11:13. Here, are we saying that "Zm X Zn has to have order of mn or multiple of mn, but since the (a,b) has to be a generator as was shown above, |(a,b)| has to be greater or equal to mn. Thus, it's a contraction to (a,b)
The set Zm has m elements, the set Zn has n. So Zm X Zn has mn elements. If Zm X Zn is cyclic, then there exists an element g in Zm X Zn with order mn. But assuming gcd(m,n) > 1, he showed that every element in Zm X Zn has order smaller than mn. A contradiction to Zm X Zn = Zmn (because the right side is cyclic), so gcd(m,n) = 1.
I can understand this, until the
okay? Sick. hahaha love it !
11:13. Here, are we saying that "Zm X Zn has to have order of mn or multiple of mn, but since the (a,b) has to be a generator as was shown above, |(a,b)| has to be greater or equal to mn. Thus, it's a contraction to (a,b)
The set Zm has m elements, the set Zn has n. So Zm X Zn has mn elements. If Zm X Zn is cyclic, then there exists an element g in Zm X Zn with order mn. But assuming gcd(m,n) > 1, he showed that every element in Zm X Zn has order smaller than mn. A contradiction to Zm X Zn = Zmn (because the right side is cyclic), so gcd(m,n) = 1.
micheal penn the GOAT
Is that property "Cross product of two cyclic group is cyclic iff gcd(m,n)=1" true for more than two cyclic groups ?
I think it requires pairwise coprimeness. So gcd(a, b) = 1, gcd(a, c) = 1 and gcd(b, c) = 1. But yes if that works I think you're right.
good lecture sir from Pakistan
Can you clean the blackboard first?
Otherwise, great info! 😊