The following is taken from Derek Robinson's abstract algebra text
Let $\{G_\lambda, \lambda\in \wedge \}$ be a set of groups a restricted choice function for the set is a mapping $f:\wedge \rightarrow \cup_{\lambda \in \wedge} G_\lambda$ such that $f(\lambda)\in G_\lambda$ and $f(\mu)=1_{G_\mu}$ for all but a finite number of $\mu.$ Let $G$ be the set of all restricted choice functions and define a binary operation on $G$ by $fg(\lambda)=f(\lambda)g(\lambda).$ Then $G$ is called the restricted external direct product $\text{Dr}_{\lambda\in \wedge}G_\lambda$ of the groups $G_\lambda.$
i) For $\lambda\in \wedge$ define $f_\lambda:G_\lambda \rightarrow G$ as follows: if $x\in G_\lambda,$ then $f_\lambda(x)$ sends $\lambda$ to $x$ and $\mu$ to $1_{G_\lambda}$ for $\mu \neq \lambda.$ Prove that $\bar{G_\lambda}=\{f_\lambda(x)\mid x\in G_\lambda\}$ is a normal subgroup of $G$ and that $\bar{G_\lambda}=G_\lambda.$
Question: I don't understand the notation for $f_\lambda(x),$ it states $f_\lambda$ sends "$\lambda$ to $x$ and $\mu$ to $1_{G_\lambda}$ for $\mu \neq \lambda.$". Does it mean the following: $f_\lambda(x)=x$ if $\mu=\lambda$ and $f_\lambda(x)=1_{G_\mu}$ if $\mu\neq\lambda$? Basically if $x\in G_\lambda$, then does $x$ take the form: $x=(x_1,x_2,\ldots, x_\mu, x_\lambda, x_\lambda,\ldots)=(x_1, x_2, \ldots, x_\mu,1,1,\ldots).$
Thank you in advance
You have misunderstood the notation. $f_\lambda(x)$ is an element of $G$, so $f_\lambda(x)$ is a restricted choice function; that is, $f_\lambda(x)$ is itself a function from $\Lambda$ to $\cup_{\lambda \in \Lambda}G_\lambda$.
So $f_\lambda(x)$ is defined by its action on $\Lambda$ which is $f_\lambda(x)(\lambda) = x$ and $f_\lambda(x)(\mu) = 1_{G_\mu}$ for $\mu \ne \lambda$.
So intuitively you could think of the element $f_\lambda(x)$ of $G$ as $(\ldots,1,1,1,x,1,1,1,\ldots)$ with the $x$ being in position $\lambda$ (but that is not intended to imply that $\Lambda$ is countable).
Note that it should be $\bar{G}_\lambda \cong G_\lambda$ rather than $\bar{G}_\lambda = G_\lambda$