yes, there are 60 elements, as the underlying set is the cartesian product.
naturally, to prove it is not cyclic, you must show that no element is of order 60.
use the fact that
Let G = . How many elements are there in the group G? Also show that G is not cyclic.
I know that no. of elements in = 6 and no. of elements in = 10. So can i conclude that the total number of elements is then 60?
As for the cyclic part, i dont have any idea where to start from. I know that a group is cyclic if for every element , , where r is the generator of the cyclic group. But how do i show that such a generator does not exist?
Thanks.
interneti, the group in question is the direct product of the groups of units of Z7 and Z11, U(7)xU(11).
for example, U(7) = {1,2,3,4,5,6} and U(11) = {1,2,3,4,5,6,7,8,9,10}.
in Z7xZ11, (1,1) has order 77, but in Z7*xZ11*, (1,1) has order 1, it is the identity element.
in Z7*, 3 is a generator: <3> = {3,2,6,4,5,1}. this gives an isomorphism of (Z7*,*) with (Z6,+):
3^k <---> k
you can show Z11* is likewise cyclic, and isomorhpic to (Z10,+).
can you see why Z6 x Z10 is not cyclic?
no will always be (1,1) as a direct consequence of Lagrange.
it is still the case that the order of every group element will be a divisor of 60, the challenge is to show that some number LESS than 60 will work for any group element.
THANK YOU DEVENO You have right.
Look at this answer please..
The direct product of groups G= (Z/7Z)*(Z/11Z) the cadrinaly of G |G| is infinite,for example
take ab from G(a from Z/7Z and b to Z/11Z are different from 1 ) ab*ab*ab*...ab=abab....ab is different from 1.because
a and b are in different groups,
So the cardinaly of G must be infinite.G is not cyclic because if we take ab we have that for every z from Z
never a^z is b, so b is not at <a>. from this G is not cyclic.
THANK YOU