Re: Number of cyclic subgroups of order 15 in Z_90 (+) Z_36
- To: mathgroup at smc.vnet.net
- Subject: [mg37987] Re: Number of cyclic subgroups of order 15 in Z_90 (+) Z_36
- From: "Diana" <diana53xiii at earthlink.remove13.net>
- Date: Sat, 23 Nov 2002 19:16:06 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
I meant:
> Z_90 is the additive group {0, 1, 2, 3, ..., 89}. Z_36 is the additive
group
{0, 1, 2, 3, ..., 35}, not 36.
Diana
"Diana" <diana53xiii at earthlink.remove13.net> wrote in message news:...
> Mathematica groupies,
>
> I just got my copy of Mathematica tonight. I understand permutations of
S_5
> and A_5, etc., but I am having a little difficulty figuring out how to
> calculate the order of elements of Z_90 and Z_36, and the order of
elements
> of the external direct product of Z_90 (+) Z_36.
>
> I am wanting to calculate the number of cyclic subgroups of order 15 in
Z_90
> (+) Z_36 with Mathematica.
>
> Z_90 is the additive group {0, 1, 2, 3, ..., 89}. Z_36 is the additive
group
> {0, 1, 2, 3, ..., 36}.
>
> So, elements of Z_90 (+) Z_36 would be 2-tuples of the form: (a, b), where
a
> is an element of Z_90, and b is an element of Z_36. If you add (a, b) to
> itself 15 times, you will get {0, 0}. Note that the operation of adding
the
> first component of the 2-tuple is modulo 90, and the operation of adding
the
> second component of the 2-tuple is modulo 36.
>
> Can someone help?
>
> Thanks,
>
> Diana M.
>
>