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Just for clarification, here is the definition for a presentation for a group:

Let $G$ be a group and $S$ be a set.

Let $F(S)$ be the free group on $S$ and $R\subset F(S)$ and $\overline{R}$ be the normal closure of $R$.

Then, $(S|R)$ is a presentation for $G$ if and only if $G\cong F(S)/\overline{R}$.

(And for convenience, we write $r=1$ if $r\in R$)

With this definition, let's consider an example, $(x|x^6=1)$ is a presentation for $\mathbb{Z}_6$. BUT Why?

To make that assertion, I have to show the existence of an isomorphism $\phi:F(\{x\})/\overline{\{x^6\}}\rightarrow \mathbb{Z}_6$ and THIS IS NOT TRIVIAL to me.

Even with a really basic group, I have a trouble with visualizing a presentation. I cannot even guess what would presentations would look like fo complicated groups.

How do I formally get informations from a presentation? Please help me with details..

This is how I feel what others do: One concludes $(x|x^6=1)$ is a presentation since $\mathbb{Z}_6$ is cyclic so that generated by a single element and since $6•1=0$. This doesn't really seem legit to me.

Here's an illustration how I view this: By the definition of free group, we know that $\mathbb{Z}$ is a free group with a basis $\{1\}$. Since $|\{1\}|=|\{x\}|$, $\mathbb{Z}\cong F(\{x\})$. Now, we have to take the normal closure of $\{x^6\}$.. Hmm what would the closure look like...?

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Rubertos
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  • When I first read the title, I thought "Hm. This sounds like a Workplace.SE question--how to give a presentation in front of a group of people." :P Now I know there's a different definition of "presentation"... – apnorton Oct 06 '14 at 18:14
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    My answer would be... with difficulty. Finding a formal proof for things you can see obviously isn't always as easy as might be hoped. Part of the answer is that once you've proved you know enough maths you are not always required to give the full formal proof. – Jessica B Oct 06 '14 at 18:17
  • @Jessica The problem is, I don't see a presentation obvious.. That's why I posted this. I cannot understand why $D_2n,\mathbb{Z}_n\times\mathbb{Z}_m$ have such presentations. I cannot even make a tiny little clue why those have such presentations... – Rubertos Oct 06 '14 at 18:20
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    @anorton There are (probably apocryphal) stories of library books being badly mis-shelved along these lines. – Jessica B Oct 06 '14 at 20:59

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To get an intuition of what a presentation is saying, maybe it helps to think about the names we give: generators and relations.

You have a set of generators (and their inverses). The idea is that they should generate that group: any element can be written as a product of the generators, a bit like any element of a vector space being a linear combination of the basis vectors.

The problem is, that basis vectors don't interact (that's the point of linear independence), but you can't get away with that in general for groups. So if you just take generators, and don't say anything about them (other than that they cancel with their inverses) you only get free groups. Free groups are groups with as little interaction between elements as possible.

To get more types of groups, we therefore specify relations, telling us how the generators interact with each other. There will in general be infinitely many actual relationships, but we don't want to write them all down. Instead, we write down a (hopefully) meaningful set of relations, that 'generate' all the real ones. And what we write down is which elements are in fact the same.

So, for example, take your presentation $(x|x^6=1)$. That says we take one element ($x$), and we start multiplying it by itself. This gives us $x, x^2,x^3,x^4,x^5$. But when we get to $x^6$, we declare that we've in fact got back to $1$. So we have an element of order $6$, and every element of the group is a power of that element. That is the definition of the cyclic group of order $6$.

To get $\mathbb{Z}_6\times\mathbb{Z}_7$, we take the two things we know: $(x|x^6=1)$ and $(y|y^7=1)$. We know everything we want is given by combining these, so we take the two generators $x$ and $y$. As $x$ and $y$ still interact with themselves as before, the relations $x^6=1$ and $y^7=1$ still hold. But we also need to say something about how $x$ and $y$ interact. By definition of the product, we want $x$ and $y$ to commute, so we add in the relation $xy=yx$. In fact, that is enough, so $\mathbb{Z}_6\times\mathbb{Z}_7$ has a presentation $(x,y|x^6=1,y^7=1,xy=yx)$.

To get a more visual understanding, find out about Cayley graphs.

Jessica B
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  • +1. How do you conclude that $xy=yx$ is enough? It seems like a presentation for a group is being used only for groups that we already know well, not for groups that we are about to know. Would you please please give an explanation on a presentation for $D_2n$? – Rubertos Oct 06 '14 at 18:52
  • To prove that $xy=yx$ is enough, note that this tells you that any element of the $\mathbb{Z}_6$ commutes with any element of the $\mathbb{Z}_7$. eg. $y^3 x^2=y^2 xyx=y^2x^2y=\ldots=x^2 y^3$. – Jessica B Oct 06 '14 at 20:47
  • For $D_{2n}$, you have two things you can do: rotate, and reflect. From this we get two generators, one of order $n$, and one of order $2$. Thus we have generators $x$ and $y$, plus relations $x^n=1$ and $y^2=1$. It remains to say how the two generators interact. In this case, reflecting, rotating, and reflecting again, has the overall effect of rotating in the other direction. We therefore get a relation $yxy=x^{-1}$. You'll have to convince yourself that's all that's needed. Thus the presentation is $(x,y|x^n,y^2,yxy=x^{-1})$. – Jessica B Oct 06 '14 at 20:56
  • On your other point, group presentations most certainly are used for groups we want to find out about. However, it's done by people who have first understood the definition well enough. – Jessica B Oct 06 '14 at 20:57
  • Thank you for helping me in details. I'm really satisfied. Thank you :) – Rubertos Oct 06 '14 at 23:37