A basic question on group presentation This is likely a pretty basic question, but I couldn't find answers anywhere. It's to do with how group presentations are represented.
It's said that, for example, the cyclic group of order $n$ may be represented as
$$\langle a \mid a^n = e\rangle$$
But, this makes sense only if $n$ is the smallest possible positive integer such that $a^n = e$. Naively, the fact that $a^n = e$ doesn't rule out the possibility that $|a|$ is some divisor of $n$, which would make this an inaccurate presentation of the group.
So, is the convention to assume that $n$ is the smallest possible integer making $a^n = e$ true? Is this 'convention' true for all presentations, where one of the relations given is something like $x^n = e$?
Thank you! Apologies if this is basic.
 A: It's not a convention. It's what the meaning of a presentation is.
When we give a presentation,
$$\langle x_1,x_2,\ldots\mid R_1,R_2,\ldots\rangle$$
(where $R_i$ are the relations satisfied by the generators), we are describing "the most general group that is generated by elements $x_1,x_2,\ldots$ satisfying the relations $R_1$, $R_2,\ldots$". As such, when we write
$$\langle a\mid a^n=e\rangle$$
we are saying "the most general group that is generated by an element $a$ subject to the condition $a^n=e$." The most general such group is the one in which the order is exactly $n$, and not any divisor of $n$.
This idea of "most general group" is captured by von Dyck's Theorem, and by the construction of a group given a presentation.

von Dyck's Theorem. If $G$ is the group given by the presentation
$$\langle x_1,x_2,\ldots\mid R_1,R_2,\ldots\rangle$$
then given any group $H$ and any elements $h_1,h_2,\ldots\in H$ such that, replacing $x_i$ with $h_i$ in the relations $R_1,R_2,\ldots$ results in statements that are true in $H$, there exists a unique group homomorphism $\phi\colon G\to H$ such that $\phi(x_i)=h_i$ for $i=1,2,\ldots.$

That is: any group generated by elements satisfying the relations $R_1,R_2,\ldots$ must be a quotient of $G$.
This is true for the cyclic group of order $n$, but not for the cyclic group of order $k$ for $k\neq n$, $k\mid n$: because, for example, the cyclic group of order $10$ is generated by an element satisfying $x^{20}=1$, but so does the cyclic group of order $20$, and the latter is not a quotient of the former.
Construction. To construct the group given by the presentation given above,  first we rewrite all relations so that they are in the form $w_i(x_1,x_2,\ldots)=1$; for example, if the first relation is $x_1x_2=x_2x_1$, then we rewrite it as $x_1x_2x_1^{-1}x_2^{-1}=1$. Then we take the free group $F$ on $x_1,x_2,\ldots$, and let $N$ be the smallest normal subgroup of $F$ that contains all the relations $w_1(x_1,x_2,\ldots), w_2(x_1,x_2,\ldots),\ldots$. Then the group we want is $G=F/N$.
Caveat. Note that just because a presentation has a relations of the form $x^n=e$, that by itself does not mean that the order of $x$ is $n$; it is possible that, when combined with other relations, the order of $x$ will be smaller. For example, as discussed here (and you may want to take a look at that answer anyway), the group
$$G = \langle a,b\mid a^5 = b^4 = 1, aba^{-1}b=1\rangle$$
will actually yield a group in which the order of $b$ is two, not $4$. That's because the other relations, together with the relation $b^4=1$, imply that $b^2=1$ must also hold. So just because you see $x^n=1$ in the presentation, it does not mean that $x$ will definitely have order $n$.
In the case of the cyclic group, $\langle a\mid a^n=1\rangle$, there are no other relations to interact with $a^n=1$, so that one can show that you get the group of order $n$; but this is not a matter of "convention" or "understanding", but of what the presentation requires and means.
