Suppose that * is an associative operation on a set S.

Question

Suppose that * is an associative operation on a set S. Define $x^n$ to mean $x*x*x*...*x$, $n$ times. (so, for example, $x^3=x*x*x.$) Suppose further that an elements $a$ of S is such that all of $a,a^2,...,a^9$ are different but $a^{10}=a^3$. Then there are some elements $b$ belonging to S such that $b=b^2$. One such elements b is:

a) $a^4$ .$\qquad$ b) $a^5$ $\qquad$ c) $a^7$ $\qquad$ d)$a^9$ $\qquad$ e) $a^{13}$

Try to do this question, but still no idea, $a^{10}=a^3\Longrightarrow a=0 \, or\, a=1$ , why this question mentioned associative operation, I don't get this.

Answer 1

By associativity: $$a^{10}=a^3 \implies a^{14} = a^{10}*a^{4}=a^{3}*a^{4} = a^{7}$$

Answer 2

This question is leaned towards group theory.

$a^{10}=a^3$

We can operate on both side with $a$. (This is not usual multiplication)

$a*a^{10}=a*a^3$

$a^{11}=a^4$

again operate both sides with $a$

$a^{12}=a^5$

$a^{13}=a^6$

$a^{14}=a^7$ which is clearly $b^2=b$

Hence your answer must be $a^7$ option(c)