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The presentation of the dihedral group $D_{2n}$ is

$$D_{2n}= \langle r,s \mid r^{n}=s^{2}=1, rs=sr^{-1} \rangle.$$

Why is it incorrect to conclude that $r=s=1$ ? In other words why doesn’t this group presentation describe the trivial group.

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Call $G$ the group given by that presentation. Then you can show there exists an homomorphism $\phi:G\to\mathrm{GL}_2(\mathbb R)$ which maps $s$ to $S=\left(\begin{smallmatrix}-1&0\\0&1\end{smallmatrix}\right)$ and $r$ to $R=\left(

\begin{smallmatrix}

\cos (\theta ) & -\sin (\theta ) \\

\sin (\theta ) & \cos (\theta )

\end{smallmatrix}

\right)$ with $\theta=2\pi/n$; to check this, you have to use the characteristic property of group presentations, and verify that matrices $S$ and $R$ satisfy the same relations as the generators $s$ and $r$.

Now, since $\phi(s)\neq I$ and $\phi(r)\neq I$, it follows of course that $s\neq 1_G$ and $r\neq 1_G$.

(You can also easily find two non-identity permutations $\rho$, $\sigma\in S_n$ which satisfy the relations, and construct a map $g:G\to S_n$ such that $g(s)=\sigma$ and $g(r)=\rho$, giving another proof that $r\neq 1_G$ and $s\neq 1_G$.)

**NB** This is another canonical example of the things we want group representations for.

A group given by presentation $G=\langle X|R\rangle$ is, by definition, the *most general* group with a generating set $X$ and satisfying the relations $R$. Say, for simplicity, that $X=\{x_1,\ldots,x_n\}$ and $R=\{r_1(x_1,\ldots,x_n),\ldots,r_m(x_1,\ldots,x_n)\}$. Being the “most general” group with $n$ generators $x_1,\ldots,x_n$ that satisfy the $m$ relations $r_1,\ldots,r_m$ means that given *any* group $H$ with elements $h_1,\ldots,h_n$ for which $r_i(h_1,\ldots,h_n)=1$ for $i=1,\ldots,m$, there will be a (unique) group homomorphism from $G$ to $H$ that maps $x_i$ to $h_i$. It’s like the “free-est group” with elements $x_1,\ldots,x_n$, subjected only to the relations in $R$. (The magic words are **Van Dyck’s Theorem**)

What you are noting is that if we pick $H$ to be the trivial group, and $h_1,\ldots,h_n$ to be the trivial element, then all the relations are satisfied; but this does not tell you that the trivial group is the most general group that satisfies the given conditions, it only tells you that the trivial group is a *quotient* of the most general group that satisfies the given conditions (as it must, since the trivial group is a quotient of everything).

In the case of the dihedral group $D_{2n}$ you are looking for the most general group that has an element of order dividing $n$, an element of order dividing $2$, and such that those elements satisfy the relation $sr = r^{-1}s$ (which can be written as $srsr=1$). The trivial group certainly satisfies those relations, but it is not the most general one that does (e.g., if $n$ is even, you can look at $H=C_2\times C_2$, and identify $r$ with $(1,0)$ and $s$ with $(0,1)$). Every time you find a group and elements that satisfy the given relations, you have found a map from the group presented to the group you found, but you have not necessarily found the group presented.

To really know that you have found the “most general group” that satisfies the given conditions is not always trivial. One way is to show that the group you found (call it $G$, with elements $g_1,\ldots,g_n$ playing the roles of the $x_i$) has the property ascribed to the group presented by the presentation: given any group $H$, and any elements $h_1,\ldots,h_n$ in $H$ that satisfy $r_1,\ldots,r_m$ in $H$, there exists a unique group homomorphism from your group $G$ to $H$ that maps $g_i$ to $h_i$. The trivial group does not have this property for the presentation given in $D_{2n}$, so it cannot be $D_{2n}$.

Another method that works sometimes is to find a “normal form” for the elements of the group presented, allowing you to count exactly how many elements there are. This is not hard to do in $D_{2n}$: every element of $D_{2n}$ can be written as a product of powers of $r$ times powers of $s$. Using the rules that say that $r^n=1$, $s^2=1$, and that $sr=r^{-1}s = r^{n-1}s$, you can “move” all the $r$s that occur in such a product all the way to the left, and all the $s$’s to the right, so that every element of $D_{2n}$ can be written in the form $r^is^j$, with $0\leq i\lt n$ and $0\leq j\lt 2$. We do not know if each element corresponds to one and only one such expression, but certainly each element corresponds to at least one such expression. That means that $D_{2n}$ has, at most, $2n$ elements. If you can find a group $G$, generated by elements $g_1$ and $g_2$ such that $g_1^n=1$, $g_2^2=1$, and $g_2g_1=g_1^{-1}g_2$, *and* it turns out that $G$ has exactly $2n$ elements, then you know there is a homomorphism from $D_{2n}$ to $G$ that sends $r$ to $g_1$ and $s$ to $g_2$, so that $G$ will be a quotient of $D_{2n}$; since we are assuming that $G$ has exactly $2n$ elements, and we know that $D_{2n}$ has at most $2n$ elements, then you can conclude that the map has to be injective and the kernel is trivial, and so that $D_{2n}$ is in fact isomorphic to the group $G$ you found. This is one way of “realizing” a presentation as a group that you can get your hands on.

**Edit:** Note that by the property ascribed to the group given by presentation, the group must be generated by $X$; otherwise, there is a map from $G$ onto $\langle X\rangle$ that sends each $x_i$ to itself, so you get two maps from $G$ to itself sending each $x_i$ to itself (the identity, and the composite map $G\to\langle X\rangle \hookrightarrow G$); but the definition requires that there be a unique homomorphism to accomplish this.

In the question

“$r^{n}=s^{2}=1, rs=sr^{-1}$ … why is it incorrect to *conclude* that $r=s=1$ ? … why doesn’t this group presentation *describe* the trivial group”

the key words are “conclude” and “describe”. You are in effect asking what logic applies to statements about $r$ and $s$.

The answer is that assertions about $r$ and $s$ are understood as valid in this context only if they hold in ALL groups. The assertion “if a pair of group elements $r$ and $s$ satisfies the defining relations of $D_{2n}$, then $r=s=1$” is false in some groups and true in others. It is true in groups with no subgroup of order $2n$, such as large cyclic groups of prime order. It is false in other groups, such as the dihedral group of order 2n, or the group of order 2 where $r=1, s=-1$ is a pair satisfying the equations.

Truth in all groups is a semantic definition and can be replaced by an equivalent, syntactic definition that gives the combinatorial rules for making all valid deductions from the defining relations. $r=s=1$ is not a valid algebraic consequence of the algebraic rules. As in all problems of logic, showing that something does NOT follow from some rules or axioms is a matter of constructing an example (a model) where the rules apply but the assertion is false. Here the models are groups with a pair of elements $r,s$ satisfying the relations, so for showing what is not true in the logic of generators and relations the algebraic approach doesn’t provide an alternative to constructing models. For assertions that *do* follow from the defining relations, the algebraic definition is useful, because instead of surveying a possibly infinite collection of models to show that the statement is correct, one can perform a finite combinatorial derivation that works for all models.

(Added in response to comment from Matt E about normal forms: showing that any element can be placed in the normal form follows from the algebraic relations alone. However, to use normal forms as a method of calculating in the group or understanding it, you need to show that distinct elements in the group have different normal forms. This is again the non-algebraic (or not purely algebraic) problem of showing that relations do not hold in a structure, which is done by constructing models such as actions of the group on particular spaces.)

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