I've just read the first few pages of [*Combinatorial Group Theory*](http://books.google.com/books/about/Combinatorial_Group_Theory.html?id=1LW4s1RDRHQC) 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.

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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 $$G = \langle a, b, c, \dots; P, Q, R, \dots \rangle\,.$$