Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could you explain to me why these two presentations present the same group?

$$\begin{align*} &\bigl\langle a,b\ :\ aba=bab\bigr\rangle\\ &\bigl\langle x,y\ :\ x^2=y^3\bigr\rangle \end{align*}$$

Note: This is the braid group on three strands.

share|cite|improve this question
Are you looking for a ‘conceptual’ reason, or do you just want us to give you an isomorphism? – Zhen Lin Jan 23 '12 at 17:57
Actually I want to modify the relations and the generators of one to obtain the other – John Jan 23 '12 at 18:01
up vote 8 down vote accepted

Note that $(ab)(ab)(ab) = (aba)(bab) = (aba)^2$; so there is a homomorphism from $H=\langle x,y\ :\ x^2=y^3\rangle$ to $G=\langle a,b\ :\ aba=bab\rangle$ that maps $x$ to $aba$ and $y$ to $ab$. However, the subgroup of $G$ generated by $aba$ and $ab$ is all of $G$, since $a = (ab)^{-1}(aba)$, and if the subgroup contains $a$ and $ab$, then it contains $b$. So the induced map $H\to G$ is onto.

Now notice that $y^{-1}x$ and $x^{-1}y^2$ satisfy the given relations for $a$ and $b$: $$\begin{align*} (y^{-1}x)(x^{-1}y^2)(y^{-1}x) &= x\\ (x^{-1}y^2)(y^{-1}x)(x^{-1}y^2) &= x^{-1}yy^2 = x^{-1}y^3 = x^{-1}x^2 = x. \end{align*}$$ So there is a homomorphism from $G$ to $H$ that maps $a$ to $y^{-1}x$ and $b$ to $x^{-1}y^2$. But the subgroup generated by $y^{-1}x$ and $x^{-1}y^2$ is all of $H$, since it contains $y^{-1}xx^{-1}y^2 = y$, and hence also $x$. So the induced map $G\to H$ is onto.

Finally, what about the composite maps? $G\to H\to G$ gives $$\begin{align*} a&\longmapsto y^{-1}x &&\longmapsto b^{-1}a^{-1}aba = a\\ b&\longmapsto x^{-1}y^2 &&\longmapsto a^{-1}b^{-1}a^{-1}abab = b. \end{align*}$$ Since the composition is the identity, the first map is one-to-one. Since it was already onto, it is an isomorphism, as desired. (Or check that the other composition is also the identity.)

share|cite|improve this answer

Map the second group to the first by $x \longrightarrow aba$, $y \longrightarrow ba$. You can check that the relations in the second presentation are satisfied by $aba$ and $ba$, so you get a homomorphism.

Now define a map from the first to the second by $a \mapsto xy^{-1}$ and $b \mapsto y^2 x^{-1}$. Again, these elements of the second group satisfy the relations in the first, so you have a homomorphism.

Now these homomorphisms are mutually inverse, e.g $x \mapsto aba \mapsto xy^{-1} y^2 x^{-1} xy^{-1} = xyy^{-1}=x$. Therefore the groups are isomorphic.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.