Non-cyclic group order 49?

[ArcanaNoir]

Give an example of a non-cyclic group of order 49.
I have a couple problems like this.
I think there are only two isomorphism class for order 49, Z_49 and Z_7 x Z_7.
since Z_49 is cyclic, I'm guessing Z_7 x Z_7 is non-cyclic. Right?

A related question is "is there a non-cyclic group of order 39? there are two classes of isomorphism here, Z_39 which is cyclic, and Z_13 x Z_3 which I don't know if is cyclic. how can I tell?

 

[I like Serena]

Hi Arcana! :)

What is the definition of a cyclic group?

Consider an element of Z7xZ7, say (0,1). What is its order?
What about (1,1)?
Can the group be generated by 1 element?

Same questions for Z13xZ3...

 

[ArcanaNoir]

The order of (0,1) is 7. The order of (1,1) is 7. I'm not sure if the group can be generated by one element, but it seems doubtful. Is there a rule to answer that for groups that are direct products?

 

[I like Serena]

We're getting there. I think that is basically the point of the exercise.

Note that for any element (a,b) in Z7xZ7 holds that a has order 7 and that b has order 7.
So any element (a,b) also has order 7.
Can you find an element that has order 49? Which is what is required for the group to be cyclic.
(Did you look up the definition of a cyclic group?)

How does that work out for Z13xZ3?

 

[ArcanaNoir]

I already know the definition of a cyclic group.

I see I forgot that a finite group with prime order is generated by each of it's non identity elements, thus each non identity element has the order p=the order of the group. So this means any element of Z_7 x Z_7 has order 7, thus no element has order 49 thus it is non cyclic.

As for Z_13 x Z_3, (a,b), all a have order 13 and all b have order 3. I'll have to work that out on paper but I think I know what to look for now.

Thanks!

 

[I like Serena]

Good!