Order of Group. Direct Product of Cyclic Group

Let G = $\displaystyle {(\mathbb{Z}/7\mathbb{Z})^*} \times {{(\mathbb{Z}/11\mathbb{Z})^*} $. How many elements are there in the group G? Also show that G is not cyclic.

I know that no. of elements in $\displaystyle {(\mathbb{Z}/7\mathbb{Z})^*}$ = 6 and no. of elements in $\displaystyle {(\mathbb{Z}/11\mathbb{Z})^*}$ = 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 $\displaystyle g \in G$, $\displaystyle g=r^i$, where r is the generator of the cyclic group. But how do i show that such a generator does not exist?

Thanks.

Re: Order of Group. Direct Product of Cyclic Group

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 $\displaystyle (a,b)^d = (a^d,b^d)$

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Re: Order of Group. Direct Product of Cyclic Group

Re: Order of Group. Direct Product of Cyclic Group

Quote:

Originally Posted by

**interneti**

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?

Re: Order of Group. Direct Product of Cyclic Group

Quote:

Originally Posted by

**Deveno** 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 $\displaystyle (a,b)^d = (a^d,b^d)$

So I take $\displaystyle a \in (\mathbb{Z}/7\mathbb{Z})$ and $\displaystyle b \in (\mathbb{Z}/11\mathbb{Z})$ and d = 60. So $\displaystyle (a,b)^{60} = (a^{60}, b^{60})$. I am not very sure about the exact working but a rough idea is that $\displaystyle (a^{60}, b^{60})$ will never be equal to $\displaystyle (1,1)$. Is the reason due to the fact that 7 and 11 are relatively prime?

Re: Order of Group. Direct Product of Cyclic Group

no $\displaystyle (a,b)^{60} = (a^{60}, b^{60})$ 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.

Re: Order of Group. Direct Product of Cyclic Group

Quote:

Originally Posted by

**Deveno** no $\displaystyle (a,b)^{60} = (a^{60}, b^{60})$ 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.

If I take d = lcm(6,10) = 30, then $\displaystyle (a,b)^{30} = (a^{30}, b^{30})$ will always be (1,1). Hence, there is no element of order 60 and thus the group is not cyclic. Is that correct?

Re: Order of Group. Direct Product of Cyclic Group

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

Re: Order of Group. Direct Product of Cyclic Group

Quote:

Originally Posted by

**H12504106** If I take d = lcm(6,10) = 30, then $\displaystyle (a,b)^{30} = (a^{30}, b^{30})$ will always be (1,1). Hence, there is no element of order 60 and thus the group is not cyclic. Is that correct?

that's the idea, now prove it.

Re: Order of Group. Direct Product of Cyclic Group

Just a remark, which should be apparent from [b]Deveno[/tex] and your discusussion, that you can prove in general that for finite groups $\displaystyle A,B$ then $\displaystyle A\times B$ is cyclic if and only if $\displaystyle A,B$ are cyclic and $\displaystyle (|A|,|B|)=1$.