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.

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$[G:H]=4$ 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
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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
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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
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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 $\implies Z(\mathrm{Aut}(G)) = \{1\}$....
David C. Huang's user avatar
1 vote
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233 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
46 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
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6 votes
3 answers
441 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
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1 vote
0 answers
26 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
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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
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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
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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
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1 vote
1 answer
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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
176 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
84 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
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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
71 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
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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
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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
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8 votes
1 answer
191 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
127 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
256 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
141 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
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40 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
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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
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4 votes
0 answers
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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
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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
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1 answer
88 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
71 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
302 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
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1 vote
1 answer
219 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
3 votes
0 answers
63 views

Equivalent of simple groups in topology

I read about the classification theorem of finite simple groups, and I was wondering if there is a topological meaning to spaces that have a simple fundamental group? In addition, is there something ...
Kirill Sm's user avatar
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1 answer
106 views

On Subgroups of Same order of Simple group

We know that, If there exists none other than only subgroup $H$ of order $m$, then $H$ is normal in $G$. When It comes to simple group, There should exist at the least $2$ subgroups of same order. To ...
user avatar
1 vote
1 answer
22 views

Image of minimal degree representation of quasisimple group unique up to conjugacy.

Let $ G $ be a quasisimple finite group. Let $ d_{min} $ be the minimum dimension of a nontrivial irrep of $ G $. Must it be the case that the image of all (nontrivial) dimension $ d_{min} $ irreps of ...
Ian Gershon Teixeira's user avatar
1 vote
2 answers
127 views

What is the relation of word "simple" with math-word "simple group" of group theory?

Is there any relation with the meaning of word "simple" with what "there are groups $G$ in which the only normal subgroups are the trivial ones: $1$ and $G$. Such groups are called ...
Sensebe's user avatar
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5 votes
1 answer
222 views

Is every finite non-abelian simple group generated by involutions?

Let $ G $ be a finite non-abelian simple group. Is it true that the set of involutions $$ \{ g: g\in G, g^2=1 \} $$ generates $ G $? For example consider the group $ G=A_5 $ of order $ 60 $. The ...
Ian Gershon Teixeira's user avatar
9 votes
0 answers
155 views

Why is studying centralizers the/a key to classifying finite groups?

In this MO thread https://mathoverflow.net/questions/38161/heuristic-argument-that-finite-simple-groups-ought-to-be-classifiable, Borcherds says One problem, as least with the current methods of ...
D.R.'s user avatar
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11 votes
0 answers
325 views

Why do we trust the Classification of Finite Simple Groups?

It seems to me there are a two main reasons to believe a theorem/conjecture to be true: Because it has a correct proof (e.g. the Feit-Thompson Theorem, Dirichlet's Theorem) Because there is an ...
Zoe Allen's user avatar
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0 votes
1 answer
200 views

Let $G$ be a simple group. Show that any homomorphism from $ G$ to $G'$ (arbitrary $G'$) must be either injective or the trivial homomorphism. [duplicate]

I've spent a lot of time thinking about this question but I can't seem to come up with anything of substance. I tried to use some basic lemma for group homomorphisms, but nothing seemed to work. My ...
Joseph_Kopp's user avatar
1 vote
1 answer
49 views

Simplicity of a subgroup of $S_p$

The following is an exercise, to prove simplicity of a group $G$, which satisfies conditions - i) $G$ is transitive subgroup of $S_p$ (symmetric group, $p$ is prime) ii) $|G|=p\cdot m\cdot k$, iii) $m\...
Maths Rahul's user avatar
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0 answers
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Statement explanation of a paper about finite simple groups

I was trying to understand the statement of the main theorem of this paper https://arxiv.org/pdf/1510.03665.pdf The statement says that if $S$ is a finite non-abelian simple group and $r$ is an odd ...
W4cc0's user avatar
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0 votes
1 answer
48 views

Are complex reflection groups never perfect?

This is a follow-up to Conceptual reason why Coxeter groups are never simple A complex reflection group is a finite subgroup of $ U_n $ that is generated by pseudo reflections. A pseudo reflection is ...
Ian Gershon Teixeira's user avatar
1 vote
1 answer
72 views

Group mod center is perfect structure result

Suppose that $$ 1 \to Z(G) \to G \to P \to 1 $$ is a short exact sequence of groups where $ P $ is perfect and $ Z(G) $ is the center of $ G $. Must it be the case that $ G $ is the direct product of ...
Ian Gershon Teixeira's user avatar
2 votes
1 answer
151 views

A question about the classification of compact Lie groups

I was able to prove that every compact Lie group is isomorphic to $$(T^n\times G_1\times \cdots \times G_m)/\Gamma$$ where the $G_i$ are compact, simply connected, simple Lie groups and $\Gamma$ is a ...
slowspider's user avatar
  • 1,047
2 votes
1 answer
169 views

Prove that $A_5$ is simple.

Prove that $A_5$ is simple. The solution presented is as follows: Let $N$ be a non trivial normal subgroup of $A_5.$ If $(12)(34)\in N,$ then $$(25)(34)(12)(34)(34)(25) = (15)(34)$$ and $$(12)(34)(15)...
Arthur's user avatar
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0 votes
1 answer
113 views

For $n\ge 5,\forall e\neq\rho\in A_n,\exists\sigma=\alpha\rho\alpha^{-1}$ for some $α\in A_n$ s.t. $ρ(i)=σ(i)$ for some $i\in\{1,2,...,n\}$, but $ρ≠σ$

For $n\ge 5$, for all $e\neq \rho\in A_n, \exists \sigma=\alpha\rho\alpha^{-1}$ for some $\alpha\in A_n$ such that $\rho(i)=\sigma(i)$ for some $i\in\{1,2,...,n\}$, but $\rho\neq\sigma$. The solution ...
Arthur's user avatar
  • 2,564
2 votes
0 answers
59 views

Is every finite group a subgroup of this type of generalization of almost simple groups?

Recall, a group $G$ is an almost simple group if there is an non-abelian simple group $S$ such that $S \leq G \leq \operatorname{Aut}(S)$ We will define a $k$-almost simple as follows. When $0$--...
JoshuaZ's user avatar
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2 votes
0 answers
55 views

random generation of lie algebras

It is well known that a nonabelian finite simple group, say $\mathrm{PSL}_n(\mathbf{F}_p)$, can be generated by two elements. In fact, the probability that two elements generate it tends to $1$ as the ...
darko's user avatar
  • 1,191
1 vote
1 answer
107 views

Why can't a group of order 132 contain 3 Sylow $2$-subgroups and be simple?

In the question titled Prove that if |G|=132 then G cannot be simple, it is shown a group of order $132$ cannot be simple. My summary of the proof: assume group is simple deduce the number of Sylow $...
zabop's user avatar
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2 votes
1 answer
53 views

Minimal normal subgroups of normal closure

This question arose from my attempt at understanding the answer in this post - the comments below it, to be precise. Everything revolves around the following problem: Let $S$ be a simple, nonabelian ...
Gauss's user avatar
  • 2,289
4 votes
1 answer
104 views

Simple subgroups and the Wielandt subgroup

My goal is to prove the following: Let $G$ be a finite group and let $S \leq G$ be a simple subgroup. Suppose that $SH = HS$ for all subnormal subgroups $H$ of $G$. Show that $S$ is contained in the ...
Gauss's user avatar
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