order problem

[Herb]

Hello,

I have following problem.

What is order of (1,2,7)+<(3,2,7)> in Z_9 x Z_4 x Z_49 / <(3,2,7)> ?

my answer is 9.
-Because : Let t = (1,2,7)+<(3,2,7)>
t = (-2,0,0)+(3,2,7)+<(3,2,7)> = (-2,0,0)+<(3,2,7)>
= (7,0,0)+<(3,2,7)>
So, n*t = (0,0,0)+<(3,2,7)>
=> (7n,0,0)+<(3,2,7)> = (0,0,0)+<(3,2,7)>
=> 7n = 0 (mod 9)
=> minimum n in N = 9.

my book answer is 3.
-book explanation : Let n=ord((1,2,7)+<(3,2,7)>)
Then n=min{k in N | k in <3>, 2k in <2> , 7k in <7>}
=min{k in N | k in <3>}
=3.

But I don't know my solution's wrong point.
If anybody know wrong point of my trying, please post reply.

Thanks in advance for your help~

[William Elliot]

On Thu, 2 Sep 2004, Herb wrote:

> What is order of (1,2,7)+<(3,2,7)> in Z_9 x Z_4 x Z_49 / <(3,2,7)> ?
>

3((1,2,7) + <(3,2,7)>)
= (3,6,21) + <(3,2,7)>
= (3,6,21) + 11(3,2,7) + <(3,2,7)>
= (0,0,0) + <(3,2,7)>

[Brian M. Scott]

On Thu, 2 Sep 2004 06:41:44 -0700, William Elliot
<ma...@privacy.net> wrote in
<news:2004090205...@agora.rdrop.com> in
alt.math.undergrad:

He knows that; he wants to know what's wrong with *his*
argument.

Brian

[Brian M. Scott]

On 2 Sep 2004 03:53:20 -0700, Herb <fuzz...@nate.com> wrote
in <news:cd698bc7.04090...@posting.google.com> in
alt.math.undergrad:

> Hello,

> I have following problem.

> What is order of (1,2,7)+<(3,2,7)> in Z_9 x Z_4 x Z_49 / <(3,2,7)> ?

> my answer is 9.
> -Because : Let t = (1,2,7)+<(3,2,7)>
> t = (-2,0,0)+(3,2,7)+<(3,2,7)> = (-2,0,0)+<(3,2,7)>
> = (7,0,0)+<(3,2,7)>
> So, n*t = (0,0,0)+<(3,2,7)>
> => (7n,0,0)+<(3,2,7)> = (0,0,0)+<(3,2,7)>
> => 7n = 0 (mod 9)

This step is the problem. What you know at this point is
that (7n,0,0) is in <(3,2,7)>. This means that there is a k
such that

3k = 7n mod 9
2k = 0 mod 4
7k = 0 mod 49;

it does not necessarily mean that 7n = 0 mod 9, because k
need not be a multiple of 3. And indeed n = 3, k = 28 is a
solution.

> => minimum n in N = 9.

[...]

Brian