20 votes
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 ...
Berci's user avatar
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19 votes
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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 ...
Mariano Suárez-Álvarez's user avatar
15 votes
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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 $1=i^4$.
Wojowu's user avatar
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14 votes
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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 $y$....
Arturo Magidin's user avatar
12 votes

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 ...
Qiaochu Yuan's user avatar
12 votes
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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 ...
user1729's user avatar
  • 31k
11 votes

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 ...
Derek Holt's user avatar
  • 90.2k
11 votes

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 ...
lhf's user avatar
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11 votes
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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 ...
Derek Holt's user avatar
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11 votes
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Why is $\langle a,b,c,d\mid abcda^{-1}b^{-1}c^{-1}d^{-1}\rangle\cong \langle a,b,c,d\mid aba^{-1}b^{-1}cdc^{-1}d^{-1}\rangle$?

The following map works: $$ \begin{align*} \phi: a&\mapsto ac^{-1}d^{-1}\\ b&\mapsto dcb\\ c&\mapsto dc^{-1}d^{-1}\\ d&\mapsto dcd^{-1}c^{-1}d^{-1} \end{align*} $$ This map satisfies $\...
user1729's user avatar
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11 votes
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Can you completely determine a finitely presented finite group?

In fact, in this case you can determine the group! Finiteness is absolutely crucial here, however. Let's say our generators are $a_1,...,a_n$, our relators are $R_1,...,R_m$, and every element of our ...
Noah Schweber's user avatar
10 votes
Accepted

Every group has a presentation

Every group $G$ has a presentation, yes, since $G$ is spanned by $G$ and by all relations $g.h=gh$, with $g,h\in G$.
José Carlos Santos's user avatar
10 votes

If I kill any element $g$ of a group $G$ and as a result $G/\langle\langle g\rangle\rangle$ is killed, is $G$ then cyclic?

The answer is "no". The issue is that the normal subgroup $\langle\langle g \rangle \rangle < G$ that is generated by $g$ can be much larger and more complicated than the ordinary subgroup $\langle ...
Lee Mosher's user avatar
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10 votes
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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 ...
lulu's user avatar
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10 votes
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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 $a=tx^{-1}t$, $b=t^{-1}x$. ...
YCor's user avatar
  • 17.9k
10 votes
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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 ...
Arturo Magidin's user avatar
10 votes
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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, $xy\neq yx$...
kabenyuk's user avatar
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9 votes
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Group presentation of $A_5$ with two generators

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 $(...
Dune's user avatar
  • 7,397
9 votes
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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 ...
pisco's user avatar
  • 19k
9 votes
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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 $Y$, ...
Eric Wofsey's user avatar
9 votes
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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}\...
YCor's user avatar
  • 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\!\...
Moishe Kohan's user avatar
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 ...
Arturo Magidin's user avatar
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 ...
Parcly Taxel's user avatar
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 ...
Arturo Magidin's user avatar
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 $yxy$.
Captain Lama's user avatar
  • 25.8k
8 votes
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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}...
Derek Holt's user avatar
  • 90.2k
8 votes
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How to write the commutator subgroup in terms of the generators of the group?

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 ...
Derek Holt's user avatar
  • 90.2k

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