Is this statement true?
-statement-
Let $A \simeq A_1 \times A_2$ with $B \lhd A$ (A is a either a ring or a group)
then, $A/B \simeq (A_1/B) \times (A_2 / B)$
ex) When the ring case
Let the Ring $R$ s.t. $R \simeq R_1 \times R_2$ with its ideal I ($I \lhd R$)
Then the $R/I \simeq (R_1/I) \times (R_2 / I)$
On the other case when the group
For the group $G$ s.t. $G \simeq G_1 \times G_2$ with its normal subgroup N ($N \lhd G$)
Then the $G/N \simeq (G_1/N) \times (G_2 / N)$
I tried to prove this by taking the mappings like the below.
When the ring case $(R, +, \bullet)$
$\phi_R$ : $R \to (R_1/I) \times (R_2 / I)$ by $\phi_R(r) = (r_1 + I, r_2 + I)$ for $r (\in R) \simeq r_1 \times r_2$
With the same method for group, $(G, \bullet)$
$\phi_G$ : $G \to (G_1/N) \times (G_2 / N)$ by $\phi_G(g) = (g_1 \bullet N, g_2 \bullet N)$ for $g (\in G) \simeq g_1 \times g_2$
As far as I knew, kernel of the each mappings are $I$ and $N$. So all we need to do check the well-defined, homomorphism, surjectivity.
Then my conclusion is the statement is true.
But still I don't have any confidence that my knowledge and proof is right or not. What do you think about that?
Any help would be appreciated.
For a quotient group $G/N$ to be defined, $N$ must be a normal subgroup of $G$.
Similarly, for a quotient ring $R/I$ to be defined, $I$ must be an ideal of $R$.
For $A\cong A_1 \times A_2$ with $B \lhd A$, it is not necessarily that $B\lhd A_1$ and $B\lhd A_2$.
For example, take $A=\langle x \rangle \times \langle y \rangle$, $B=A, A_1=\langle x \rangle, A_2=\langle y \rangle$.
In this case, $B\lhd A$ but $B \not\lhd A_1$ and $B\not\lhd A_2$.
So $A_1/B$ and $A_2/B$ are not defined and hence $A/B\cong A_1/B\times A_2/B$ is not true.
However, there is a result that is similar, that is, if $B=B_1\times B_2$ where $B_1\lhd A_1$ and $B_2\lhd A_2$, then $$A/B\cong A_1/B_1\times A_2/B_2$$