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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.

4 votes
2 answers
287 views

When looking at proofs for why no group $G$ of some order is simple, one frequently mentioned trick is to define a homomorphism from the group to $S_n$, and show that the kernel is a non-trivial ...
Cs_J's user avatar
  • 65
9 votes
0 answers
219 views

Let $A \subset \Bbb N$ be the set of natural numbers for which there is a nonsolvable group of order $n$; this set is enumerated in OEIS A056866. What is the natural density $$a := \lim_{n \to \infty}...
Travis Willse's user avatar
4 votes
2 answers
193 views

There is a theorem by Wang and Chen that says: when the finite group $A$ acts via automorphisms on the finite group $G$ with $|A|$ and $|G|$ coprime, and $C_G(A)$ is either odd-order or nilpotent, ...
Steve D's user avatar
  • 6,060
0 votes
0 answers
67 views

I’m interested in transfers in group theory. It was I. Schur who first introduced the notion of a “transfer homomorphism” to study the simplicity of a group $G$. Let $G$ be a finite group and $H \leq ...
UGOBEL's user avatar
  • 83
2 votes
0 answers
57 views

It is known that there exists an isomorphism $\mathfrak{su}(2)\oplus\mathfrak{sl}(2,\mathbb{R})\cong\mathfrak{so}^\ast(4)$. Is there some surjection from $\mathrm{SU}(2)\times\mathrm{SL}(2,\mathbb{R})$...
kringelton4000's user avatar
2 votes
0 answers
66 views

I know that the following simple groups have 2-transitive actions: The alternating groups $A_n$ for $n\geq5$ The linear groups $L_n(q)$, except $L_2(2)$ and $L_2(3)$ The unitary groups $U_3(q)$, ...
Core Silverman's user avatar
0 votes
0 answers
55 views

Is there any classification of finite non simple, non solvable group whose all non trivial normal subgroups are non solvable. I am aware of $S_n$ for $n \geq 5$. Is there any other such groups. More ...
cryptomaniac's user avatar
  • 1,007
6 votes
1 answer
309 views

I've been reading up about character theory of finite groups and one result I see popping up a lot is that, if $S$ is a nonabelian simple group, then elements of $\operatorname{Aut}(S^n)$ act as ...
Neil M's user avatar
  • 75
3 votes
1 answer
332 views

I'm looking for an example of a nontrivial group $G$ with the property that for all nontrivial proper subgroups $H$ of $G$, there does not exist a nonidentity automorphism $\alpha:G\to G$ such that $\...
user avatar
2 votes
0 answers
59 views

I'm trying to prove that no group of order $p^2 q$ ($p,q$ distinct prime numbers) is simple. Is my idea correct? Call $\ell_p$ the number of $p$-Sylows in $G$. By the Third Sylow Theorem, the number $\...
hdecristo's user avatar
  • 1,277
2 votes
1 answer
116 views

If a group of order $168$ has more than one sylow $7$ subgroups then is it simple? I am relatively new to solving problems like this. I tried solving this problem as follows: Let us assume that $G$ is ...
Thomas Finley's user avatar
0 votes
1 answer
97 views

I realise that this question has been posted before here, but while trying to solve it myself, I ran into a block and I'm not sure how to proceed. This is how the question has been laid out in a ...
yomama's user avatar
  • 63
2 votes
1 answer
115 views

Can a finite non-abelian simple group be expressed as a product of two it's maximal subgroups with trivial intersection? I think it's not possible. The heuristic behind the thought is that if such a ...
cryptomaniac's user avatar
  • 1,007
5 votes
1 answer
236 views

In "Finite Simple Groups" by Wilson, on page 30, he described the construction of the triple cover of $A_{7}$. He added 18 images of $(2,0,0,0,0,0)$ to the set of 45 vectors which was ...
H.Shahsavari's user avatar
4 votes
1 answer
125 views

Definition 1: A group $G$ is quasisimple if it is perfect (i.e., $G=[G,G]$, its derived subgroup) and $\operatorname{Inn}(G)$ is simple. NB: I know that $\operatorname{Inn}(G)\cong G/Z(G)$. ...
Shaun's user avatar
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