# Let $N_1, . . . , N_n$ be normal subgroups of $G$, consider $G/N_1\times ··· \times G/N_n$

I’m reading Hans Kurzweil ‘s “The Theory of Finite Groups”, where it says

1.6.4 Let $$N_1, . . . , N_n$$ be normal subgroups of $$G$$. Then the mapping $$α: G→G/N_1\times ··· \times G/N_n$$ given by $$g \mapsto (gN_1,...,gN_n)$$ is a homomorphism with $$\operatorname{Ker}α = \cap_i N_i$$. In particular, $$G/\cap_i N_i$$ is isomorphic to a subgroup of $$G/N_1 \times ··· \times G/N_n$$.

I’m confused here: can we write $$G/N_1\times \cdots \times G/N_n$$ ? To write a product of groups as this, it’s required that each $$G/N_i$$ has only $$e$$ as common element.

What if $$G=C_2 \times C_3 \times C_5 \times C_7$$

$$N_1=C_2 \times C_3$$

$$N_2=C_2 \times C_5$$

$$N_3=C_2 \times C_7$$

, shouldn’t $$G/N_1 \cong C_3 \times C_5$$

$$G/N_2 \cong C_2 \times C_7$$

$$G/N_3 \cong C_5 \times C_7$$

, and they have common elements besides $$e$$?

• I believe this is the Chinese Reminder theorem for groups. Jan 9, 2019 at 13:21
• @IAmNoOne but there’s no solution formula? Jan 9, 2019 at 23:55

I’m confused here: can we write $$G/N_1\times ··· \times G/N_n$$ ? To write a product of groups as this, it’s required that each $$G/N_i$$ has only $$e$$ as common element.
Note that $$G/N_i$$ is not a subgroup of $$G$$. So here we are not considering the internal direct product, which requires the condition you mentioned above to be a group. Here $$G/N_1\times ··· \times G/N_n$$ represents the external direct product , which is a group under the componentwise operation.
Given two groups $$G_1$$ and $$G_2$$, you can form their direct product: $$G_1×G_2$$, to be $$\{(g_1,g_2)\mid g_1\in G_1,g_2\in G_2\}$$, with the group operation defined as $$(g_1,g_2)+(h_1,h_2)=(g_1+h_1,g_2+h_2)$$.
As an example, consider the Klein four group, $$V_4=C_2×C_2$$.