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I am just reading the book "Algebra" by Hungerford and on one page it says that

if $G_i$ is a family of groups $\forall i\in I$ then $\prod_{i\in I}^{w}G_i\unlhd\prod_{i\in I}G_i$

where $\prod_{i\in I}^{w}G_i$ is the set of all $f\in\prod_{i\in I}G_i$ such that $f(i)=e_i$, the identity in $G_i$, for all but a finite number of $i\in I$

I think this is a very easy proof but I am not sure how to write down nicely:

Let $g\in\prod_{i\in I}G_i$ and $f\in\prod_{i\in I}^{w}G_i$, I want to show that $gfg^{-1}\in\prod_{i\in I}^{w}G_i$

I know that $f(i)=e_i$ => $gfg^{-1}=geg^{-1}=egg^{-1}=e\in\prod_{i\in I}^{w}G_i$ Correct?

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1 Answer 1

up vote 1 down vote accepted

What you have is not quite correct, though it contains the essential idea and can be fixed without too much trouble.

Let $f\in\prod_{i\in I}^wG_i$ and $g\in\prod_{i\in I}G_i$. Let $J=\{i\in I:f(i)\ne e_i\}$; by hypothesis $J$ is finite. For each $i\in I\setminus J$ we have $$\left(gfg^{-1}\right)(i)=g(i)f(i)g^{-1}(i)=g(i)e_ig^{-1}(i)=e_i\;,$$ so

$$\left\{i\in I:\left(gfg^{-1}\right)(i)\ne e_i\right\}\subseteq J$$

and is therefore finite. Thus, $gfg^{-1}\in\prod_{i\in I}^wG_i$, as desired.

By the way, what Hungerford calls the weak direct product is more often called the direct sum.

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