Recent content by lola1990

Homework Statement Homework Equations The Attempt at a Solution

Wait, so (a,b) isn't (0,0), it isn't (x,y), and it isn't (2x,2y). I think all others are fine, so that there are 6 possibilites for (a,b). So there are 48 bases, and |Aut(P)|=48. But, 7 doesn't divide 48, so that is a contradiction. Is that right?

Right, so there are 8 possibilities for the first vector (x,y). We know that the second vector (a,b) can't be a multiple of the first or 0. I am stuck on how to find out how many possibilities there are for the second vector.

Also, I'm not sure how to find out how many bases of Z3x Z3 there are. Clearly, there are 9 possibilities for choosing the first vector (x,y). How can we then choose the second vector?

oh ok, got it. So then we have that a bijective linear map from Z3xZ3 to Z3xZ3 is defined by choosing a basis of Z3xZ3. I'm confused about this point- why is that true?

No, I do see why a bijective linear map is an automorphism. I know that the automorphism induces a linear map from Z3 x Z3 to Z3 x Z3, but do I know that it is bijective? Also, does a 2x2 matrix define a bijective linear map or just a linear map?

Also, I'm not sure how to find the number of bases of Z3 x Z3. It would have to consist of 2 linearly independent elements of Z3 x Z3.

ooh... so ax=x+x+..+x (a times), s0 f(x,y)+f(x,y)...+f(x,y) (a times)=f((x,y)+..+(x,y) (a times))=f(a(x,y)). Is that right? So then, af(x,y)+bf(z,w)=f(a(x,y))+f(b(z,w))=f(a(x,y)+b(z,w)). Thus, if f is an automorphism, it is a linear map in Z3xZ3. Do I also need to show that a linear map in Z3x...

Oh, sorry a 2x2 invertible matrix

Ok, so just to recap, we want to find the number of automorphisms from Z3 x Z3 to Z3 x Z3. We want to show that f(x,y) is such an automorphism iff it is a linear transformation of Z3 x Z3 (represented by a 3x3 invertible matrix with entries from Z3). Ok, so if f(x,y) is an automorphism, and...

I guess if P=Z3 x Z3, then we'd have to define f((x,y))=(g(x), h(y)). Then, f is Z3 linear iff g and h are Z3 linear. Is this right? Also, I am talking about group isomorphisms, so I'm not sure how to deal with the constants.

Ok, so suppose f:P -> P is an automorphism. I guess I'm confused about how to go about showing that something is linear in Z3 if the function is defined on P, which has order 9.

Just want to check that 0 is not an element of Z/9Z, right? Because then the order would be 7 and the proof would go out the window. What do you mean by Z3 linear? I guess if f is an automorphism, then f(x+y)=f(x)f(y). Not quite sure if this is what you mean...

Oh, I do have a theorem that the automorphism group of the cyclic group of order n is isomorphic to (Z/nZ)^x. So, since the co-prime elements of Z/9Z are 1,2,4,5,7,8 the automorphism group of Z9 has order 6. Likewise, the automorphism group pf Z3 has order 3. Does this mean that the automorphism...

Oh wait, I messed up the first part. The two possibilities for the kernel are 1 and 7, and if it is 1 then |C/kernel|=7. Since the number of permutations of P is 9!, I'm unclear on why this leads to contradiction. Is there way to know |Aut(P)|<9! ?