Finite abelian group of size p-1

[sid_galt]

Suppose we have two finite abelian groups $$G,G^{\prime}$$ of size $$n=pq$$, $$p,q$$ being primes. $$G$$ is cyclic.

Both $$G,G^{\prime}$$ have subgroups $$H,H^{\prime}$$, both of size $$q$$. The factor groups $$G/H,\ G^{\prime}/H^{\prime}$$ are cyclic and since they are of equal size, they are isomorphic. Are $$G,G^{\prime}$$ also isomorphic?

Edit: The title is wrong. p-1 has nothing to do with this problem. Sorry about that.

 

[ThirstyDog]

I don't think the groups G and G' are required to be isomorphic. You have already defined G so consider
$$G' = <a,b|a^{p}=b^{q}=e>$$
where e is the identity. This should be a suitable counter example -try working through it.

If I am wrong about it being a counter example then this should prove that G and G' are required to be isomorphic as there are only two non-isomorphic groups of order pq with p and q being prime.

Hope this helps.

 

[Ancient_Nomad]

EDIT: sorry for that earlier post

Let p > q. Now, if q does not divide p-1 then any group of order p*q must be cyclic. (This is a result which can be proved easily with sylow's theorems)
A simple example are groups of order 15 = 5*3.

If on the other hand, q does divide p-1 then this group must be non-abelian. (The proof for this is a bit involved, but it can be found in some textbook like herstein).

Now since, you have already stated that G is cyclic. This implies q does not divide p-1. Then G must be cyclic and so must be G'. Hence they must be isomorphic.

Suppose p = q. In this case, its really difficult to say anything, because G' has two possible structures:
1. a cyclic group of order p^2.
2. a group isomorphic to C_p x C_p. (C_p is the cyclic group of order p)

I'll try to find the references containing the proofs for the results I have stated above and post them later, but I hope this was useful.

 

[matt grime]

Thirsty dog, the group you defined for G' is infinite: it is the free product of C_p and C_q.Do you know an structure theorems for finite abelian groups? Are you supposed to assume that p and q are distinct primes?

 

[ThirstyDog]

Hey Matt,

I forgot to say Abelian... I am guessing you realized this by your next comment.

I was under the impression that any finite Abelian group is expressible as the Cartesian products of a finite number of cyclic groups whose orders are prime powers.

I think that if p and q are distinct then the cyclic group of order pq is isomorphic to the Cartesian of cyclic groups of order p and q. If p and q are equal then no such isomorphism exist.

This says in terms of the original question G and G' are definitely isomorphic if p is distinct from q. But if p=q then G and G' are not always isomorphic. Ancient_Nomad has pointed this out.

 

[PhysicalAnomaly]

Intuitively, I would guess that H,H' are isomorphic => G,G' are isomorphic...

 

[ThirstyDog]

Consider
$$C_{4} = <c|c^{4}=e> \mbox{ and } C_{2}\times C_{2} = <a,b|a^{2}=b^{2}=e,ab=ba>$$
These are clearly not isomorphic as the first has an element of order 4 while the latter does not.

We can choose the subgroups
$$H = <c^{2}> \mbox{ and } H' = <a>$$
Thus H and H' are isomorphic. Also
$$C_{4}/H \cong C_{2} \cong C_{2}\times C_{2}/H'$$

This shows that just because an two Abelian groups can have isomorphic subgroups and their quotients are also isomorphic doesn't imply the original groups are.