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Questions tagged [group-presentation]

For questions concerning groups defined via a presentation by generators and relations. Should probably be used along with the general (group-theory) tag.

0 votes
1 answer
42 views

I am currently taking a course on Free groups and we have the following proposition from B. Neumann, 1937: if $ G = ⟨x_1,...,x_n|r_1,...,r_m⟩ = ⟨y_1,...,y_k|S⟩$, then there exists a finite subset $S_0=...
Cactus's user avatar
  • 55
3 votes
1 answer
160 views

Let $G$ be a group with presentation $\langle S \ \vert \ R \rangle$. Then any $g \in G$ has (in case $R$ not trivial in general non unique) the form $g=s_1^{a_1}s_2^{a_2}...s_n^{a_n}$ with $s_i \in S$...
user267839's user avatar
  • 10.1k
3 votes
1 answer
124 views

Let $G$ be a group with presentation $\langle S \ \vert \ R \rangle$. Then any $g \in G$ has (in $R$ not trivial in general non unique) the form $g=s_1^{a_1}s_2^{a_2}...s_n^{a_n}$ with $s_i \in S$ and ...
user267839's user avatar
  • 10.1k
1 vote
0 answers
40 views

The Todd-Coxeter algorithm is a useful tool to study groups defined by generators and relations. It computes the action of the group on cosets, giving a homomorphism of the group to the group of ...
liyiontheway's user avatar
0 votes
0 answers
52 views

Let $\mathbb{Z}_{p}$ be the ring of $p$-adic integers and let ${\rm SL}_{2}(\mathbb{Z}_{p})$ be the two dimensional special linear group over $\mathbb{Z}_{p}$. Note that ${\rm SL}_{2}(\mathbb{Z}_{p})$ ...
stupid boy's user avatar
2 votes
1 answer
109 views

I would like to know if there is some literature (or low hanging results) on the construction below, or related constructions for that matter. I'm interested about what can be said in each direction (...
augustoperez's user avatar
  • 3,388
0 votes
2 answers
134 views

Consider the group $G = \langle a, b : a^4 = 1, a^2 = b^2, bab^{-1} = a^{-1}\rangle$. This is a standard presentation of the quaternion group. But why could it not be the Klein four group? My ...
hdecristo's user avatar
  • 1,265
1 vote
0 answers
98 views

Let's say $S_3=\langle x,y \mid x^2,y^3,xyxy \rangle =\langle A\mid R \rangle$. Let $X$ be the subgroup of $F(A)$ that is generated by the element in $R$. $\phi$ is a homomorphism from $F(A)$ to $S_3$....
AmazingBBoy's user avatar
6 votes
1 answer
137 views

In the group of isometries of the plane, let $r$ be a rotation by $\frac{2\pi}5$ radians and let $s$ be a translation. I'd like to find a finite presentation of the subgroup $G=\langle r,s\rangle$. ...
Karl's user avatar
  • 13.5k
3 votes
0 answers
98 views

It is well known that the symmetric group $S_n$ ($n\geq 3$) is $2$-generated. For example, one may take as generators $(1\ 2)$ and $(1\ 2\ \ldots \ n)$. This allows us to write every element of $S_n$ ...
Jacob's user avatar
  • 3,430
0 votes
1 answer
102 views

I'm trying to understand this statement. The free group $F_2=\langle a,b\rangle$ is freely generated, so by the definition there is a homomorphism from $F_2$ to $\mathbb{Z}\oplus\mathbb{Z}=\{c^m,d^n\...
itkyitfbku's user avatar
2 votes
1 answer
112 views

I am looking for a reference that proves that $SL(2,\mathbb{Z}) = \langle x,y | x^4 , (xy)^3 = x^2 \rangle$. It sounds like a basic question, but I did not find a proof of this claim. It is mentioned ...
Johana T's user avatar
  • 187
2 votes
1 answer
73 views

A group $G$ is said to be algebraically fibred if there exists an epimorphism $G\to\mathbb{Z}$ with finitely generated kernel. I want to study wheter $G=\langle a,b\mid a^2=1,abab=baba\rangle$ fibres ...
Marcos's user avatar
  • 2,171
1 vote
0 answers
100 views

The free product $ \mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/3\mathbb{Z}$ contains free nonabelian subgroups. Let's see presentation of a groups: $ \mathbb{Z}/2 = \langle a \mid a^2 = e \rangle$ $ \mathbb{...
Miganyshi's user avatar
-3 votes
1 answer
336 views

Using Tietze transformations show that the group $G = \langle x, y, z \mid (xy)^2xy^2\rangle$ is a free group of rank 2. It means that $G = \langle \left \{ w_1, w_2 \right \} \mid \varnothing \...
Antony's user avatar
  • 85

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