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What I'm asking comes from Bosch, Algebra, the first chapter on elementary group theory.

1) Let $X$ be a set and $G$ a group. Then the set $G^X$ of maps from $X$ to $G$ is a group in a natural way, that is for every $f,g\in G^X$, set $(f\cdot g)(x):=f(x)\cdot g(x)$.

2) Let $X$ be a set of indexes and $\left(G_x\right)_{x\in X}$ a collection of groups. Then the product set $\prod_{x\in X}G_x$ becomes a group if we define the product component-wise, that is $(g_x)_{x\in X}\cdot (h_x)_{x\in X}:=(g_x\cdot h_x)_{x\in X}$.

Now suppose that the groups $G_x$ are copies of the same group $G$. Then, using notations in 1), we have $$\displaystyle\prod_{x\in X}G_x=G^X$$ Well, I can't understand this equation. I mean, an element in the lefthandside is a collection $(g_x)_{x\in X}$ of elements of $G$, whereas on the righthandside we have functions taking values in $G$. How can the two objects be the same?

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2 Answers 2

up vote 3 down vote accepted

Strictly speaking, what we have is only an isomorphism of groups, given by $(g_x)_{x\in X}\mapsto (x\mapsto g_x)$ with inverse map $f\mapsto(f(x))_{x\in X}$. But this isomorphism is so canonical, the difference so shallow that actually calling it equality seems to be one of the lesser sins. For example, would you really object to the equation $(A\times B)\times C=A\times (B\times C)$ for cartesian products of sets?

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Well, the indexing is simply a function on $X,$ yes? That is, $g=(g_x)_{x\in X},$ where $g_x\in G_x=G$ for all $x\in X,$ so in fact, $g$ is in effect a function $X\to G$ given by $g(x)=g_x$.

More generally, if we have a collection of sets $\{A_x\}_{x\in X},$ then $\prod\limits_{x\in X}A_x$ can be thought of as the set of all functions $f:X\to\bigcup\limits_{x\in X}A_x$ such that $f(x)\in A_x$ for all $x\in X$.

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