Nontrivial subgroups of Zp x Zp

[carrie.lugo]

Am I right in saying that Zp x Zp (the product of the integers mod p, a prime, with itself) has only 3 nontrivial subgroups?

By Lagrange's theorem, we know any nontrivial subgroup would have order p, since the order of Zp x Zp is p^2.

So I am looking at it like this:
Picture all elements of Zp x Zp filling a p x p matrix, with (0,0) in the top left and (p-1, p-1) in the bottom right. The only nontrivial subgroups I see that you can form are by taking all the elements in the first column, all the elements in the first row, and all the elements on the main diagonal.

Is this correct?

 

[DonAntonio]

Am I right in saying that Zp x Zp (the product of the integers mod p, a prime, with itself) has only 3 nontrivial subgroups?

By Lagrange's theorem, we know any nontrivial subgroup would have order p, since the order of Zp x Zp is p^2.

So I am looking at it like this:
Picture all elements of Zp x Zp filling a p x p matrix, with (0,0) in the top left and (p-1, p-1) in the bottom right. The only nontrivial subgroups I see that you can form are by taking all the elements in the first column, all the elements in the first row, and all the elements on the main diagonal.

Is this correct?

No, it's not. The number is $\,\frac{p^2-1}{p-1}=p+1\,$ , which is three iff $\,p=2\,$ .

The trick? Since $\,\Bbb Z_p\times \Bbb Z_p\,$ is a vector space over the field $\,\Bbb Z_p\,$ , of dimension $\,2\,$ , what

you want is all the 1-dimensional subspaces, so just count how many possible basis are there that yield different 1-dimensional subspaces...!

DonAntonio

 

[carrie.lugo]

Thanks, DonAntonio! I see it now. :)