[GAP Forum] direct sums and counting subgroups

[Dmitrii Pasechnik]

Dear Laura,
you¹re right, I wasn¹t thinking straight when I replied (sitting in front of
230 exam papers on calculus 1 that I have to mark in 2 days :( )

anyhow, the count is easy (the only facts needed are essentially that in a
finite cyclic group of order n a subgroup of given order k is unique, k
divides n, and some obvioups properties of the direct sum):

first, there is only one subgroup isomorphic to Z_p x Z_p.

second, a subgroup H=<(x,y)> isomorphic to Z_{p^2} must have the projection
isomorphic to  Z_{p^2} to either the first or to the second summand. (At
this point I rushed to conclusion that there are no more cases to count...)
However, in both cases we have to deal with the projection of H to the other
summand.

First, let x be of order p^2. Then the order of y must divide p^2.
The first case is when the order of y equals p^2.
In the cyclic group of order p^2 there are p^2-p elements of order p^2.
This gives p^2-p possibilities.
The second case is when the order of y equals p.
In the cyclic group of order p there are p-1 elements of order p^2.
This gives p-1 further possibilities.
The third case is when the order of y equals 1.
Such an element is unique.
This gives 1 further possibility.

Now, it remains to consider the case when y is order p^2, and x is of order
either p or 1. Just as above, this gives (p-1)+1=p more possibilites.

Thus, we get p^2+p+1 possibilities in total.
Please let me know if you need to know more details in the above.

On 11/25/07 9:37 PM, "laurawicklund at comcast.net" <laurawicklund at comcast.net>
wrote:

> I have the answer.  There are p^2+p+1 subgroups of order p^2 in the abelian
> group Z_{p^3}\oplus Z_{p^2}.  I found the same results using GAP.  For
> example, I found that there are 13 subgroups of order 9 in the group
>  
> G := DirectProduct(CyclicGroup(3^3), CyclicGroup(3^2));
> Using GAP is not part of my course work.  It is just an interest of mine.
>  
> I assume you are speaking up to an isomorphism.  Is this correct?  If so, I
> thought there were only two.  Namely,
>  
> Z_{p^2} and Z_p X Z_p
>  
> I am not aware that one of these groups is realized twice.  I don't really
> understand what you mean by this.  My text books only state that there are two
> (up to  an isomorphism).  If you have a chance, please explain.
>  
> Thank you for your time.
>  
> -Laura
>  
>> -------------- Original message --------------
>> From: Dmitrii Pasechnik <dima at ntu.edu.sg>
>> 
>>> surely you don't need GAP for solving this problem.
>>> The answer is 3. There are 2 possibilities for an abelian group to be of
>>> order p^2; one of them is realised only once, the other - twice.
>>> I don't like giving more details, as this could be your homework :)
>>> 
>>> 
>>> On 11/25/07 6:09 AM, "laurawicklund at comcast.net"
>>> wrote: 
>>> 
>>>> I recently installed GAP on my computer and am trying to figure out how to
>>>> solve the following problem using GAP (I spent a few hours reading through
>>>> the 
>>>> manual but haven't been successful).
>>>> 
>>>> Let p be prime. Find the number of subgroups of order p^2 of the additive
>>>> abelian group $G:=Z_{p^3}\oplus Z_ {p^2}$
>>>> 
>>>> The group G is the direct sum of additive groups ZmodnZ(p^3) and
>>>> ZmodnZ(p^2) 
>>>> 
>>>> I would be content solving by problem for a specific p, say p=3.
>>>> 
>>>> Thank you for your time.
>>>> 
>>>> -Laura 


-- 
Dima Pasechnik
http://www.ntu.edu.sg/home/dima/