# Thread: Cyclic group question - from Herstien

1. ## Cyclic group question - from Herstien

This is Q#9, Section 2.9, pg 75 from Herstien
Q: order(G) = pq, where p and q are distinct primes. G has a normal subgroup of order p and a normal subgroup of order q. Prove G is cyclic.

My attempt:

Let N1 be the subgroup of order p
Let N2 be the subgroup of order q

Clearly, N1 and N2 are cyclic (as for every group with prime order). Let a, b be generator of N1 and N2 respectively.

Consider c=ab

I can show that order(c)=pq. And, hence G is cyclic (with c as generator)

Help I need:
1. Is my attempt correct?
2. I never used the fact that N1 and N2 are normal (something mentioned in the question) - So am I missing / mis-understanding something?

Thanks

2. If you've never used the fact that $\displaystyle N_1$ and $\displaystyle N_2$ were normal subgroups:

$\displaystyle S_3$ is a group of order $\displaystyle 6,$ it contains an element of order $\displaystyle 2$ and another one of order $\displaystyle 3,$ which of course generate subgroups of orders $\displaystyle 2$ and $\displaystyle 3.$
But $\displaystyle S_3$ is not cyclic...

The hypothesis "$\displaystyle N_1,N_2$ normal" allows you to prove $\displaystyle \text{order}(ab)=pq,$ for instance, consider the commutator $\displaystyle [a,b]=aba^{-1}b^{-1}$ and prove it is the identity element, that will mean $\displaystyle a,b$ commute and then you will be able to conclude $\displaystyle \text{order}(ab)=pq.$

3. Originally Posted by clic-clac
If you've never used the fact that $\displaystyle N_1$ and $\displaystyle N_2$ were normal subgroups:

$\displaystyle S_3$ is a group of order $\displaystyle 6,$ it contains an element of order $\displaystyle 2$ and another one of order $\displaystyle 3,$ which of course generate subgroups of orders $\displaystyle 2$ and $\displaystyle 3.$
But $\displaystyle S_3$ is not cyclic...

The hypothesis "$\displaystyle N_1,N_2$ normal" allows you to prove $\displaystyle \text{order}(ab)=pq,$ for instance, consider the commutator $\displaystyle [a,b]=aba^{-1}b^{-1}$ and prove it is the identity element, that will mean $\displaystyle a,b$ commute and then you will be able to conclude $\displaystyle \text{order}(ab)=pq.$
Absolutely. Thanks very much. I missed the fact that to prove order(ab)=pq, I am implicitly using the fact that the sub-groups are normal.
Thanks again !!

4. Originally Posted by aman_cc
This is Q#9, Section 2.9, pg 75 from Herstien
Q: order(G) = pq, where p and q are distinct primes. G has a normal subgroup of order p and a normal subgroup of order q. Prove G is cyclic.

My attempt:

Let N1 be the subgroup of order p
Let N2 be the subgroup of order q

Clearly, N1 and N2 are cyclic (as for every group with prime order). Let a, b be generator of N1 and N2 respectively.

Consider c=ab

I can show that order(c)=pq. And, hence G is cyclic (with c as generator)

Help I need:
1. Is my attempt correct?
2. I never used the fact that N1 and N2 are normal (something mentioned in the question) - So am I missing / mis-understanding something?

Thanks
Of course you used the fact that both sbgps. are normal (otherwise the claim is false: there are non-cyclic, and even non-abelian groups of order pq, when one of the primes divides (the other one minus one)) ,but you didn't pay attention: how did you prove ord(c)=pq??

Tonio