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Suppose $X=Y_1\cup Y_2\,\cup \dots \cup\, Y_n$ where the $Y_i$s are disjoint. In the notes provided by the professor he mentioned that the symmetric group $\operatorname{Sym}(Y_i)$ can be viewed as a subgroup of $\operatorname{Sym}(X)$ by extending the functions from $\operatorname{Sym}(Y_i)$.

Precisely: if $f \in \operatorname{Sym}(Y_i)$ then let $g(x)=f(x),\, x\in \operatorname{Sym}(Y_i)$ else $g(x)=x$.

Then the external direct product $\operatorname{Sym}(Y_1)\times \operatorname{Sym}(Y_2)\times \dots\times \operatorname{Sym}(Y_n)$ can be embedded into $\operatorname{Sym}(X)$ by: $(f_1,f_2,..,f_n)\mapsto f_1\circ f_2\circ \dots \circ f_n$ where the $f_i$s are extended on the right.

He proceeds to say if $\operatorname{Sym}(Y_i)$ is viewed as a subgroup of $\operatorname{Sym}(X)$ then the product of these subgroups is an internal direct product. This is the statement which I do not get. Don't these subgroups have to be normal before the internal direct product can be considered. However, this need not be the case here.

Could someone please explain what he meant by internal direct product.

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$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\DeclareMathOperator{\Sym}{Sym}$The subgroups $G_{i} = \Sym(Y_{i})$ have only to be normal in the subgroup they generate.

Now clearly for $i \ne j$ you have that $G_{i}$ and $G_{j}$ commute elementwise, as their supports $Y_{i}, Y_{j}$ are disjoint.

Hence each $G_{i}$ is normal in the subgroup $\Span{G_{i} : i = 1, \dots, n}$, and the latter coincides with the product $G_{1} \cdots G_{n}$.

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  • $\begingroup$ Hello just to clarify $<G_{i} : i = 1, \dots, n>$ is the same as $G_1..G_n$ because the subgroups commute pairwise right where $G_1..G_n=\{g_1..g_n|g_i\in G_i\}$ $\endgroup$ – Jhon Doe Jul 17 at 16:15

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