# Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?

I've just read the first few pages of Combinatorial Group Theory by Magnus, Karrass, and Solitar, and based on their definitions there, and more specifically, the reasoning given in the hint to exercise $5$(c) (on p. 9), I don't see why $\langle a ; a^2 \rangle$ (or even $\langle a ; a \rangle$, for that matter) shouldn't be regarded as a presentation of $C_4$.

Exercise $4$ (also on p. 9), asks the reader to establish that "if the word $K$ is derivable from the relators $M, N, \dots,$ and $M, N, \dots$ are derivable from the relators $P, Q, R, \dots,$ then $K$ is derivable from $P, Q, R, \dots\,$".

Exercise $5$(b) asks the reader to establish that $\langle a; a^4 \rangle$ is a presentation of $C_4$. The sequence below

$$a^4,a^4a^{-2},a^2,a^2a^{-2},1$$

...shows that the word $a^4$ is derivable (using the operations stipulated on p. 6) from the relator $a^2$ (and is hence itself a relator).

Of course, I realize that $\langle a; a^2 \rangle$ is also a presentation of $C_2$, and that $C_2 \neq C_4$.

And pretty much the same reasoning would lead one to conclude that $\langle a; a^2 \rangle$ is a presentation of $C_{2n}$, and $\langle a; a \rangle$ is a presentation of $C_n$, for any $n \in \{2, 3, \dots\}$. If so, a presentation of a group $G$ could not be considered (necessarily) a description, or specification, of $G$; so what's its purpose?

Therefore, I conclude that something must be missing, either from the authors' definition of a group presentation (p. 7)1, or from my grasp of it.

EDIT: To clarify, let me say more about the origin of my question, namely the hint to exercise $5$(c) (already alluded to at the beginning of this post).

Exercise $5$(c) asks

Give a presentation for [the cyclic group of order 4] using one generating symbol $b$ and two defining relators, neither of which is $b^4$ or $b^{-4}$.

The hint for this question says:

...use $b^8$ and $b^{12}$ as defining relators, and show that $b^4$ can be derived. Hence, by [exercise 4, cited above], $b^8$ and $b^{12}$ are a set of defining relators.

I have no problem following this hint, but I can just as well carry out exactly the same steps not just for $(b^8, b^{12})$, but for any pair $(b^n, b^{n+4})$ with $n \in \mathbb{Z} \backslash \{-1, 0, 1\}$, like so: $$b^4,b^4b^n,b^{n+4},b^{n+4}b^{-(n+4)},1$$

And if this maneuver somehow renders $\langle b; b^n, b^{n+4}\rangle$ (for any $n \in \mathbb{Z} \backslash \{-1, 0, 1\}$) a presentation for $G$, then a similar maneuver would render $\langle a; a^2\rangle$, and even $\langle a; a\rangle$ a presentation for $G$.

I realize that this conclusion is nonsense. I'm trying pinpoint the misstep in the reasoning.

1 For completeness, this definition goes like this: "A word $R(a, b, c)$ which defines the identity element $1$ in [a group] $G$ is called a relator. … If every relator [of a group $G$] is derivable from the relators $P, Q, R, \dots,$ then we call $P, Q, R, \dots$ a set of defining relators or a complete set of relators for the group $G$ on the generators $a, b, c, \dots .$ If $P, Q, R,\dots$ is a set of definitng relators fro the group G on the generators $a, b, c,\dots$ we call $$\langle a, b, c, \dots; P(a, b, c, \dots), Q(a, b, c, \dots), R(a, b, c, \dots), \dots\rangle$$ a presentation of $G$ and write $$G = \langle a, b, c, \dots; P, Q, R, \dots \rangle\,.$$

• @studiosus: I don't have that confusion, nor I see anything in my post that would suggest that I do. (And, BTW, if you're going to insult a poster's intelligence, please at least be more explicit about your reasoning.) – kjo Feb 6 '14 at 12:22
• The relevant word "if" appears in your quotation of Exercise 4 (2nd paragraph of your post). The point is that the relation $a^2=1$ implies the relation $a^4=1$, but not vice verse ("if" is not the same as "if and only if"). – Moishe Kohan Feb 6 '14 at 12:27
• I think the problem is this. What you've shown in your derivation is that if $a^2$ were a relator, then $a^4$ is also a relator. That does not show that $a^2$ is a relator in $C_4$. So, while $a^4$ is derivable from $a^2$, $a^2$ is not a relator, so doe not give a presentation of $C_4$. – Callus Feb 6 '14 at 12:30
• Touche - I didn't notice the link. All the same, it is good practice to give the authors names (for there are people like me out there who do not notice links, for whatever unknown reason...). – user1729 Feb 6 '14 at 15:49
• Your post is getting a little long. We should take the conversation out of the comments. Try one of the chat rooms. To answer though, what you're showing is that $a^4$ is a relator if $a^2$ is a relator, but just because $a^4$ is a relator doesn't mean the group is $C_4$. In other words, $a^4$ is a relator in $\langle a; a^2\rangle$, but that doesn't mean that $\langle a; a^4\rangle$ – Callus Feb 7 '14 at 2:55

When you write $G=\langle a;a^2\rangle$, this is meant that $G$ is the group generated by an element $a$ such that $a^2=1$. Notice that this means that $a=a^{-1}$. So $G$ is just $\{1,a\}$. For any $n\in\mathbb{N}$, the group $G=\langle a;a^n\rangle$ is just the group, $\{1,a,a^2,\ldots,a^{n-1}\}$, which is $C_n$. In your case you have if $a^2=1$, then $(a^2)^2=a^4=1$, but not the other way around. This is what @studiosus was getting at (I don't think his/her intent was to insult).

• The thing is that it's not obvious to me that anywhere I've used the implication that $a^2$ is derivable from $a^4$. BTW, I realize that there's something wrong with my argument, but this realization relies on too much prior knowledge about group theory. The same is true of most of your answer. I'm trying to think through this problem as if I did not know any other group theory beyond what the book has presented so far (i.e. very, very little). Also, next time someone (a professor maybe, or a colleague) makes a error of arithmetic in a blackboard derivation, explain to them that the... – kjo Feb 7 '14 at 2:20
• ...reason is that they are "confused about addition", or subtraction, or whatever the operation was where the misstep happened, and see if they don't take your "explanation" as an insult. – kjo Feb 7 '14 at 2:21

The full statement of question $1.1.4$ is:

Show that if the word $K$ is derivable from the relators $M,N,...$, and $M,N,...$ are derivable from the relators $P,Q,R,...$, then $K$ is derivable form $P,Q,R,...;$ hence if $M,N,...$ is a set of defining relators, then so is $P,Q,R,...$.

And here is the relevant part of the hint of $1.1.5(c)$, which will be the main example I will focus on (since that seems to be where your questions came from):

Hint: For $(c)$ use $b^8$ and $b^{12}$ as defining relators, and show that $b^4$ can be derived.

If we follow the hint we need to note that we are assuming that $b^4$ is a set of defining relators (this is basically what is shown in part $(b)$ of the same problem) and show that we can get $b^4$ from the given $b^8,b^{12}$. But we do not know a priori that $b^8$ and $b^{12}$ are relators of the group, so we can not apply problem $1.1.4$ yet. Of course it is a simple matter to show this, but this is where your analysis of the case of $b^n$ and $b^{n+4}$ fails, you never show that $b^n,b^{n+4}$ are relators, and for many $n$ they are not relators. A similiar problem occurs when you try to use $b^2$, or $b$ as relators, you don't know they are (and they obviously are not).

If I had to guess as to where you went wrong is that you were defining $b^2$ to be a relator, but when discussing relators we are working in a group , so there is a fixed set of relators, and just because you can get all the relators from $b^2$, that does not mean that it itself is a relator.