Cyclic group question - from Herstien

aman_cc · #1 $Old$ November 4th, 2009, 08:53 AM

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

clic-clac · #2 $Old$ November 4th, 2009, 10:49 AM

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

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

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

aman_cc · #3 $Old$ November 4th, 2009, 11:05 AM

Quote:

Originally Posted by clic-clac $View Post$

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

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

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

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 !!

tonio · #4 $Old$ November 4th, 2009, 11:23 AM

Quote:

Originally Posted by aman_cc $View Post$

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