Consider the product of cyclic groups $G = C_4 \times C_2$. We see that $H = C_2 \times \{ 0 \}$ is normal in $G$ since $G$ is abelian. The cosets are $H = \{(0, 0), (1, 0)\}$, $(0, 1)H = \{(0,1), (1, 1)\}$, $(2, 0)H = \{(2, 0), (3, 0)\}$, and $(2, 1)H = \{(2, 1), (3, 1) \}$. All three nonidentity cosets have order $2$, so $G/H \cong C_2 \times C_2$. On the other hand, $H \cong J = \{0\} \times C_2 \triangleleft G$, yet $G/J = \{ J, (1, 0)J, (2, 0)J, (3, 0)J \} \cong C_4$. (There's some notation abuse: the $C_2$ and $\{0\}$ in $J = \{0\} \times C_2$ are different from the $C_2$ and $\{0\}$ in $H = C_2 \times \{0\}$.) Therefore, two normal subgroups of $G$ may be isomorphic despite having nonisomorphic quotients.
It's tempting to say the following:
$$(C_4 \times C_2)/(C_2 \times \{0\}) \cong C_4/C_2 \times C_2 /\{0\} $$ and $$(C_4 \times C_2)/(\{0\} \times C_2) \cong C_4/\{0\} \times C_2/C_2 \text.$$ But does this trick actually work in general? In other words, is the proposition below true?
Let $G$ be an external direct product $G_1 \times G_2 \times \dots \times G_s$ for some groups $G_1, G_2, \dots, G_s$. Let $H = H_1 \times H_2 \times \dots \times H_s$ be a normal subgroup of $G$ with $H_i \triangleleft G_i$ for each $i = 1, 2, \dots, s$. Then $$ G/H = (G_1 \times G_2 \times \dots \times G_s)/(H_1 \times H_2 \times \dots \times H_s) \cong (G_1/H_1) \times (G_2/H_2) \times \dots \times (G_s/H_s) \text.$$
This result would be like ordinary division of real numbers: $(a \cdot b)/(c \cdot d) = (a/c) \cdot (b/d)$.
Best Answer
There are natural homomorphisms $G_i\to G_i/H_i$, which gives rise to the homomorphism $f:\prod G_i\to \prod(G_i/H_i)$. This is clearly surjective, since each $G_i\to G_i/H_i$ is surjective. We wish to know the kernel.
Let $(g_i)\in\prod G_i$ be such that $f((g_i))=0$. This is equivalent to, for each $i$, $g_i\in\ker(G_i\to G_i/H_i)=H_i$. This means that $\ker f=\prod H_i$.
Thus, by the isomorphism theorem, $\prod G_i/\prod H_i\cong\prod(G_i/H_i)$.