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Suppose that $G$ is a cyclic group of order $10$ with generator $a\in G$.Order of $a^{8}$ is _______
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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$_2$ . $\frac{p}{p} = \frac{K_2.p}{p} = \frac{n}{p} = \frac{n}{gcd(n,k)}$

by Junior (585 points)
edited
+4

I found another method  to do that this type of question

Suppose that $G$ is a cyclic group of order $n$ with a generator $a\in G.$Then the order of $a^{k},$ where $k\in Z^{+}$ is:
order of $a^{k}=\frac{n}{gcd(n,k)}$

In given question order of cyclic group $n = 10$ with given generator $a$ and $k=8$

$a^{8} = \frac{10}{gcd(10,8)}$

$a^{8} = \frac{10}{2}$

$a^{8} = 5$

see this pdf,this is resource to above concept

https://gateoverflow.in/?qa=blob&qa_blobid=5282079886704787639

+1
+1
Great 👌
0

You can easily see that,
a^8*5=a^40=(a^10)^4=e^4=e.

Can you explain this line, please$?$

In the exam how we solve?

+1
edited the answer, you may check it !