Show that $(G:\sum_{\lambda\in\Lambda} N_\lambda )={\bigcap}_{\lambda\in\Lambda}(G :N_\lambda)$

Question

(N and G are submodules of M ,R-module and $(N_\lambda)_\Lambda$ is a family of submodules of M ,R-module) Here's what ı know $N=\langle J\rangle$ then $(G:J)=(G:N)$ and $\sum_{\lambda\in\Lambda} G_\lambda=\langle {\cup}_{\lambda\in\Lambda} G_\lambda\rangle$

I know ı need to show but $(G:$$\sum_{\lambda\in\Lambda} N_\lambda $$)\subset{\bigcap}_{\lambda\in\Lambda}(G :N_\lambda)$ and $(G:$$\sum_{\lambda\in\Lambda} N_\lambda $$)\supset{\bigcap}_{\lambda\in\Lambda}(G :N_\lambda)$

first $x\in (G:$$\sum_{\lambda\in\Lambda} N_\lambda $$)\Rightarrow x\in R,xn\in G,\forall n\in {\cup}_{\lambda\in\Lambda} N_\lambda \\ \Rightarrow x\in R,xn\in G,\forall n\in N_\lambda,\exists\lambda\in\Lambda$

second $x\in {\cap}_{\lambda\in\Lambda} (G:N_\lambda)\Rightarrow x\in(G:N_\lambda)\forall\lambda\Lambda \\ \Rightarrow x\in R,xn\in G,\forall n\in N_\lambda,\forall\in\Lambda $

How can ı manipulete from here?.Thanks for all advices

Answer 1

You almost have it, it is just a matter of being careful with quantifiers.

$$\begin{align} x\in\left(G:\sum_{\lambda\in\Lambda}N_\lambda\right)&\Leftrightarrow x\in\left(G:\bigcup_{\lambda\in\Lambda}N_\lambda\right)\\ &\Leftrightarrow \forall n\in\bigcup_{\lambda\in\Lambda}N_\lambda,\ xn\in G\\ &\Leftrightarrow\forall\lambda\in\Lambda,\ \forall n\in N_\lambda,\ xn\in G\\ &\Leftrightarrow\forall\lambda\in\Lambda,\ x\in(G:N_\lambda)\\ &\Leftrightarrow x\in\bigcap_{\lambda\in \Lambda}(G:N_\lambda) \end{align}$$