Are group presentations useful in studying the pure math of a group as a whole?

Question

Group presentations are often used by computers to quickly identify an element of a group or multiply two elements. However, they don't give a full understanding of the mathematical properties of a group. Can group presentations be used to find some of the mathematical properties of a group, or are group presentations only useful when trying to get an understanding of the elements and their orders rather than the mathematical properties of the group as a whole (for example, its representations or automorphism group)?

Answer 1

Let's say we want to know the automorphisms of the dihedral group of order $2n$. We want do this using the presentation $G = \langle r,s | r^n = 1,s^2 = 1, sr = r^{-1}s\rangle$, where $n \geq 3$. Note that first some kind of construction is needed to be sure there really is such a group of order $2n$: from the presentation it's easy to see that group has order at most $2n$, but how do you know the actual order isn't smaller and maybe some hidden aspect makes it all collapse down to the trivial group? I will ignore this issue and just take for granted we know the group has order $2n$.

If $f \colon G \to G$ is an automorphism then $f$ is determined by $f(r)$ and $f(s)$. Since $$ G = \{1,r,\ldots,r^{n-1},s,rs,\ldots,r^{n-1}s\rangle $$ and the last $n$ elements $s,rs,\ldots,r^{n-1}s$ have order $2$, $f(r)$ must be a power of $r$. Its order is $n$, so $f(r) = r^a$ where $a \in (\mathbf Z/(n))^\times$. Then $f(s) = r^bs$ where $b \in \mathbf Z/(n)$.

Conversely, suppose $a \in (\mathbf Z/(n))^\times$ and $b \in \mathbf Z/(n)$. We want to define an automorphism $f_{a,b} \colon G \to G$ where $f_{a,b}(r) = r^a$ and $f_{a,b}(s) = r^bs$. First we'll show there is a homomorphism $G \to G$ with those values at $r$ and $s$, and then we'll show it is an automorphism.

There is a unique homomorphism $F_2 = \langle x,y\rangle \to G$ where $x \mapsto r^a$ and $y \mapsto r^bs$. Let $K$ be its kernel. We have $x^n \mapsto (r^b)^n = (r^n)^b = 1$ and $y^2 \mapsto (r^bs)^2 = 2$, so $x^n$ and $y^2$ are in $K$. Also $yx \mapsto r^bsr^a = r^{b-a}s$ and $x^{-1}y \mapsto r^{-a}r^bs = r^{b-a}s$. Since $yx$ and $x^{-1}y$ are mapped to the same value, $(yx)(x^{-1}y)^{-1} \in K$. Since $K$ is normal, $K$ contains the normal subgroup $N$ that is generated by $x^n$, $y^2$, and $(yx)(x^{-1}y)^{-1}$, so the homomorphism $F_2 \to G$ induces a homomorphism $F_2/N \to G$. From the presentation of $G$, we have $F_2/N \cong G$ by $\overline{x} \mapsto r$ and $\overline{y} \mapsto s$. Thus we have a homomorphism $f \colon G \to G$ where $r \mapsto r^a$ and $s \mapsto r^bs$. This has to be $f_{a,b}$ (it has the right values on the generating elements $r$ and $s$ of $G$, so we have proved $f_{a,b} \colon G \to G$ exists as a homomorphism.

Why is $f_{a,b}$ an automorphism? Since $G$ is finite, it suffices to prove $f_{a,b}$ is either injective or surjective. We'll show its surjective. The image $f_{a,b}(G)$ is a subgroup of $G$ that contains $\langle f(r)\rangle = \langle r^j\rangle = \langle r\rangle$ and $r^is$, so $f_{a,b}(G)$ contains $r$ and $s = r^{-i}(r^is)$. Thus $f_{a,b}(G) = G$.

We have proved ${\rm Aut}(G) = \{f_{a,b} : a \in (\mathbf Z/(n))^\times, b \in \mathbf Z/(n)\}$. Check $f_{a,b} \circ f_{a',b'} = f_{aa',b + ab'}$. That is exactly how the matrices $(\begin{smallmatrix}a&b\\0&1\end{smallmatrix})$ multiply when $a \in (\mathbf Z/(n))^\times$ and $b \in \mathbf Z/(n)$. Each $f_{a,b}$ determines $a$ and $b$ since $r$ has order $n$ in $G$ (so $r^a$ determines $a \bmod n$ and $r^bs$ determines $b$ mod $n$). Thus $$ {\rm Aut}(G) \cong \left\{\begin{pmatrix}a&b\\0&1\end{pmatrix} : a \in (\mathbf Z/(n))^\times, b \in \mathbf Z/(n)\right\}. $$

Answer 2

Here is an example of a situation where you need a presentation of a group.

Suppose that we are given a group $G$, perhaps as a subgroup of $S_n$ for some $n$, and we also have a normal subgroup $N$ of $G$. A common question then is "does $N$ have a complement in $G$"?

We could solve this in a naive way by brute force by finding all subgroups of $G$ and then checking whether any of them are complements of $N$.

A more efficient way is to use a cohomological condition. There is a complement of $N$ in $G$ if and only if the identity map on $N$ extends to a derivation (crossed homomorphism) $G \to N$, and in that case the kernel of the derivation is a complement.

That condition can be checked moderately easily if we know a presentation of $G/N$, and if we do not know one already, then it is a good idea to compute one.

Incidentally, if we did want to find all of the subgroups of $G$, then presentations play a role in algorithms to do that.

A more general application is that a presentation of $G$ is useful/necessary for finding homomorphisms (or, in the example above, crossed homomorphisms) of $G$ to another group, and computing ${\rm Aut}(G)$ is a special case of that.