how to compute product in the given ring

Question

$(-3,5)(2,-4)$ in $\mathbb{Z}_4 \times \mathbb{Z}_{11}$

I get the answer as $(2,3)$. The answer given in the solution is $(2,2)$. Can someone explain how?

Answer 1

In the second component, $5\times(-4)=-20\equiv2\bmod11$.

Answer 2

From the definition of how we define the operation of, $$G \times G'$$ Suppose $$(a,b) \in G,(a',b') \in G'$$ then $$G \times G'=\{(aa',bb'), \ for \ \ a,b \in G \ \& \ a',b' \in G'\}$$.

$\textbf{Additional}$: Just compute, $(-3)(2) = 2$ in $\mathbb{Z_4}$, $(5)(-4) = 2$ in $\mathbb{Z_{11}}$.

Answer 3

It looks like OP is good to go, but just for completeness:

Suppose $R_1$ and $R_2$ are rings. Then the direct product of two rings, $R_1 \times R_2$ is defined as follows:

$(a, b) \in R_1 \times R_2 \iff a \in R_1$ and $b \in R_2$. And further, for any $(a, b)$ and $(c, d) \in R_1 \times R_2$:

$$(a, b)+(c, d) = (a+c, b+d)$$ $$(a, b)\cdot (c, d) = (a\cdot c, b \cdot d)$$

Where $a+c$ and $a \cdot c$ are carried out as they would be in $R_1$, and likewise for $b$ and $d$ in $R_2$.