Math Help Forum - View Single Post - Cyclic group question

aman_cc · #3 November 4th, 2009, 10:05 AM

Quote:

Originally Posted by clic-clac

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 !!