Skip to main content

Questions tagged [simple-groups]

Use with the (group-theory) tag. A group is simple if it has no proper, non-trivial normal subgroups. Equivalently (for finitely generated groups), its only homomorphic images are itself and the trivial group. The classification of finite simple groups is one of the great results of modern mathematics.

Filter by
Sorted by
Tagged with
10 votes
1 answer
359 views

No simple group of order 756 : Burnside's proof

I'm interested in a proof of the non-simplicity of groups of order 756. W.R. Scott, Group Theory, p. 392, exerc. 13.4.9, gives it as an easy exercise, but depending on rather advanced results. I have ...
Panurge's user avatar
  • 1,827
8 votes
3 answers
960 views

"Are there any simple groups that appear as zeros of the zeta function?" by Peter Freyd; why is this consternating to mathematicians?

I would like to understand the "upsetting"-to-mathematicians nature of this question Freyd poses to demonstrate that "any language sufficiently rich that to be defined necessarily ...
Hooman J's user avatar
  • 247
3 votes
2 answers
70 views

Use of correspondence theorem for groups to prove that $o(N) = 2$

Let $G$ be a group and $H \triangleleft G$ simple such that $[G : H] = 2$. I have to prove that if $N \neq \{1\}$, $N \triangleleft G$ and $N \cap H = \{1\}$ then $o(N) = 2$. I know by third ...
Cyclotomic Manolo's user avatar
2 votes
1 answer
86 views

Proof that a group of order $180$ is not simple without Burnside p-complement theorem

A proof that a group of order $180$ is not simple is given here. However, the proof uses Burnside $p$-complement theorem. If you know a proof without Burnside $p$-complement theorem, please let me ...
Akasa's user avatar
  • 71
1 vote
0 answers
27 views

On the number of invariant Sylow subgroups under coprime action - Antonio Beltrán and Changguo Shao article

This is an article that Antonio Beltrán and Changguo Shao wrote. Lemma 2.5. states: [All groups are supposed to be finite (this is mentioned before)] Lemma 2.5. Let $A$ be a group acting coprimely on ...
math_survivor's user avatar
2 votes
1 answer
80 views

A question on an isomorphism between ${\rm PSL}_2(9)$ and $A_6$ [closed]

I found a nice argument proving that ${\rm PSL}_2(9)\cong A_6$ on page 52 of The finite simple groups by Prof. R.A. Wilson. Let $f:A_6\to S_{{\rm PL}(9)}$ with $(123)^f= z\mapsto z+1$, $(456)^f= z\...
Probability enthusiast's user avatar
6 votes
1 answer
87 views

Probability that two elements commute in a noncommutative simple finite group

Good afternoon ! Let $G$ a finite non-abelian group. Let $p_G$ the probability that two elements randomly chosen commute. It is well known that : $$p_G \leqslant \frac{5}{8}$$ The upper bound is ...
LexLarn's user avatar
  • 825
2 votes
1 answer
73 views

Automorphism group of ${\rm PSL}_2(p)$ [closed]

This question is related to this answer by Prof. Holt. I can see why ${\rm PGL}_2(p)$ induces the full automorphism group of a Sylow $p$-subgroup $S$ of ${\rm PSL}_2(p)$. Let $a$ be any automorphism ...
Probability enthusiast's user avatar
3 votes
0 answers
49 views

Which alternating groups are characteristic 2 type?

First, let's recall the definition of characteristic 2 type (I think this is correct, but I am not 100% sure). For any 2-local subgroup $H$ of $G$, if $H$ contains the entire Sylow 2-subgroup, all ...
Core Silverman's user avatar
5 votes
1 answer
237 views

Problem 5C.3 Isaacs' Finite Group Theory

I have a question about the following problem [Finite Group Theory, Martin Isaacs, Chapter 5]: Let $G$ be simple and have an abelian Sylow 2-subgroup $P$ of order $2^{5}$. Deduce that $P$ is ...
Elianna 's user avatar
0 votes
0 answers
29 views

Groups of order $2^n p$ for $n\geq 1$ and $p$ prime with $2^n> (p-1)!$ are non-simple. Is my proof correct?

I'm doing my homework in Group Theory and as part of an exercise, I want to show the following Lemma: Let $n\geq 1$, $p$ a prime, s.t. $2^n > (p-1)!$ and $G$ a group of order $2^n p$. Then G has a ...
Joachim's user avatar
3 votes
2 answers
89 views

Show $\mathrm{Inn}(G)\,\operatorname{char}\,\mathrm{Aut}(G)$ for $G$ a non-abelian simple group

Let $G$ a non-abelian simple group and let $A=\mathrm{Inn}(G)$ and $B=\mathrm{Aut}(G)$. I would like to know the solution to $A\,\operatorname{char}\, B$. However, I know the following. Let $\phi \in \...
Akasa's user avatar
  • 71
2 votes
1 answer
125 views

Prove the index of a proper subgroup of a simple group of order 17971200 is at least 14.

I didn't find a solution for this problem or other usual approaches that could directly work. So, here is my attempt. I am self-studying and reviewing group theory recently, and would like to know if ...
William Chuang's user avatar
1 vote
1 answer
84 views

Isomorphic two abelian subgroups which lies in the some finite union of conjugacy classes of simple group [closed]

Let $G$ be a simple group and $H_1$ and $H_2$ be abelian subgroups of $G$ such that $H_1\cong H_2$ and $H_1,H_2\subseteq Cl_G(e_G)\cup Cl_G(x)$ for some $x\in G$, where $Cl_G(\cdot)$ denotes the ...
pharazphazel's user avatar
0 votes
0 answers
65 views

"$S$ is the unique irreducible $\operatorname{End}(S)$-module" [duplicate]

Let $S$ be a finite-dimensional vector space. Then $S$ is an irreducible $\operatorname{End}(S)$-module. Furthermore I was told that $S$ is the only irreducible $\operatorname{End}(S)$-module. I guess ...
Filippo's user avatar
  • 3,584
3 votes
0 answers
68 views

An example of non simple group which is also a Lie group such that $G$ is a connected Lie group and has no non trivial normal Lie subgroup

I want to know if there exist a non simple group (as abstractl group) $G$ such that $G$ is a connected Lie group and has no non trivial normal Lie subgroup. I have tried some obvious examples like ...
Eloon_Mask_P's user avatar
4 votes
1 answer
155 views

Let $K\lhd G$ be s.t. both $K$ and $G/K$ are simple. Show that either $K$ is the only proper normal subgroup of $G$, or $G \cong K \times (G / K)$.

Sorry about the title, I couldn't fit the whole exercise (Exercise 8.1.6, Nicholson Introduction to Abstract Algebra 4th edition): Let $K \triangleleft G$ be such that both $K$ and $G/K$ are simple. ...
iwjueph94rgytbhr's user avatar
3 votes
2 answers
118 views

No simple group of order $p^nq^m$, with barely invoking Sylow theorems

It is a well known fact that for two distinct primes $p$ and $q$, and natural numbers $m, n \geq 1$, there can be no simple group of order $p^nq^m$. Most proofs I have seen of this statement either ...
paulina's user avatar
  • 695
2 votes
1 answer
100 views

Elements of $\mathbb{Z}/p\mathbb{Z}$ and prove it's simple group

I'm trying to understand properties of groups. Since $\mathbb{Z}/p\mathbb{Z}$ ($p$ is prime) is quotient group It should be set of all left cosets of $p\mathbb{Z}$ in $\mathbb{Z}$ which is, $ \{ m+pn ...
user avatar
2 votes
2 answers
114 views

$G$ is a finite, simple, nonabelian group. If $|G:H| = p$, then $|\{gHg^{-1}\}|=p$

I want to show that if $G$ is a finite, simple, nonabelian group with $H \leq G$ so that $|G:H| = p$ for $p$ is prime, then $|\{gHg^{-1}\}|=p$, i.e. the number of conjugates of $H$ in $G$ is $p$. ...
Grigor Hakobyan's user avatar
0 votes
2 answers
143 views

$[G:H]=4$ where $H\neq 1$ means that $G$ is not simple.

The question is as follows: Suppose $G$ is a finite group with a nontrivial subgroup of index $4$. Prove that $G$ is not simple. I am on the hunt for a non-trivial normal subgroup. This proof ...
Chris Christopherson's user avatar
4 votes
1 answer
115 views

Diagonal in the power of a group

Let $L$ be a simple group and $G=L^t$. The diagonal $D=\lbrace(x,x,\ldots,x), x\in L\rbrace$ is a subgroup of $L^t$. I can prove that if $D$ is maximal then $t$ is a prime. If $t=mn$ then the subgroup ...
marcos's user avatar
  • 149
-1 votes
1 answer
65 views

Are commutator subgroups simple?

I am interested in the following setting: $$ 0\rightarrow [G,G]\rightarrow G \rightarrow \mathbb{Z}^r\rightarrow 0, $$ in particular in the case $r=1$. Is $[G,G]$ simple in this case? I am aware of ...
MathBug's user avatar
  • 404
0 votes
0 answers
55 views

If $G$ is simple, then $Z({\rm Aut}(G)) = \{1\} \iff G$ is non-abelian [duplicate]

NOTE: I am aware of the link at For a Simple Group G, Z(Aut(G)) Is Trivial if and only if G is Non-Abelian. I am looking for proof verification of $G$ non-abelian $\implies Z(\mathrm{Aut}(G)) = \{1\}$....
Grigor Hakobyan's user avatar
2 votes
1 answer
397 views

"Simple" group of order $1004913$ problem, fixed point part

Let $G$ be a group of order $1004913 = 3^3 \cdot 7 \cdot 13 \cdot 409$. We suppose that $G$ is simple. We want to obtain a contradiction. This is the Exercise 29 in Chapter 6.2 of Dummit-Foote. As ...
Kazune Takahashi's user avatar
1 vote
1 answer
50 views

A reference request for $SL(2,q)$ being quasisimple for prime powers $q\ge 4$.

Note: This is a reference-request question and thus does not need the usual type of context. The Question: What is a reference for $SL(2,q)$ being quasisimple for prime powers $q\ge 4$? Background: ...
Shaun's user avatar
  • 45.7k
6 votes
3 answers
494 views

Has every infinite simple group a faithful irreducible representation?

Has every infinite simple group a faithful irreducible representation? This question solves the finite case. However, the proof requires a non-trivial linear representation of a finite group. I want ...
wer's user avatar
  • 399
1 vote
0 answers
34 views

Splitting of an extension

Let $G$ be a group which is the extension of a free abelian group $A$ of finite rank by a finite simple group $S$. Does $G$ splits over $A$? (that is, $G=F\ltimes A$ for some finite subgroup $F\simeq ...
W4cc0's user avatar
  • 4,160
4 votes
0 answers
92 views

Motivation for triple cover of $A_6$ (and $A_7$)

In finite simple groups by Wilson, he constructs the triple cover of $A_6$ by considering the action of the subgroup of $A_6$ preserving the partition $\{12, 34, 56\}$ on the two vectors $(0, 0, 1, 1, ...
Micose's user avatar
  • 41
4 votes
0 answers
92 views

Is there a way to computationally verify that the sporadic groups are simple?

I'm trying to understand the "easy" direction of the CFSG: namely, the proofs that the 18 infinite families and 26/27 sporadic groups are indeed simple. I'm working through Simple Groups of ...
Max Packer's user avatar
0 votes
1 answer
76 views

The only proper normal subgroups of a nonabelian quasisimple group are subgroups of its centre.

This is part of a bunch of exercises set by my academic supervisor. As such, I'm not sure whether all the hypotheses are needed for the conclusion. Please note that hints are preferred. The Question: ...
Shaun's user avatar
  • 45.7k
1 vote
1 answer
90 views

Does every perfect finite group have a fixed point free representation?

Fixed point free representations of finite group are important for the spherical space form problem and also show up in other contexts for example Perfect semi direct products A representation $ \pi $ ...
Ian Gershon Teixeira's user avatar
2 votes
2 answers
235 views

Show $G$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$

Let $G$ be a finite group with at least two (distinct) subgroups of index $2$, and suppose that at least one of the index-$2$ subgroups of $G$ is simple. Prove that $G\cong \mathbb{Z}_2 \times \mathbb{...
Important_man74's user avatar
2 votes
1 answer
90 views

Is there an explicit maximal simple group?

It is not hard to prove the following lemma: LEMMA: Let $(G_i)_{i \in I}$ be a chain of simple groups. Then $G = \bigcup_{i \in I} G_i$ is a simple group. Let $N$ be a normal subgroup of $G$ and ...
Jan Matula's user avatar
0 votes
1 answer
95 views

Show that $ SL(2,5) $ has no subgroup isomorphic to $ A_5 $

I'm trying to show that there's no subgroup of $ SL(2,5) $ isomorphic to $ A_5 $. I've already shown that $ A_5 $ is simple, and my strategy is to show that $ SL(2,5) $ contains no simple subgroups of ...
Grotto Box's user avatar
3 votes
1 answer
83 views

Finite non abelian simple groups whose order is not divisible by 8

Is $A_5$ the only finite non abelian simple group whose order is not divisible by 8? If not, what is the complete set of finite non abelian simple groups whose order is not divisible by 8?
cryptomaniac's user avatar
0 votes
0 answers
42 views

Simplicity of the $A_5$ [duplicate]

In the proof given in Abstract Algebra by Dummit and Foote, page 128, it states that $(1~2)(3~4)$ commutes with $(1~3)(2~4)$ but does not commute with any element of odd order in $A_5$. So it follows ...
Sam's user avatar
  • 25
4 votes
1 answer
156 views

Let $G$ be a finite simple non abelian group and $\{1\}\neq H\le G$ be such that $\vert C_{G}(H)\vert=\vert G:H \vert$. Then $H=G$.

Let $G$ be a finite simple non abelian group and $\{1\}\neq H\leq G$ be such that $\vert C_{G}(H)\vert=\vert G:H \vert$. Show that $H=G$. I can find only $[G,G]=G$ and $Z(G)=\{1\}$ and if $H$ is a ...
user371231's user avatar
  • 2,511
8 votes
1 answer
231 views

Is every finite simple group a quotient of a braid group?

Question: Is every finite simple group a quotient of a braid group? Context: The braid group on two strands $ B_2 $ is isomorphic to $ \mathbb{Z} $ and so the infinite family of abelian finite simple ...
Ian Gershon Teixeira's user avatar
2 votes
1 answer
157 views

There are no simple groups of order $480$

Before this writing the smallest order for which the nonexistence of simple groups of that order is not explicitly demonstrated on this site is 480; this self-answered question aims to fill that gap. (...
Travis Willse's user avatar
2 votes
1 answer
524 views

There are no simple groups of order $336$

Before this writing the smallest order for which the nonexistence of simple groups of that order is not explicitly demonstrated on this site is $336$; this self-answered question aims to fill that gap....
Travis Willse's user avatar
3 votes
1 answer
271 views

There are no simple groups of order $264$

Before this writing the smallest order for which the nonexistence of simple groups of that order is not explicitly demonstrated on this site is $264$; this self-answered question aims to fill that gap....
Travis Willse's user avatar
0 votes
0 answers
45 views

a doubt on the simplicity of the alternating group $A_n$ in Dummit&Foote's Abstract Algebra

I have a doubt about the simplicity of the alternating group $A_n$ in Dummit&Foote's Abstract Algebra on page150 : It's to prove that the alternating group $A_n=G$ is simple for $n\geq5$ . Let $H\...
Dian Wei's user avatar
  • 351
4 votes
1 answer
120 views

To what extent can we determine the simplicity or non-simplicity of groups based on their prime decompositions?

This question may have more of a vague, less objective answer than usual for this site, so I apologise if it difficult to answer definitively. Below, $p$, $q$ and $r$ are distinct primes. A group of ...
Robin's user avatar
  • 3,940
4 votes
0 answers
143 views

Every proper subgroup of a simple group containing an order 45 element has an index of at least 14

Every proper subgroup of a simple group containing an order $45$ element has an index of at least $14$. So far I've supposed that $G$ is a simple group with an element of order $45$. Then $45$ ...
math's user avatar
  • 93
0 votes
1 answer
127 views

Automorphisms of $ SL(n,q) $ coming from $ GL(n,q) $

For which values of $ n \geq 2 $ and $ q $ a prime power are all automorphisms of $ SL(n,q) $ induced by conjugation by elements of $ GL(n,q) $? Old version of the question: $ \Sigma L(n,q) $ is an ...
Ian Gershon Teixeira's user avatar
0 votes
1 answer
120 views

Natural group extension constructed from Schur cover and its outer automorphism group

Let $ S $ be a finite (non-abelian) simple group. Then there always exists a natural extension of $ S $ by the outer automorphism group $ Out(S) $ with elements of $ Out(S) $ acting as outer ...
Ian Gershon Teixeira's user avatar
1 vote
0 answers
82 views

Nontrivial extension of cyclic group by simple group

Let $ G $ be a (non-abelian) finite simple group. An extension $ G\cdot m $ is nontrivial if it is not isomorphic to the direct product $ G \times m $. Suppose that there exists a nontrivial extension ...
Ian Gershon Teixeira's user avatar
5 votes
3 answers
313 views

No simple group of order 1040

Burnside (Proceedings of the London Mathematical Society, vol. 26; Collected Papers, vol. 1, p. 601) gave the following proof of the non-simplicity of groups od order 1040 : "If simple, the group ...
Panurge's user avatar
  • 1,827
2 votes
1 answer
385 views

Show that a finite simple group $G$ of even order must have order divisible by $4$.

It is requested to proof exercise $3.30$, $(ii)$ from J. Rotman, An introduction to the theory of groups, which states that a finite simple group $G$ of even order greater than $4$ must have order ...
Blue Tomato's user avatar

1
2 3 4 5
14