Groups of order pq

[abhishekkgp]

Let $\displaystyle o(G)=pq$, $\displaystyle p, \, q$ are primes with $\displaystyle p>q$. Prove that if $\displaystyle q\not | (p-1)$, then $\displaystyle G$ is cyclic.

ATTEMPT: I can show that $\displaystyle G$ has a subgroup of order $\displaystyle p$ and a subgroup of order $\displaystyle q$. This can be done using the class equation as follows:
$\displaystyle o(Z(G))$ has four possible values, viz, $\displaystyle 1, \, p, \, q, \, pq$. The only we need to consider is $\displaystyle o(Z(G))=1$ since otherwise $\displaystyle G/Z(G)$ is cyclic and hence $\displaystyle G$ is abelian.

Assume no subgroup of $\displaystyle G$ is of order $\displaystyle q$. Then the class equation gives $\displaystyle pq=1+kq$ for some positive integer $\displaystyle k$, which is clearly a contradiction. Thus $\displaystyle G$ has a subgroup of order $\displaystyle q$. Similarly $\displaystyle G$ has a subgroup of order $\displaystyle p$.
Can someone tell me what to do now??

 

[daon2]

You should probably be posting these in the advanced math forum.

Let $\displaystyle n_p, n_q$ denote the number of sylow p and q subgroups. Since $\displaystyle q < p,$ and $\displaystyle n_q | p \implies n_q = 1$. Hence there is one sylow q-subgroup, Q.

$\displaystyle n_p = pk+1$ and since $\displaystyle q \not | (p-1)$, $\displaystyle n_p \neq q$ so there is one sylow p-subgroup as well, P.

So P & Q are both normal and cyclic (why?), and p,q being distinct |P||Q| = pq = |G|. Take a generator x from P, a generator y from Q. Show <xy>=G.

 

[abhishekkgp]

You should probably be posting these in the advanced math forum.

Let $\displaystyle n_p, n_q$ denote the number of sylow p and q subgroups. Since $\displaystyle q < p,$ and $\displaystyle n_q | p \implies n_q = 1$. Hence there is one sylow q-subgroup, Q.

$\displaystyle n_p = pk+1$ and since $\displaystyle q \not | (p-1)$, $\displaystyle n_p \neq q$ so there is one sylow p-subgroup as well, P.

So P & Q are both normal and cyclic (why?), and p,q being distinct |P||Q| = pq = |G|. Take a generator x from P, a generator y from Q. Show <xy>=G.

I posted this in this forum because I guess these belong to the category of "advanced algebra". Anyways, from now on i will use the "advanced math forum" for these.

Daon, thanks for your reply.. but your solution is too advanced for me!! Sylow theorems are yet not covered in the text i am reading. Even the use of class equation is not required according to the author.. but i used it since it at least gave me some lead. Can you please do it with out the use of sylow theorems?? And yes.. Cauchy's and Sylow's theorems for ABELIAN groups ARE allowed.

 

[daon2]

I posted this in this forum because I guess these belong to the category of "advanced algebra". Anyways, from now on i will use the "advanced math forum" for these.

Daon, thanks for your reply.. but your solution is too advanced for me!! Sylow theorems are yet not covered in the text i am reading. Even the use of class equation is not required according to the author.. but i used it since it at least gave me some lead. Can you please do it with out the use of sylow theorems?? And yes.. Cauchy's and Sylow's theorems for ABELIAN groups ARE allowed.

This board used to be for all the beginning college and high-school type math, when later they added an "advanced math" section. Advanced algebra here could mean, I guess, Pre-calc and a first year college algebra course.

If you can't use Sylow, then you know by Cauchy there are still cyclic subgroups of order p and q. Since p and q are distinct they intersect trivially, as subgroups must have order dividing the orders of the group containing them. Hence the order of the intersection is 1. Now if x in P, y in Q are generators, we have PQ = <x><y> =G because the order of PQ is |P||Q|/|P intersect Q| = pq = |G|. We are still at the crossroads of showing <xy>=G.

If you know it (or want to prove it), you may use that for x,y in any finite group G, |xy| divides lcm(|x|,|y|)

 

[abhishekkgp]

If you know it (or want to prove it), you may use that for x,y in any finite group G, |xy| divides lcm(|x|,|y|)

this will solve it right?? i didn't know this beforehand.. will try to prove it.
How do you know so many "sneaky" things about groups?? Does this come with experience??

 

[daon2]

this will solve it right?? i didn't know this beforehand.. will try to prove it.
How do you know so many "sneaky" things about groups?? Does this come with experience??

Yes, tons and tons of "experience" . But no that will not necessarily solve it. You would then have to show that xy does not have order p or q, and hence must have order pq. You still have not yet used one of your assumptions.

 

[abhishekkgp]

If you know it (or want to prove it), you may use that for x,y in any finite group G, |xy| divides lcm(|x|,|y|)

I don't see how to prove this: $\displaystyle o(xy)|lcm(o(x),o(y))$.
I can do this for abelian groups but i can't see how to prove this for non-abelian groups.
Please help.

 

[daon2]

I don't see how to prove this: $\displaystyle o(xy)|lcm(o(x),o(y))$.
I can do this for abelian groups but i can't see how to prove this for non-abelian groups.
Please help.

It is definetly not true when G is not abelian. So nevermind, good catch. Though in this case it wouldn't give any more information since we do already know that |xy| must be p,q, or pq.