Dumb question: Is $\langle e \mid e\rangle$ a presentation of the trivial group?
Presentation of the trivial group
abstract-algebragroup-presentationgroup-theory
Related Solutions
There is no general approach that will work for all presentations but, for examples such as this one, and many others, a useful ad hoc strategy that can work has two main steps.
First, notice that you can find a normal form for the elements of the group, and count them to get an upper bound on the order. Second, get a lower bound on the order by finding a homomorphism from the group defined by the presentation onto a "concrete" group of known order. (For example, a group of permutations or of matrices.) The trick is to get these upper and lower bounds to agree.
For your example, notice that, since $a$ and $b$ generate the group $G$, each element of $G$ can be written as a word of the form $a^{\alpha_{1}}b^{\beta_{1}}a^{\alpha_{2}}b^{\beta_{2}}\cdots a^{\alpha_{k}}b^{\beta_{k}}$, for some $k$ and integers $\alpha_{i}$ and $\beta_{j}$. From the power relations $a^{16} = b^{6} = 1$, it follows that we can take $0\leq\alpha_{i}\leq 15$ and $0\leq\beta_{j}\leq 5$. From the commutation relation $bab^{-1} = a^3$ (or, equivalently, $ba = a^{3}b$), it follows (induction) that we can put every such word into the form $a^{\alpha}b^{\beta}$ where, again, $0\leq\alpha\leq 15$ and $0\leq\beta\leq 5$. However, these are not all distinct. Again, from the relation $bab^{-1} = a^{3}$ we get (by induction), $b^{k}ab^{-k} = a^{3^{k}}$. Putting this together with $b^{6} = 1$, and taking $k = 6$, we find that $a = b^{6}ab^{-6} = a^{3^{6}} = a^{729}$, so $1 = a^{728} = a^{16\cdot 25 + 8} = a^{8}$. That is, the order of $a$ divides $8$, so we can actually impose the restriction $0\leq\alpha\leq 7$. This gives us the upper bound $48 = 8\cdot 6$ for the order of $G$. This really is still just an upper bound (so far as we have proved), because we have not yet ruled out the possibility that the $48$ words $a^{\alpha}b^{\beta}$, with $0\leq\alpha\leq 7$ and $0\leq\beta\leq 5$ are still not all distinct. Doing this is the job of the second step.
To implement the second step, we need to find a "concrete" group $P$ of known order, and a homomorphism from $G$ onto $P$. One such example is the permutation group $$P = \langle (1,2,3,4,5,6,7,8), (2,4)(3,7)(6,8)(9,10,11) \rangle .$$ It can be shown that $P$ is a group of order $48$. Since the generators $\alpha = (1,2,3,4,5,6,7,8)$ and $\beta = (2,4)(3,7)(6,8)(9,10,11)$ of $P$ satisfy $\alpha^{16} = \beta^{6} = 1$ and $\beta\alpha\beta^{-1} = \alpha^{3}$, it follows that the assignment $a\mapsto\alpha, b\mapsto\beta$ extends to a homomorphism from $G$ to $P$, which is surjective because $\alpha$ and $\beta$ generate $P$.
Remark. Had I not gotten the answer from Maple beforehand, I might have wasted a lot of time trying to prove that the group had order $96$, which is the bound we get before noticing that $a$ had order dividing $8$. Even having done so, I might also have wasted time trying to further improve the upper bound to $24$ or even $12$.
Remark. How did we pull the permutation group $P$ from a hat? In general, this can be quite difficult, but in this case, it's not too bad once we know what we're looking for. We knew we wanted a group with generators $\alpha$ and $\beta$ such that $\alpha$ had order $8$ and $\beta$ had order $6$. The most obvious permutation of order $8$ is an $8$-cycle, so we write that down: $$\alpha = (1,2,3,4,5,6,7,8).$$ Since $3$ is prime to $8$, we know that $\alpha^{3}$ is also an $8$-cycle, so we write down $\alpha^{3}$: $$\alpha^{3} = (1,4,7,2,5,8,3,6).$$ Now we needed a permutation $\beta$ conjugating $\alpha$ to $\alpha^{3}$. But, having displayed $\alpha$ and $\alpha^{3}$ above, we can just read this off: $$\beta = (2,4)(3,7)(6,8).$$ The problem is that this $\beta$ has order $2$, not $6$. So we just tack on a disjoint $3$-cycle to get $$\beta = (2,4)(3,7)(6,8)(9,10,11).$$
If we hadn't noticed the smaller upper bound of $48$ before, we might have tried constructing a group of order $96$ along these lines by writing down a $16$-cycle, but we'd then get a hint we were on the wrong track since the $\beta$ we find has $4$-cycles.
Remark. A more systematic approach is to use the Todd-Coxeter algorithm, which is what most computer algebra systems will use to answer a question like this. It will fail if the group is not finite, but (given sufficient time and memory) will complete if the group is indeed finite. For problems that are not too large, this can be done by hand. See the book "Topics in the theory of group presentations", by D.L. Johnson (Cambridge Univ. Press, 1980) for a good description.
This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.
The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)
So we have to prove that in a group satisfying $$ xyx^{-1} = y^2 \qquad (R_1)$$ $$ yzy^{-1} = z^2 \qquad (R_2)$$ $$ zxz^{-1} = x^2 \qquad (R_3)$$ the elements $x$, $y$ and $z$ are trivial.
By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x \qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. \qquad (R_2^{[i]})$$
If we now conjugate $(R_3)$ by $y$, the left-hand side becomes $$\begin{align}yzxz^{-1}y^{-1} &= z^2y\cdot x \cdot y^{-1}z^{-2}\\ & = z^2y^{-1}xz^{-2}\\ & = z^2 y^{-1}z^{-2}\cdot z^2xz^{-2}\\ & = y^{-1}z^2\cdot x^4 \end{align}$$ (the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).
On the other hand, the left side becomes $$\begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \\ &= y^{-1}xy^{-1}x \\ &= y^{-3}x^2 \end{align}$$ (the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).
Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives $$z^2 = y^{-2}x^{-2}.\qquad (R^*)$$
If we conjugate $y$ by $z^{-2}$, we now get on the one hand $$\begin{align}z^{-2}y z^2 &= x^2 y^2 \cdot y \cdot y^{-2} x^{-2} \\ &= x^2 y x^{-2} \\ &= y^4 \end{align}$$ (the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to $$z^2 = y^3.$$
This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.
Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.
Best Answer
Yes. What exactly $\langle X|R\rangle$ means? We start with a free group $F(X)$ on $X$ and then we take the quotient $F(X)/N(R)$ where $N(R)$ is the normalizer of some subset $R\subseteq F(X)$.
So what is $\langle e| e\rangle$? While the "$e$" notation may be misleading (a typical symbol for identity) this is the same as $\langle g|g\rangle$, just by simple relabelling. And formally this is the same as $\langle \{g\}|\{g\}\rangle$ which by definition is the same as $F(g)/N(g)$. But since $g$ generates $F(g)$ and $\langle g\rangle\subseteq N(g)$ then $N(g)=F(g)$. And so the quotient $F(g)/N(g)$ is trivial.
Side note: When we write presentations, e.g. $\langle x,y\ |\ xy\rangle$ we often write it in a more readable way, e.g. $\langle x,y\ |\ xy=e\rangle$. And so we introduce the neutral element "$e$" symbol to see it better, but formally we don't actually need this neutral element directly. And so care has to be taken when you use the same symbol for a generator and the neutral element. Ideally you would never do that. Otherwise you are likely to run into problems, for example we can write $\langle g|g\rangle$ as $\langle g|g=e\rangle$ but we cannot write $\langle e|e\rangle$ as $\langle e|e=e\rangle$ because of the ambiguity. But we can write $\langle e|e\rangle$ as $\langle e|e=1\rangle$ if we denote the neutral element by "$1$". However you won't see such presentations often, it doesn't feel natural.