00:01
In this ocean, we need to prove that the group zm cross zm is a cyclic group if and only is gcd of m and n is 1.
00:22
That is if and only if m and n are open.
00:25
Now the proof is given by, let us consider if the gcd of mnn is 1, then zm cross zn is a cyclic group.
00:48
Now let zm is generated by a for some a belonging to z of n.
01:00
And zn is generated by b for some b belonging to z of n.
01:07
Now let us suppose that the gcd of mn is generated by v for some b belonging to z of n.
01:17
1 and the gcdo a and b is g.
01:27
Then g raised to par mn is given by a times b raised to par mn which is further equal to a rest to par mn comma b raised to par mn which gives e z of n and ez of n.
01:51
That is identity of zm and identity of z -a.
01:56
Now for the suppose order of g is t, then a comma b raised part t is given by ezm, ezm.
02:19
Comparing the terms we get a is equal to ezm and b is equal to ez of l.
02:30
This implies that m divides t and n also divides t.
02:39
Because a rest to par t is and b rest to party have the furling value.
02:45
Now, since we have supposed that the gcd of mn is 1, therefore their product will also divide the value t.
02:55
This implies that mn is the smallest integer such that g -race to par m -n is equal to e.
03:17
Thus order of g is mn...