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First of it all, sorry about that horrible title, if you know how to refine it please be my guest and do so. This question is of the same caliber as $\hom(V,W)$ is canonic isomorph to $\hom(W^*, V^*)$ that I used to ask a long while ago.

Unfortunately this comes with two things. 1st being that I do understand as good as nothing about this problem, 2nd that I need to write a lot of things down in order to even make this post look like a question, although the most honest kind of question I could ask would be "Whaaaaaa?"


Problem: Let $I \neq \emptyset$ be a set and $(G_i, \circ_i)_{i \in I}$ be a family of groups. Let $G:= \times_{i \in I} G_i$ and consider the mapping $\circ : G \times G \to G$ be defined by $$ (x_i)_{i \in I} \circ (y_i)_{ i \in I} := (x_i \circ_i y_i)_{i \in I} $$ Show that $(G, \circ)$ is a group.

Wow, okay. I'd love to understand this problem, but I don't. Let me show to you however that I've done my homeworks and at least know what is expected of me.

My approach: First I want to introduce all the symbols, starting by the most simple ones

  • $(x_i)_{i \in I}$ is a family of elements in $M$. This makes more sense to me when I consider $M^I := \text{map}(I,M)$ the mappings from $I \to M$ where $I$ is the set of indices. For $I= \mathbb{N}$ I would get the regular sequences with values in $M(=\mathbb{R}/\mathbb{C}$)

  • $\times_{i \in I} G_i:= \lbrace x:I \to \bigcup_{i \in I} G_i \mid x_i \in G_i , \forall i \in I\rbrace $ is the direct product of family of sets $(G_i)_{i \in I}$ for $I= \lbrace 1,2 \rbrace$ I get the cartesian product, I don't quite see it to be honest but I can take it as granted.

next for $(G, \circ)$ to be a group I need to verify several things:

  • First the operation $\circ$ must be associative, that is for three elements $x,y,z \in G$ I must show that $x \circ (y \circ z) = (x \circ y) \circ z$. So the elements in $G$ are families $(x_i)_{i \in I}$ with elements in $M$, so I believe I have to show that $$((x_i)_{i \in I} \circ (y_i)_{i \in I}) \circ (z_i)_{i \in I} :\overset{?}= (((x_i)_{i \in I} \circ (y_i)_{i \in I}) \circ_i z_i)_{i \in I}\overset{?}=(x_i \circ_i y_i \circ_i z_i)_{i \in I} \\ = ???? = (x_i)_{i \in I} \circ ((y_i)_{i \in I} \circ (z_i)_{i \in I}))$$ although I have no idea if I am associating my parenthesis correctly here, the expression is just too complicated for me to handle.

  • then I need to show that there is an neutral element in $G$, that means that for all $(x_i)_{i \in I}$ there is an element such that $$(x_i)_{I \in I} \circ e = e \circ (x_i)_{i \in I} = (x_i)_{i \in I} $$ since we're talking about family of functions maybe the right guess would be the identity mapping, but I am guessing here.

  • finally, I need to show that there exists an inverse element for every element of $G$ so that when I compose it with $\circ$ I get the identity again.


Here my journey ends, or begins, there is nothing I can add. I would appreciate some initial kicks that show me how to get going, how to even start formulate a 'proof' or how one can think about all this stuff above. Sorry for the mess.

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  • $\begingroup$ This is not a linear algebra construction since we're not dealing with vector spaces :) $\endgroup$
    – hjhjhj57
    Oct 12, 2014 at 20:33

2 Answers 2

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I think it's easier to consider a specific index set, work through a proof with that, and then generalize. Take, for example, $I = \mathbb{N}$:

Let $\{ (G_1, \circ_1), (G_2, \circ_2), \dots \}$ be a family of groups. Let $G:= G_1 \times G_2 \times \dots$. An element $x$ of $G$ is the sequence $x = (x_1, x_2, \dots)$ where $x_i \in G_i$. Define the product in $G$ as $$ (x_1, x_2, \dots) \circ (y_1, y_2, \dots) := (x_1 \circ_1 y_1, x_2 \circ_2 y_2, \dots).$$ Show that $(G, \circ)$ is a group.

(If that's too confusing, start with $I = \{1,2\}$.) Now it should be a little easier to see what's going on. For example, it's straightforward to prove that the identity element in $G$ is $e = (e_1, e_2, \dots)$ where $e_i$ is the identity element in $G_i$.

All an index set does (aside from adding obscurity) is add a way to talk about the same problem when $I$ is uncountable.

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  • $\begingroup$ that's a very good advice @Derek, thanks a lot. I will give it some more thought with that approach. $\endgroup$
    – Spaced
    Oct 12, 2014 at 20:08
  • $\begingroup$ I hope it's okay if I can put forward another question to see if I understand your method. It is indeed straightforward and easy to proof that $e=(e_1,e_2, \dots )$ where $e_i$ is the identity element in $G_i$. Same argumentation holds for $h=(h_1, h_2, \dots )$ where $h_i$ is the inverse element in $G$ right? So the only thing left would be associativity. $\endgroup$
    – Spaced
    Oct 12, 2014 at 20:14
  • $\begingroup$ Yes, that's correct. $\endgroup$
    – Derek
    Oct 12, 2014 at 20:45
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You're almost there. The key idea is the definition you give: $$ \times_{i \in I} G_i:= \lbrace x:I \to \bigcup_{i \in I} G_i \mid x(i) \in G_i , \forall i \in I\rbrace $$ So, this new space you're defining is a set of functions (you can call one this functions $x$ or $(x)_{i\in I}$ as long as you know what they are and where they indices run), where each function is defined by it's value in each $i\in I$.

Next you define the 'product' of two of these functions pointwise, i.e., we define the function $$ (x\circ y)(i) := x(i)\circ_iy(i),\quad \forall i\in I $$ It's straightforward to see this is again an element of $\times_{i \in I} G_i$, since each component is again an element of its respective group.

Finally, let's see that all the group axioms hold for this set with the operation we defined:

Associativity: Consider the functions $x,y,z\in \times_{i \in I} G_i$. Then, $$ ((x\circ y)\circ z)(i) = (x\circ y)(i)\circ_i z(i) = (x(i)\circ_i y(i))\circ_i z(i) =\cdots $$ now we use that the product of $G_i$ is associative, then it follows that $$ \cdots = x(i)\circ_i (y(i)\circ_i z(i)) = x(i)\circ_i (y\circ z)(i) = (x\circ (y\circ z))(i) $$ Since this is true for all $i\in I$, we have that $((x\circ y)\circ z) = (x\circ (y\circ z))$ (as functions!).

Neutral element: As you've seen we want to get all the desired properties of a group using that each $G_i$ is a group, so to get the neutral element in the product we define the function $e$ such that $e(i)=e_i$, where $e_i$ is the neutral element of the group $G_i$.

Consider an arbitrary $x\in \times_{i \in I} G_i$. Then, $$ (e\circ x)(i) = e(i)\circ_i x(i) = e_i \circ_i x(i) = x(i) $$ Since this holds for every $i\in I$, we have that $e\circ x = x$, you can prove that $x\circ e = x$ in exactly the same way.

Inverses: I'm sure by now you'll know how are we defining the inverses. Consider any $x\in\times_{i \in I} G_i$, then we define $x^{-1}\in \times_{i \in I} G_i$ such that $$ x^{-1}(i) = x(i)^{-1}\,\, \text{(this is the inverse in the group $G_i$).} $$ So we have that $$ (x\circ x^{-1})(i) = x(i)\circ_i x^{-1}(i) = x(i)\circ_i x(i)^{-1} = e_i = e(i) $$ The other product is done in exactly the same wa and we're done.

The first time dealing with this kind of objects it's hard to keep in mind that we are dealing with a space of functions, not elements of the group per se. As it was pointed out in another answer in some cases this allows you to think of the elements of the product as $n$-tuples of elements of the groups (when $I$ is finite), but that can be misleading when dealing with the general case.

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  • $\begingroup$ I didn't have the chance to read all of your work yet, but I still want to say that I truly appreciate your effort, I adore your readability of the steps. Thanks a lot! $\endgroup$
    – Spaced
    Oct 12, 2014 at 21:02

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