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I'm trying to solve a question of Hungerford's Algebra:

Show that the group defined by generators $a,b$ and relations $a^2=b^3=e$ is infinite and nonabelian.

I guess a good approach would be to find an infinite and nonabelian group with two generators satisfying the same conditions and let Van Dyck do all the work. I've been thinking about the group of permutations $S_{\mathbb N}$, some groups of matrices, but I couldn't get it. Could anyone give me a hint? Thanks in advance!

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  • $\begingroup$ what about it? Notice that also the group $\{e\}$ satisfies it. The OP is talking about the group with that presentation $\endgroup$ – Jorge Fernández Hidalgo Jun 1 '17 at 23:07
  • $\begingroup$ In other words the quotient $F_2/ \langle\langle a^2,b^2 \rangle \rangle $ $\endgroup$ – Jorge Fernández Hidalgo Jun 1 '17 at 23:07
  • $\begingroup$ here is a similar question. $\endgroup$ – lulu Jun 1 '17 at 23:08
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    $\begingroup$ I think the group with that presentation would be isomorphic to the group coproduct of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ which is relatively easy to describe the elements of. $\endgroup$ – Daniel Schepler Jun 1 '17 at 23:09
  • $\begingroup$ With the OP's approach, the dihedral group $D_6$ would probably suffice for the nonabelian part. $\endgroup$ – Daniel Schepler Jun 1 '17 at 23:15
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Your approach works out, although I don't think its the best route, here is what I tried:

Consider the group $S_\mathbb N$ and consider the permutation $b$ with cycles $(0,1,2),(3,4,5),(6,7,8)\dots$ and the permutation $a$ with cycles $(2,3),(5,6),(8,9)\dots, (3n-1,3n)\dots$. ( the permutation $a$ "connects" the cycles of $b$)

We can see by inspection that $a$ and $b$ do not commute.

We can also see that $ab$ has infinite order (Notice that $(ab)^2$ sends $3n$ to $3(n+1)$ )

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  • $\begingroup$ Dear Jorge, thank you for your gentle answer! I guess I was wrong by thinking Van Dyck could do all work. In fact, by his theorem, there is an epimorphism $f:G\rightarrow H$, where $G=\langle \alpha, \beta: \alpha^2=\beta^3=e_G\rangle$ is the group I defined, and $H$ is the subgroup of $S_{\mathbb N}$ that you set. Then, we have so far only the infinitude of $G$. To show that $G$ is abelian, one can take the function $\beta \mapsto b$, $\alpha \mapsto a$, which is a isomorphism. Am I right? $\endgroup$ – rgm Jun 2 '17 at 11:32
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    $\begingroup$ Nope, we are already done, quotients of abelians are abelians. $\endgroup$ – Jorge Fernández Hidalgo Jun 2 '17 at 15:32
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There is a simple geometric way to produce an infinite nonabelian quotient of this group: let $a$ and $b$ be $180^{\circ}$ resp. $120^{\circ}$ rotations in appropriate planes in $\mathbb{R}^3$. Explicitly, we can let

$$a = \left[ \begin{array}{cc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right], b = \left[ \begin{array}{cc} - \frac{1}{2} & - \frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & - \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{array} \right].$$


Incidentally, the original group is very well-known: it's the modular group $\Gamma = PSL_2(\mathbb{Z})$.

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Hint: consider the set $G$ of strings of symbols $a$ and $b$ that contain no substrings of the form $aa$ or $bbb$. Can you define a group operation on $G$ that satisfies your requirements?

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Here's an argument based on something I learned as "van der Waerden's trick" in the context of free groups: let $X$ be the set of (possibly empty) words in $a,b$ with no substring of $aa$ or $bbb$. Now, consider the group $S(X)$ and the two elements $\phi_1$ and $\phi_2$ defined as follows: $\phi_1$ prepends $a$ to a word, except if it began with $a$, in which case it deletes the $a$. Similarly, $\phi_2$ prepends $b$ to a word, except if it began with $bb$, in which case it deletes the leading $bb$. It is straightforward to check that $\phi_1^2$ and $\phi_2^3$ are the identity, and therefore in fact $\phi_1$ and $\phi_2$ are permutations of $X$. However, $\phi_1 \circ \phi_2 \ne \phi_2 \circ \phi_1$ since the first takes the empty word to $ab$, and the second takes the empty word to $ba$ - which implies the original group is nonabelian. Similarly, $(\phi_1 \circ \phi_2)^n$ are all distinct since the $n$th term takes the empty word to $(ab)^n = abab \cdots ab$ - which implies the original group is infinite.

(And in fact, it is easy to see based on this that the interpretation of any "reduced" word in $X$ gives a nontrivial element of the original group - and with some more work that's not very hard, you can prove that group is isomorphic to $X$ with the multiplication operation being "concatenate and reduce", and the proof that this operation is associative reduces to the associativity of composition in $S(X)$.)

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This group is isomorphic to $\mathbb Z_2\ast \mathbb Z_3$ (for example see Showing $\mathbb{Z}_{2} * \mathbb{Z}_{3} \cong\ (a, b\ |\ a^2 = b^3 = e)$). Also, the free product of two nontrivial groups is infinite as you can readily check. It is as well nonabelian.

(This answer is mentioned in Hungerford’s algebra book.)

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