I have a question about group presentations (in terms of generators and relations). It's been really bugging me for ages. Would really appreciate any thoughts on this. Cheers, Michael

You are 'given' a group $G$. First, we find a list of generators for the group. Let's be lazy and take all of the elements as generators. We know how to explicitly construct the free group on this set of generators. Denote this free group $F$. Now we know there is a homomorphism, call it $\phi$, from the free group $F$ onto our group $G$. By the first isomorphism theorem we know that G is isomorphic to the quotient group $F / \ker\phi$.

To define my group $G$, I could therefore state what the generators are, and also a set of generating elements of $\ker\phi$. So the question now is this:

What's in the kernel of $\phi$?

We could take all the relations from the Cayley table and rewrite them in 'standard form', i.e., rewrite $g_1*g_2 = g_3$ as $g_1*g_2*g_3^{-1} = e$. All such expressions must be elements of $\ker\phi$. Let's denote this set of elements $R$. Let's also denote the normal subgroup generated by $R$ as $N$.

We know that $\ker\phi$ is a normal subgroup of $F$ which contains the set $R$. Because $N$ is the smallest normal subgroup of $F$ which contains the set $R$, it is obvious that $\ker\phi$ contains $N$. However textbooks always assert more than this, namely that $\ker\phi = N$.

I know how to prove that $\ker\phi=N$ for specific examples. For example, I attach below a proof for $G=D_4$. However I can't construct a general proof to understand properly why this works for any group. It makes intuitive sense, but that never satisfies!enter image description hereenter image description here


Perhaps it would be clearer to denote the elements of the group $G$ by $g_i'$ and the corresponding generators of your free group $F$ by $g_i$. So $\phi(g_i)=g_i'$ for each $g_i' \in G$.

Note that the relations $g_1g_2=g_3$ imply all of the associated products involving inverses, like $g_1^{-1}g_3=g_2$, etc.

Let $g \in \ker \phi$. We can write $g$ as a product $a_1a_2\cdots a_n$, for some $n$, where each $a_i$ is some $g_j$ or $g_j^{-1}$. If $n=0$, then $g=1 \in N$, so we can assume that $n>0$.

Suppose that $n > 1$. Then one of the group relations has the form $a_{n-1}a_n=b$, where $b$ is one of the $g_j^{\pm 1}$. So $r = a_n^{-1}a_{n-1}^{-1}b \in N$ and $gr = a_1a_2\cdots a_{n-2}b$, which is a product of length $n-1$.

Repeating this argument, we can reduce the length of the product to $1$, and find $r' \in N$ with $gr' = g_i^{\pm 1}$ for some generator $g_i$.

We are assuming that $g \in \ker \phi$, and since $r' \in N \le \ker \phi$, we have $1 = \phi(gr') = \phi(g_i^{\pm 1}) = \phi(g_i)^{\pm 1} = {g_i'}^{\pm 1}$. So $g_i$ is the generator of $F$ that maps onto the identity of $G$. So one of the group relators is $g_ig_ig_i^{-1} = g_i$ and hence $g_i \in N$. So $g = g_i^{\pm 1}r'^{-1} \in N$.

  • $\begingroup$ Thanks ever so much Derek. Your proof is very clear. I see now that I was essentially asking for a proof that every finite group has a presentation. I have just posted a similar, but related, question about group presentations. It's called "Group presentations - again". I wonder whether the proof I'm looking for in my new question would be along similar lines to the one you provided above? $\endgroup$
    – user123473
    Jan 30 '14 at 21:43
  • $\begingroup$ Here's a link to the related question : math.stackexchange.com/questions/657719/… $\endgroup$
    – user123473
    Jan 30 '14 at 21:51

First summarizing the question:

We know a priori that $G$ is a group, and we got a presentation by enumerating its elements and taking a multiplication table. We call the normal subgroup generated by the multiplication table relations $N$. We call this presentation $\phi: F \to G$; you agree that $\phi$ is surjective and that its kernel contains $N$ and you complain that its kernel may be larger than $N$.

Let $w \in \ker \phi$. We will induct on the length of $w$ (this length function is well-defined in the free group $F$). If $w$ has length $0$ (i.e. $w = e$) we are done, and by our choice of presentation, no words of length $1$ are in the kernel of $\phi$. So we can assume $w$ is of the form $w = w'xy$ (where the length of $w'$ is genuinely two less than the length of $w$). By the Cayley table relation, there exists $g \in F$ with $\phi(w) = \phi(w'g)$ (namely, any $g$ such that $\phi(g)$ is the product of $\phi(x)$ and $\phi(y)$ in $G$). As $w'g$ has smaller length than $w$, we are done by induction.

  • $\begingroup$ There is a word of length one in the kernel, the generator that maps onto the identity of $G$. $\endgroup$
    – Derek Holt
    Jan 24 '14 at 10:34
  • $\begingroup$ ah true. so you need a little argument showing that phi factors through the presentation using the multiplication table without the letter $e$. or if you are just trying to prove that all groups have presentations, just start there... $\endgroup$
    – hunter
    Jan 24 '14 at 12:48

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