## New answers tagged gr.group-theory

4

Let $G$ be the group of symmetries of a square. Let $H$ be the rotations. Let $H_1$ be generated by the rotation halfway 'round and the flip in the horizontal axis, let $H_2$ be generated by the rotation halfway 'round and the flip in a diagonal. Then $H\cap H_1=H\cap H_2$, but $H_1$ and $H_2$ are not conjugate.

2

The triviality problem (input: a finite group presentation of a group $G$, output: yes/no according to whether $G$ is a trivial group) is not solvable by an algorithm. Indeed just apply the Adian-Rabin theorem to the property "being trivial".
Now we deduce the general case of your question as follows:
Suppose by contradiction that you have an ...

3

Theorem (Elek–Szabo). Let $G$ be an infinite residually finite hyperbolic group with Property (T). Then $P_G$ is a finitely generated sofic group that is not residually amenable.
That's Theorem 3 of Elek, Gábor; Szabó, Endre, On sofic groups., J. Group Theory 9, No. 2, 161-171 (2006). ZBL1153.20040 arXiv:math/0305352.
The construction is also somewhat ...

2

This is a (cw) post to illustrate Claim 2 from the answer by Sean Eberhard. Below is the plot of the list $(\frac1n,\frac{\log f(n)}{\pi\sqrt{2/3}(n/\log n)^{1/2}})$ for $n$ from $1000$ to $10000$.
I used the list of values from OEIS A023893.
As you see after $n$ about 3000 the sequence sort of changes behavior; at about $n=5000$ (to be precise, at $5226$) ...

11

Suppose $G$ is a finite simple group of order $n$ with a nontrivial representation of degree $d$. Then $G$ is isomorphic to a subgroup of $U(d)$. By Collins's sharp version of Jordan's theorem (https://www.degruyter.com/document/doi/10.1515/JGT.2007.032/html), $G$ has an abelian normal subgroup of index at most $(d+1)!$, which must be trivial since $G$ is ...

8

Claim 1: (same as my comment) Let $q_1, \dots, q_k > 1$ be prime powers. Then $G = C_{q_1} \times \cdots \times C_{q_k}$ is isomorphic to a subgroup of $S_n$ if and only if $q_1 + \cdots + q_k \leq n$.
Proof: If $q_1 + \cdots + q_k \leq n$ we can choose disjoint cycles of lengths $q_1, \dots, q_k$, so the condition is clearly sufficient. Conversely ...

8

Based on Sean Eberhard's comment, it follows that if $f(n)$ is the number of non-isomorphic abelian subgroups of $S_n$, then $f(n)\geq p_{\mathbb{P}}(n)$, where $p_{\mathbb{P}}(n)$ is the number of partitions of $n$ into prime parts. It was proved by Roth and Szekeres that
$$\log p_{\mathbb{P}}(n) = \frac{2 \pi}{\sqrt{3}}\Big(\frac{n}{\log n}\Big)^{1/2}\Big(...

3

Yes, there is a combinatorial way to see this.
Firstly, it is much easier to show a baby version of this type of phenomenon. A while ago I read a nice blog post by Qiaochu explaining the identity
$$\int_{0}^1 (2\cos \pi x)^n (2\sin^2\pi x)dx=\begin{cases} 0 &\text{ if n is odd} \\
C_{n/2} & \text{ if n is even} \end{cases}$$
by interpreting this ...

2

We denote by $\mathbb{Z}/m$ the cycle group of order $m$.
Some examples: The finite groups $\mathbb{Z}/{p_1^{a_1}}\oplus \cdots \oplus \mathbb{Z}/p_k^{a_k}$, where the $p_i$ are pairwise distinct primes and the $a_i$ are positive integers such that $\sum_i p_i^{a_i}\leq n$, can be clearly embedded into $S_n$.
Edit: I also thought to be true that whenever $n=...

4

This was not originally an answer to the question you asked but to the converse question about geodesic paths in $X$ mapping to quasigeodesic paths in $\widehat{X}$. But you asked for a reference for that in a comment, so here it is. But note that this result involves extending the given generating set of $G$ to a larger finite generating set by adjoining ...

6

The answer is yes: $\text{Rad}(kG)^s$ is generated as an ideal by $(g-1)^s$ for $G$ an elementary abelian $p$-group and $s \leq p-1$.
Lemma: Let $V$ be a $k$-vector space and let $s \leq p-1$. Then $\text{Sym}^s(V)$ is spanned by the elements $v^s$ for $v \in V$.
Proof: $\text{Sym}^s(V)$ is clearly spanned by products of the form $v_1 v_2 \cdots v_s$. We ...

answered Nov 29 at 18:03

David E Speyer

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3

Under these assumptions, note that $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of ...

6

It is a standard fact that the automorphism group of $G=\mathrm{SL}_2(\mathbf{R})$ as topological group equals $\mathrm{PGL}_2(\mathbf{R})$, which is also the automorphism group of the Lie algebra viewed as $\mathbf{R}$-algebra. In particular, the outer automorphism group as topological group is cyclic of order 2.
I claim that this is the same as abstract ...

2

If $G$ is torsion-free then the question of reversibility of $K[G]$ (that is, does $ab = 0$ imply $ba = 0$) is in fact equivalent to the zero divisor conjecture, for any field $K$.
Connell showed that such a $K[G]$ is prime, so given non-zero $a, b \in K[G]$ there exists $c \in K[G]$ such that $bca \neq 0$. If it were the case that $ab = 0$ then we have $a (...

5

It is equal.
Since the ground field is of characteristic zero and $v$ is symmetric one may as well compute the stabilizer of the polynomial $f:=\sum_{i=1}^nx_i^m$.
Assume $g\in\mathrm{GL}_n(\mathbb C)$ stabilizes $f$. Then it also stabilizes the Hessian $$\det\nolimits_{ij}(\partial_i\partial_jf)\in\mathbb C^*(x_1\cdots x_n)^{m-2}$$ up to a factor. Hence it ...

2

Okay, following the ideas from the comments, I do think there is a counterexample.
Let $G=K_4$, so $W$ is generated by involutions $v_1,\ldots,v_4$ subject to no other relations. Consider the Coxeter element $c=v_1v_2v_3v_4$. We can reverse the autonomous subset $\{v_1,v_2\}$ here to get $c'=v_2v_1v_3v_4$.
Now consider the homomorphism $\phi\colon W\to S_3$ ...

2

(Edit made in view of a comment by Richard Lyons correcting an inaccuracy in my former answer)
A useful invariant of quadratic forms is the Witt ring $W$.
The 1-dimensional form $x^2$ correspond to $1\in W$.
The standard form, which I will denote $A_1$, satisfies $[A_1]=n\cdot 1 \in W$.
It is easy to see that $[A_0]\in W$ is 0 if $n$ is even and it 1 if $n$ ...

10

In addition to all the completely correct answers to this question, it's probably worth spelling out some of the misconceptions that are common to physicists who work in this area and giving the physics answer.
The main thing that arises all the time is that physicists often conflate representations of a group and of its Lie algebra. The reason one can get ...

19

Here is a very minimalistic argument.
$SO(4)$ contains a rank $3$ elementary abelian $2$-group (product of three groups of order two), namely the group of diagonal matrices in $SO(4)$.
$SU(3)$ does not. Any abelian subgroup of $SU(3)$ is conjugate to a group of diagonal matrices, and among the diagonal matrices there are only four whose squares are the ...

1

Somewhat belatedly, here is a class of examples. Let $G$ be a nontrivial free product of two groups (nontrivial means excluding $G = Z_2 * Z_2$). Suppose that $KG$ has a zero divisor (e.g., if at least one of the groups in $G = G_2 * G_2$ contains a torsion element). Then $KG$ has a right zero divisor that is not a left zero divisor. This is a consequence ...

36

Maybe the simplest argument, if you know something about compact Lie groups, is that SO(4) and SU(3) both have rank 2, i.e., they each contain a maximal torus, which is $S^1\times S^1$. Since all maximal tori are conjugate in a compact Lie group, if SO(4) were a subgroup, then, after a conjugation in SU(3), you can assume that SO(4) contains the diagonal ...

3

This conjecture has now been proved by Valentin Bonzom, Guillaume Chapuy and Maciej Dołęga in their paper $b$-monotone Hurwitz numbers: Virasoro constraints, BKP hierarchy, and O(N)-BGW integral.

29

No. Indeed $\mathrm{SO}(4)$ satisfies the following condition (which is a first-order existential formula) but not $\mathrm{SU}(3)$:
$$\exists w,x,y,z: [x,w]\neq 1\neq [y,z],\; [w,y]=[w,z]=[x,y]=[x,z]$$
(it just says there are two commuting pairs of non-abelian subgroups).
Indeed $\mathrm{SO}(4)$ even contains a copy of the direct product of two non-abelian ...

22

It isn't too bad to describe the irreducible representations of $SO(4)$. We can realize $SO(4)$ as $SU(2) \times SU(2) / \langle (- \text{Id}, - \text{Id})\rangle$. To see this, identify $\mathbb{R}^4$ with the quaternions, and identify $SU(2)$ with the norm $1$ quaternions. For $(u,v) \in SU(2) \times SU(2)$, the map $q \mapsto uqv^{-1}$ from the ...

answered Nov 16 at 14:30

David E Speyer

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53

No. There is probably a straightforward representation-theoretic argument, but I am too ignorant of the subject to give one, so here is a topological argument.
If $H \subset G$ are Lie groups with $H$ closed in $G$, then $G/H$ has the natural structure of a smooth manifold without boundary. If $G$ is compact, so is $G/H$, as it is the continuous image of a ...

9

The factorization always exists, but in general is not unique.
Let me identify $A$ with $H$. Clearly, $K$ can be identified with $\ker(f)$. Let $f(A)$ be the image of $A$ in $B$. Then $A$ is an abelian extension of $f(A)$ by $K$. If I understand correctly, you are asking whether given an inclusion $f(A)\hookrightarrow B$, it is possible to find an extension ...

0

If you have access to a soft as Maple , Mathematica or Sage, you can proceed as follows. Let $A,B\in \mathrm{SL}_2$ be conjugate.
Step 1. Using a Grobner basis library, you solve the equation in $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ $XB-AX=0,\det(X)=1$ ($5$ equations in the $4$ unknowns $(a_{i,j}))$.
Since the dimension of the algebraic set of the ...

2

This seems to be true and, like many facts about free groups, can be proved using Stallings' famous folding technique. Here's a sketch of the argument.
Think of $F(B)$ as the fundamental group of a bouquet of circles $Y=Y_1\vee Y_2$, where $F(B_i)=\pi_1(Y_i)$. The map $f$ can now be realised as a morphism of graphs $\phi:X\to Y$, where $X$ also decomposes ...

13

$\DeclareMathOperator\BS{BS}$Since this question goes in several directions, I hope a discursive answer is appropriate.
The question fits into an important family of questions in geometric group theory which look to provide some sort of "algebraic" characterisation of hyperbolic groups. The search for this characterisation starts with two famous ...

20

The answer is 'no', the affine symplectic group cannot appear as a Lie subgroup of any compact Lie group. The reason is that the affine symplectic group contains $\mathrm{SL}(2,\mathbb{R})$ as a Lie subgroup, and $\mathrm{SL}(2,\mathbb{R})$ cannot be a Lie subgroup of any compact Lie group.
To see this, note that any compact Lie group has a bïinvariant ...

8

Let $p$ be a prime, and let $G$ be a non-abelian simple group with $p \mid |G|$. Suppose that $G \leq S_p$ and that $G$ is transitive.
By a theorem of Burnside a non-solvable transitive group of prime degree is $2$-transitive, so $G$ is $2$-transitive.
Let $B \leq G$ be such that $|B| = p$. Then $G = AB$, where $A$ is a point stabilizer. Note that $A \cap B ...

8

I think that this may be a result of H. Wielandt, and that, if so, it was probably done without CFSG. I do not remember Wielandt's proof (if my memory of the existence of such a proof is correct), but I think it can be done without CFSG if $p$ is not a Fermat prime, by the following argument, which requires some block theory, and a Theorem of E. Shult.
$\...

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