# Quotient of product group is product of quotient groups

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: $$(a \cdot b)/(c \cdot d) = (a/c) \cdot (b/d)$$.

• Yes, this is true. The proof isn’t bad using the fundamental hom theorem. – Randall Oct 16 '20 at 17:56

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, $$\prod G_i/\prod H_i\cong\prod(G_i/H_i)$$.