

Recall: One of the key properties satisfied by all groups is that its binary operation is associative on the set defined by the group.

If you are referring to Fraleigh's text, see

Definition 4.1: "A group $\langle G, *\rangle$ is a set $G$, closed under a binary operation $*$, such that the following axioms of satisfied:

$\mathcal G_1$: For all $a, b, c \in G,$ we have $$(a*b)*c = a*(b*c)\tag{associativity of *}$$

Of course, $\mathcal G_2,\;\text{and}\;\mathcal G_3\;$ are crucial, as well: the existence of an identity element $e\in G$, and for each $g\in G$, its inverse, $g^{-1}$ is also in $G$, including the definitions of the identity and how we define the inverse of a group element.

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Oh thanks, that's what i have been missing. – bigO Mar 14 '13 at 20:55

Not only, As @amWhy noted, the binary operation is associative but also depend on how you show the binary operation, the figures of composition of elements varies: $$\cdot\longrightarrow a*b=a\cdot b$$ so $$\underbrace{a\cdot a\cdot a\cdot ...\cdot a}_{\large n\;times}=a^n~$$ And $$+\longrightarrow a*b=a+ b$$ so $$\underbrace{a+ a+ a+ ...+ a}_{\large n\;times}=na$$

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Don't know why the downvote? +1 – amWhy Mar 14 '13 at 21:09
Thank you! I did not downvote, just to let you know – bigO Mar 14 '13 at 21:28
@amWhy: Thanks Amy for the edit. :-) – Babak S. Mar 15 '13 at 5:32