# Tag Info

## Hot answers tagged group-presentation

Accepted

### Can the map sending a presentation to its group be considered as a functor?

Yes, yes, yes, to all three questions. And it can be done very generally and very nicely for universal algebra $\$ [or even for first order structures]. Let's fix an algebraic signature consisting of ...
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### Trying to prove that $H=\langle a,b:a^{3}=b^{3}=(ab)^{3}=1\rangle$ is a group of infinite order.

Consider an equilateral triangle in the plane and let $r$, $s$ and $t$ be the motions of the plane given by reflection with respect to each of the sides of the triangle. Then $a=rs$ is a rotation of ...
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### Quaternion Group: Determine that $i^4 = 1$.

From $k^2=ijk$ we get $k=ij$, and from $i^2=ijk$ we get $i=jk=jij$. Hence $i=jij=jjijj=j^2ij^2=i^2ii^2=i^5$, so $1=i^4$.
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### How to verify this group presentation is contradictory

Why would we guess a group of order (at most) $9$ from the presentation $$\langle x, y : x^3 = y^3 = 1, yx = x^2y \rangle ?$$ Well, a bit of practice might help. Imagine you have a word in $x$ and $y$....
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### Can the map sending a presentation to its group be considered as a functor?

Giving a presentation $\langle S \mid R \rangle$ of a group amounts to describing it as the cokernel of a map $F(R) \to F(S)$ between free groups. There is a category $C$ whose objects are such maps ...
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### Equivalent group presentation

It is in general a hard (infact, algorithmically undecidable!) problem to determine if two given finite group presentations define isomorphic groups. On the other hand, there are some obvious things ...
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### What is an algorithm for determining if a finitely presented group is finite

No that problem is known to be undecidable in general. Even the problem of deciding whether the group is trivial is undecidable. It is semi-decidable in the sense that if the group defined by the ...
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### How to approach proofs similar to "Show a group, $G$, is infinite if $G = \langle r, s, t\mid rst = 1\rangle$"

$G$ is the set of words on $r,s,t$ subject to the relation $rst=1$. The relation $rst=1$ means that you can replace every occurrence of $t$ by $(rs)^{-1}=s^{-1}r^{-1}$. Therefore, $G$ is the set ...
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### Can we apply relations in a group presentation one by one?

Every element of $N(a,b)$ can be written as a product of terms of the form $g^{-1}a^\epsilon g$ and $g^{-1} b^\epsilon g$ for elements $g \in G$. To complete the proof, you just need to show that you ...
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### How to approach proofs similar to "Show a group, $G$, is infinite if $G = \langle r, s, t\mid rst = 1\rangle$"

One thing I often find clarifying is to try adding relations. If you still get an infinite group after you added a relation then you must have started with an infinite group. Here, for instance, set ...
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### Do you know this finitely presented group on two generators?

Note that the abelianized presentation is $\langle a,b\mid ab^2\rangle$. This suggest changing generators so that $ab^2$ is a generator. Define $x=ab^2$, $t=ab$, so that $a=tx^{-1}t$, $b=t^{-1}x$. ...
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### Is $G=\langle r, f\,:\, r^3=e , f^3=e, fr=r^2f\rangle$ well known?

If $fr= r^2f$ and $r^3=e$, then you have $frf^{-1}=r^{-1}$. But then $f^2rf^{-2}=r$ and $r = f^3rf^{-3}=r^{-1}$. So $r^2=e$; since $r^3=e$, then $r=e$, and so your group is just the cyclic group of ...
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### Suppose $G=\left\langle x, y, t\mid x^7=y^7=t^3=1, txt^{-1}=x^2, tyt^{-1}=y\right\rangle$. Show that $y\in Z(G)$.

That's not correct. For example. Let $$x=(1,2,3,4,5,6,7),\ y=(5,8,9,10,11,12,13),\ t=(1,3,4)(2,7,6).$$ It is easy to check that $x^7=y^7=t^3=1$, $txt^{-1}=x^2$, and $tyt^{-1}=y$. However, $xy\neq yx$...
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Finally, I am able to complete my sketch of the proof. We begin by proving the following: $G := \langle x,y \mid x^3=y^3=(xy)^2 = 1 \rangle$ is isomorphic to $A_4$ Proof: $A_4$ is generated by $(... • 7,397 9 votes Accepted ### Finding presentation of group of order 39 The 13-sylow, assume generated by$x$, is normal in$G$. Let$y$be a generator for a 3-sylow. Then by normality, we have $$yxy^{-1} = x^i$$ for some integer$i$. Hence$i^3 \equiv 1 \pmod{13}$, which ... • 19k 9 votes Accepted ### Presentations of Amalgamated Free Products of Two Groups. A presentation for$H*_LK$is$\langle X,Y\mid R,S,T\rangle$where$T$is as follows. For each$\ell\in L$, choose words$x_\ell$and$y_\ell$representing$\ell$using generators from$X$and$Y, ... • 331k 9 votes Accepted ### Second derived subgroup of Baumslag Solitar group BS(2,3) Yes it's true and there's an elementary proof, which works more generally for an arbitrary Baumslag-Solitar group $$\mathrm{BS}(m,n)=\langle t,x\mid tx^mt^{-1}=x^n\rangle;\quad m,n\in\mathbf{Z}\... • 17.9k 9 votes ### How to show G_{m}\cong G_n if and only if n=m, where G_m:= \langle x,y \mid x(yx)^{m}=y(xy)^{m}\rangle Good question. Every such group G_m is an Artin group of spherical type with the corresponding diagram D given by two vertices connected by an edge labeled 2m+1.$$ \begin{aligned} \circ\!\... • 98k 9 votes Accepted ### Is the presentation of the generalized quaternion group of order 16 on Groupprops wrong, or am I missing something? For the benefit of those who don’t want to go wade through another website, you are talking about the presentation of the generalized quaternion group of order 16, given as$Q_{16} = \langle a,b\mid ... • 399k 9 votes Accepted ### What is a simple (not many relators) presentation of the Monster group? If you want to work with elements of the monster group, don't use$196882×196882$binary matrices – or any explicit presentation for that matter. The main problem is that said group has no small ... • 104k 9 votes Accepted ### Find the order of the group$G = \langle a,b \mid a^5 = b^4 = 1, aba^{-1}b = 1 \rangle$Your group is cyclic of order$10$. As freakish correctly notes, from$a^5=1$you can conclude that either$a$has order$5$, or else is trivial (the order divides$5$). Likewise, from$b^4=1$you can ... • 399k 9 votes Accepted ### Question on the definition of the Dihedral groups No, you are right, it does not make a difference. I think it's there to highlight that we are taking a conjugate, which could be less obvious with$yxy$. • 25.8k 8 votes Accepted ### On groups with presentations$ \langle a,b,c\mid a^2=b^2=c^2=(ab)^p=(bc)^q=(ca)^r=(abc)^s=1\rangle $... I haven't come across a name for this family in full generality, but the special case in which$p=2$was defined and studied by Coxeter in his paper H. S. M. Coxeter, The abstract groups$G^{ m, n, p}...
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The answer is no. If $G$ is a free group of rank greater than $1$ then $[G,G]$ is not even finitely generated. If what you were asking were true, then for any $2$-generator group $[G,G]$ would be ...