1

I am trying to solve a problem to find the order of some integers with (mod 9). I understand the concept I also have the solution to the problem. My calculations are also correct except for a few mistakes.

The questions is :

"What order can an integer have (mod 9)? Find an example of each."

2 and 5 has order 6 with mod 9. As:

  • $2^6 \equiv 1\pmod 9$

  • $5^6 \equiv 1 \pmod 9$

4 and 7 has order 3 mod 9. As:

  • $4^3 \equiv 1 \pmod 9$

  • $7^3 \equiv 1 \pmod 9$.

and 8 has order of 2 mod 9. As:

  • $8^2 \equiv 1 \pmod 9$.

The books gives the solution as follows:

"1,2,3, or 6."

Because

"(2 and 5 have order 6, 4 and 7 have order 3, 8 has order 2, and 1 has order 1)."

Now what does the first line

"1,2,3, or 6." in the solution mean?

Is it just a typo? Or do I miss something?

As far as I know 3 and 6 do not have a solution as the (mod 9) to power of 3 only gives 2 values as $3^2 \equiv 0 \pmod 9$.

*Update: So by various comment i understood 1,2,3,6 are possible values of order. So how can I know only these (1,2,3,6) are values of power which are possible order?. Is there any method to calculate or know this that these are only potential order values?

(I understood the solution already just made this update so other people can understood it too in-case someone is looking at this problem for solution) Reference : Introduction to Number Theory by Underwood Dudley (2nd edition),pp 75.

wlad
  • 8,185
Deepak
  • 263
  • $1,2,3$ and $6$ are the possible values for the powers (the orders), not the elements being raised to those powers. – Daniel Apr 07 '15 at 05:26
  • See http://math.stackexchange.com/questions/127109/order-of-an-element-modulo-n-divides-phin – lab bhattacharjee Apr 07 '15 at 05:27
  • You are confusing "order" with examples! 1,2,3 and 6 are the orders an integer (in this case positive integers<9, except 3 and 6, obviously, as you have quoted in your answer) can have. – Jesse P Francis Apr 07 '15 at 06:10
  • @JessePFrancis Sow can I know only these (1,2,3,6) are values of power which are possible order?. Is there any method to calculate or know this that these are only potential order values? – Deepak Apr 07 '15 at 06:18
  • @shunya: check the answer I just posted: and I think you might have to add your doubt in previous comment in the question too! – Jesse P Francis Apr 07 '15 at 06:36
  • The question asks "what orders can an integer have..." so the numbers listed in the solution are, of course, the possible orders. – Bill Dubuque Apr 07 '15 at 15:33

2 Answers2

2

The group you are in is the group of units mod $9$, which is:

$\{[1],[2],[4],[5],[7],[8]\}$.

This is an abelian group of order $6$, and so must be isomorphic to $\Bbb Z_6$. As such the only possible orders are $1,2,3,6$, each of which occur, since $\Bbb Z_6$ is cyclic.

Since:

$[2]^2 = [4] \neq [1]$, we see $[2]$ does not have order $2$ (nor order $1$, since it is not the identity).

$[2]^3 = [8] \neq [1]$, so $[2]$ does not have order $3$. Hence $[2]$ is a generator:

$[2]^0 = [1], [2]^1 = [2], [2]^2 = [4],[2]^3 = [8], 2^4 = [7], 2^5 = [5]$.

It follows that $o([2]^k) = \dfrac{6}{\gcd(k,6)}$, so:

$[1]$ has order $1$, $[2],[5]$ have order $6$, $[4],[7]$ have order $3$, and $[8]$ has order $2$.

David Wheeler
  • 15,406
1

You are confusing "order" with examples! 1,2,3 and 6 are the orders an integer (in this case positive integers<9, except 3 and 6, obviously, as you have quoted in your answer) can have (mod 9).

Now to see they are the (only possible) values:

Order of an element modulo n divides ϕ(n)

Now, $ϕ(n) = n\prod_{p|n}({1-\frac{1}{p}})$ where the product is over the distinct prime numbers dividing n. So $ϕ(9) = 9*2/3 = 6$

Clearly, only 1,2,3 and 6 can divide $ϕ(9)$

Community
  • 1
Jesse P Francis
  • 1,469
  • 3
  • 27
  • 71