It is a well known property that: order of group = order of generator of that group.
This means Order(a)=10, ===> $a^{10}$=e.
Let Order($a^{8}$) = x,
∴ $(a^{8})^x$ = e ===> $a^{8x}$ = e ------------ (1)
we know that, e$^i$ = e ( for some i > 0 ), therefore we can substitute this property in (1)
∴ $a^{8x}$ = (e)$^i$ ==> $a^{8x}$ = $(a^{10})^i$
∴ 8x = 10 i
You can easily guess that, smallest possible value which can satisfy it is x = 5 for i = 4.
∴ Order($a^{8}$) = x = 5
Now my problem is " we should have to guess ? "
8x = 10 . i
2$^3$ x = 2$^1$.5$^1$.i ( just represent every number into product of prime numbers. )
cancel the common terms, ( indirectly canceling the GCD of 8,10 )
2$^2$ x = 5$^1$.i
as we require minimum value in i, which can equate it !
just substitute i=2$^2$, then the eqn should be like 2$^2$ x = 5$^1$.2$^2$ ==> x = 5.
How the formula, which is mentioned by lakshman is derived ?
k x = n i, then what we did ?
represent in prime form and cancel the GCD in both side and remaining part of coefficient of x is assigning to i.
let gcd of (k,n) = p
then k = K$_1$.p and n = K$_2$.p
k x = n i
K$_1$.p x = K$_2$.p.i ==> K$_1$ x = K$_2$ . i, then keeping i = K$_1$, x = K$_2$,
x = K$_2$
x = K$_2$ . $\frac{p}{p} = \frac{K_2.p}{p} = \frac{n}{p} = \frac{n}{gcd(n,k)} $