Re: Number of subgroups

On Sep 26, 10:40 am, jane <jane1...@xxxxxxxxxx> wrote:
I'm solving the following problem: Find the number of subgroups of order p^2 of the group G = Z_{p^3} x Z_{p^2}.


I assume p is prime here.

My result : there are 3 such subgroups - every abelian group of order p^2 must be either Z_{p^2} or Z_p x Z_p. In the 1'st case this one must be generated by the element of order p...

By "g is of order p", I usually mean in this context "pg = (0,0)", or
equivalently <g> has p elements. I think you want "g has order p^2".

but there is only one element in G of order p, namely (p,1).

It's true that the order of <(p,1)> is p^2; (p,1) has order p^2 (and
not order p). But what about <(p,0)>? What about <(0,1)>?

If <(x,y)> is a cyclic subgroup of G of order p^2, then p^2*(x,y) =
(0,0), and for no smaller value m < p^2 do we have m*(x,y) = (0,0).
Another way of saying this is that p^2*x = 0 mod p^3, and p^2 * y = 0
mod p^2, and no smaller value m satisfies /both/ these conditions /
simultaneously/.

Try working out, for p = 3, the 9 elelemnt subgorup generated by
(3,2).

In the second case, consider the intersection of this subgroup H of order p^2 with Z_{p^3} x {1} and {1} x Z_{p^2}. If the 1 of these intersections is a subgroup of order p, then the other one must also be a subgroup of order p, and we get that H = <(1,p)> . If none of them of order p, then 1 of them must be of order p^2 and hence either {1} x Z_{p^2} or Z_{p^3} x {1}.


Here I think you are writing {1} (which is not a subgroup) when you
mean the trivial subgroup {0}. 0 is the identity element. 1 is the
additive generator.

As your logic depends on there being only one subgroup H isomorphic to
Z_(p^2), you need to rethink it.

Cheers - Chas


.

Relevant Pages

  • Re: I need help proving a theorem!!!
    ... another group is using the 1-step, 2-step, or finite subgroup tests. ... the identity element of G is an element of H; ... have to verify and establish that suchandsuch is indeed true. ... dont know how to use this with these guidelines of our subset. ...
    (sci.math)
  • Re: I need help proving a theorem!!!
    ... The only ways I have learned to show that a group is a subgroup under ... another group is using the 1-step, 2-step, or finite subgroup tests. ... the identity element of G is an element of H; ... have to verify and establish that suchandsuch is indeed true. ...
    (sci.math)
  • Re: Sylow subgroups of Alt(4)
    ... possibly be in a Sylow ... identity element, then this subgroup has to be the ... subgroup since any other such subgroup would have to ... a product of two disjoint transpositions. ...
    (sci.math)