If $\operatorname{ord}_ma=10$, find $\operatorname{ord}_ma^6$

Solution 1:

If $$ord_ma=d, ord_m(a^k)=\frac{d}{(d,k)}$$ (Proof @Page#95)

Here $$d=10,k=6\implies ord_m(a^6)=\frac{10}{(10,6)}=\frac{10}2=5$$

Solution 2:

Hint $\ $ If $\rm\ ord\ a\, = 10\:$ then $\rm\:a^{6n}\equiv 1\iff 10\mid 6n\iff 5\mid 3n\iff 5\mid n.$

Remark $\ $ Above we invoked: $\ $ if $\rm\:ord\ a\, = k\:$ then $\rm\:a^n\equiv 1\iff k\mid n.\:$ This can be viewed much more conceptually: the set $\cal O$ of integers $\rm\,n\,$ with $\rm\:a^n\equiv 1\:$ is closed under subtraction, hence, by a fundamental theorem, every element of $\cal O$ is a multiple of the least positive element of $\cal O,\,$ which in our case is called the order.