Question about homomorphism mapping and one-to-one.

[Steven G]

I read a proof from Copilot that basically used the following to show that a mapping was 1-1.
Suppose there are two sets A and B with an equivalence relation, ~, on A. There is a homomorphic mapping T from A to B. In order to show that T is one-to-one the proof basically said suppose that T(x) = T(y), conclude that x~y, and then state T is 1-1
Is that enough to show that T is one-to-one? I really question this unless we think that everything in the same class is the same.
What do you think?

 

[Dr.Peterson]

Is that enough to show that T is one-to-one? I really question this unless we think that everything in the same class is the same.

It doesn't seem to mention an equivalence relation on B, so by default that would be "=". Does that help anything?

 

[Steven G]

It doesn't seem to mention an equivalence relation on B, so by default that would be "=". Does that help anything?

Good point. ~ was not equality.

OK, here is the entire problem:
Show that A is isomorphic to Z/32Z if A= Z^3 / MZ^3, where M=8 0 1 21 0 1 8 4 1 and elements in Z^3 = Z × Z × Z are represented by column vectors. The matrix M will is a 3x3 matrix

The proof from Copilot went like this:
If you multiply M by (x y z)^T the fist entry will be 8x+z

Now we define ~ by (x, y, z)~(x' y' z') iff (x-x', y-y', z-z') is in MZ^3

Define a mapping T from A to Z/32Z by T(x y z) = (8x+z) mod 32.
They showed that T is a homomorphism.
Then they wanted to show that T is 1-1.
They went on to suppose that T(x y z) = T(x' y' z')
They left out the details but concluded that (x y z) ~ (x' y' z') and then said so then T is 1-1
Is that last fact valid?

 

[blamocur]

This does not look right to me: $T((4,0,0)) = 0$, but $(4,0,0) \neq 0 \mod M\mathbb Z^3$ because $M^{-1} ((4,0,0)) = \frac{1}{52}(-16, -52, 336) \notin \mathbb Z^3$

I suspect that it is not true that $\mathbb Z^3/M\mathbb Z^3$ is isomorphic to $\mathbb Z/ 32 \mathbb Z$ either because $\det M = 52 \neq 32$.

 

[Steven G]

Yo

This does not look right to me: $T((4,0,0)) = 0$, but $(4,0,0) \neq 0 \mod M\mathbb Z^3$ because $M^{-1} ((4,0,0)) = \frac{1}{52}(-16, -52, 336) \notin \mathbb Z^3$

I suspect that it is not true that $\mathbb Z^3/M\mathbb Z^3$ is isomorphic to $\mathbb Z/ 32 \mathbb Z$ either because $\det M = 52 \neq 32$.

You are correct that the matrix must be wrong. If you replace the 1st row of the matrix with 3 8 1 it should work. I still have my same question.

 

[blamocur]

I wouldn't expect too much from Copilot. In the past someone posted a "proof" by ChatGPT, and it was complete garbage.
I agree that modified $M$ looks better:
$$M = \left(\begin{array}{lll} 3&8&1\\ 21 & 0 & 1 \\ 8 & 4 & 1 \\ \end{array} \right)$$Moreover, if you define $T: \mathbb Z^3 \rightarrow \mathbb Z$ as $T(\mathbf x) = \mathbf u \cdot \mathbf x$ where $\mathbf u = (13,5,-18)$ you can show that $T(M\mathbb Z^3) \in 32\mathbb Z$, which means that $T$ defines a homomorphism $T^\prime : \mathbb Z^3 / M \mathbb Z^3 \rightarrow \mathbb Z_{32}$ (where $\mathbb Z_{32} = \mathbb Z/32\mathbb Z$).

It is pretty easy to prove surjectivity of $T^\prime$ (by showing, for example that $T((0,11,3)) = 1$), but at this point I don't know how to prove the injectivity.

 

[blamocur]

at this point I don't know how to prove the injectivity.

Actually I do now. To prove injectivity of $T^\prime$ we have to prove that
$$\mathbf u\cdot \mathbf x \in 32\mathbf Z \Longrightarrow \mathbf x\in M \mathbb Z^3 \Longleftrightarrow \exists \mathbf w \in \mathbb Z^3: \mathbf x = M \mathbf w$$
and since $M$ is invertible we know that we can find such $\mathbf w =M^{-1}\mathbf x \in \mathbb R^3$, so all we have to show is that $\mathbf w$ has integer coefficients, i.e.
$$\mathbf w = M^{-1} \mathbf x \in \mathbb Z^3$$First we notice that
$$32 M^{-1} = \left( \begin{array}{rrr} 4 & 4 & -8 \\ 13 & 5 & -18 \\ -84 & -52 & 168 \end{array} \right)$$We also define a useful invertible matrix with integer coefficients:
$$A = \left( \begin{array}{rrr} 1 & 12 & 0 \\ 0 & 1 & 0 \\ -3 & 0 & 1 \end{array} \right) \;\;\;\;\;\;\;\;\;\; A^{-1} = \left( \begin{array}{rrr} 1 & -12 & 0 \\ 0 & 1 & 0 \\ 3 & -36 & 1 \end{array} \right)$$$$A \star 32 \star M^{-1} = \left( \begin{array}{rrr} 160 & 64 & -224 \\ 13 & 5 & -18 \\ -96 & -64 & 192 \end{array} \right)$$We can see that $32AM^{-1}\mathbf x \in 32 \mathbb Z^3$, and thus $AM^{-1}\mathbf x \in \mathbb Z^3$, because the top and the bottom row of $32AM^{-1}$ are divisible by 32, and the middle row is equal to $\mathbf u$ (remember that $\mathbf u\cdot \mathbf x \in 32\mathbb Z$). But since $A^{-1}$ has integer coefficients we know that $M^{-1}\mathbf x \in \mathbb Z^3$. Q.E.D.