17 votes
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What is the intuition behind simple Lie groups?

Why are finite-dimensional simple Lie groups so special? Because they admit a full classification on the level of (complex) finite-dimensional simple Lie algebras by combinatorial data, e.g., root ...
Dietrich Burde's user avatar
12 votes

What is the reason that A5 is simple and A4 is not?

In general, $A_{n+1}$ is the group of rotations of the $n$-simplex; explicitly, $A_{n+1}$ acts by permutation matrices on $$\Delta_n = \{ (x_0, \dots x_n) \in \mathbb{R}^{n+1}_{\ge 0} : \sum_{i=0}^{n+...
Qiaochu Yuan's user avatar
12 votes
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Is this "coincidence" about representations of the Monster actually a coincidence?

Griess and Smith prove in Griess, Robert L. jun.; Smith, Stephen D., Minimal dimensions for modular representations of the Monster, Commun. Algebra 22, No. 15, 6279-6294 (1994). ZBL0820.20021. that ...
Jeremy Rickard's user avatar
11 votes

How are simple groups the building blocks?

Let $G=G_0$ be a finite group. Consider the set of proper nontrivial normal subgroups of $G_0$. If this set is empty, then $G_0$ is simple. Otherwise, the set is ordered by inclusion and we may ...
David Hill's user avatar
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11 votes
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How do we compute the order of the Monster group?

After more searching, I found a satisfactory answer in the paper "A uniqueness proof for the Monster" by Griess, Meierfrankenfeld, and Segev. The main theorem states: Let $G$ be a finite group ...
Ted's user avatar
  • 33.9k
10 votes

Sort-of-simple non-Hopfian groups

Here is a nonabelian (countable) example: Let $S$ be a permutation group on a set $X$ with distinguished point. By the wreath product $G\wr S$ I thus mean $G^X\rtimes S$, where $S$ permutes the ...
YCor's user avatar
  • 17.9k
10 votes

How do I show that every group of order 90 is not simple?

This is a very very late answer. But anyway I'll post for the use of people who are learning form this site (like myself). First note that $90=2\times 3^2\times 5$ and of the form $pq^2r$ with three ...
Bumblebee's user avatar
  • 18.3k
10 votes

Proof that all abelian simple groups are cyclic groups of prime order

Let $G$ be an abelian simple group with $|G| > 1$. We want to prove that $G$ is a cyclic group of prime order. First note that because $G$ is abelian, then every subgroup of $G$ is normal. Let $x ...
Ari Royce Hidayat's user avatar
10 votes

Show that the group is not simple

Suppose $G$ is simple and let $1<m\leq 4$ be the index of a subgroup $H$. Then the action of $G$ on the set of left cosets of $H$ yields a nontrivial homomorphism $G\to S_m$, and since $G$ is ...
Matt Samuel's user avatar
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10 votes
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Discovery of the first Janko Group

I was recommended to look into the book "Finite Simple Groups - An Introduction to Their Classification" by Daniel Gorenstein, and indeed, it answers my question to its full extent! Janko's approach ...
Dune's user avatar
  • 7,397
10 votes
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On automorphisms of groups which extend as automorphisms to every larger group

How about this: A Characterization of Inner Automorphisms Paul E. Schupp Proceedings of the American Mathematical Society Vol. 101, No. 2 (Oct., 1987), pp. 226-228 https://www.jstor.org/stable/...
verret's user avatar
  • 6,699
10 votes
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Is there a simple abelian group $G$ with infinite order?

Suppose that $G$ is simple Abelian, if $x$ in $G$ has an infinite order the group generate by $2x$ is a proper normal subgroup since it does not contain $x$. If $x$ has a finite order, there exists $...
Tsemo Aristide's user avatar
10 votes

Rigorous proof of the Jordan-Hölder Theorem

Both invoke Schreier's Theorem (Lang implicitly). Lang's statement of Schreier's Theorem is: Theorem 3.4. (Schreier) Let $G$ be a group. Two normal towers of subgroups ending with the trivial group ...
Arturo Magidin's user avatar
10 votes
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Reference request: a list of (small) finite simple groups

Here is a list of orders of nonabelian simple groups up to 10000. Of course, in addition, there is a abelian simple group of order each prime. You can already see from this short list that the most ...
Derek Holt's user avatar
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10 votes
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What is the reason that A5 is simple and A4 is not?

Since we're being heuristic anyway, it might be easier to think about $S_n$. The reason that $S_n$ doesn't want to have a lot of normal subgroups is that conjugation in $S_n$ moves stuff around ...
hunter's user avatar
  • 30.4k
10 votes

Compact simple group which is not a Lie group

kabenyuk's answer is correct but we don't have to appeal to the Gleason-Yamabe theorem. By the Peter-Weyl theorem a compact (Hausdorff) group $G$ has the property that its finite-dimensional unitary ...
Qiaochu Yuan's user avatar
10 votes
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Isaacs Character Theory - exercise 4.11

For Question 1, to show that $t = |G|/q$, it is sufficient to show that all involutions in $G$ are conjugate. There is a hint on how to do that in the book. If not, choose two non-conjugate ...
Derek Holt's user avatar
  • 90.2k
9 votes
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Is a finite simple group generated by its minimal subgroups?

If $G$ is not generated by all its elements of prime order, let $H$ be the subgroup they do generate. I claim that $H$ is normal, and therefore $G$ is not simple. Given any $h \in H$, we can write $h ...
Misha Lavrov's user avatar
9 votes
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Nonexistence of a simple group of order 576

Let $G$ be simple of order $576 = 64 \times 9$. The number of Sylow $2$-subgroups is 1,3 or 9 by Sylow's Theorem, but $G$ simple implies that it cannot be 1 or 3, so it must be 9. Then the conjugation ...
Derek Holt's user avatar
  • 90.2k
9 votes
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Does every quasisimple finite group have a faithful complex irrep?

No. A quasisimple group has a faithful irreducible representation if and only if the centre is cyclic. Since there are simple groups with non-cyclic Schur multiplier (e.g., $PSL_3(4)$, which has Schur ...
David A. Craven's user avatar
9 votes
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Compact simple group which is not a Lie group

Let $G$ be a compact simple group. Since the connected component $G_0$ of the group $G$ is normal, either $G_0=1$ or $G_0=G$. If $G_0=1$, then $G$ is a totally disconnected group and hence has any ...
kabenyuk's user avatar
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9 votes
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What is a simple (not many relators) presentation of the Monster group?

If you want to work with elements of the monster group, don't use $196882×196882$ binary matrices – or any explicit presentation for that matter. The main problem is that said group has no small ...
Parcly Taxel's user avatar
8 votes
Accepted

$G \cong H$ and $G$ is simple. Then $H$ is simple as well.

Hint: Let $f:H\to G$ be an isomorphism and let $K\subseteq H$ be a normal subgroup. What can you say about $f(K)\subseteq G$?
Eric Wofsey's user avatar
8 votes

If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple.

As you state, $n_5$ is either $1$ or $16$. If it's the former, we are done. If it is $16$, then there is a homomorphism from $G\to S_{16}$ given by the fact that $G$ acts on the $5$-sylow subgroups by ...
Arkady's user avatar
  • 9,340
8 votes
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How to prove that a finite group of order $280$ is not simple?

Goal: Show that $G$ must have exactly $1$ Sylow $p$-subgroup for some prime $p$ dividing $280$. Consider the number, $n_5$, of Sylow $5$-subgroups of $G$. The third theorem tells us that $n_5$ must ...
Kaj Hansen's user avatar
8 votes
Accepted

If $G$ is a simple group of order $60$, how can we there exists some $G\to A_6$ not trivial without using the fact that $G\cong A_5$?

A simple group $G$ of order $60$ will have six Sylow $5$-subgroups. Then $G$ acts transitively on them (by conjugation), so there's a homomorphism $\phi:G\to S_6$ with transitive image. From the ...
Angina Seng's user avatar
8 votes
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Relation between the order of an element of a group and their character in a simple group

Let $\rho$ be a representation affording $\chi$. Then the eigenvalues of $\rho(g)$ are $\pm 1$. Suppose $m$ of them are $1$ and $n$ are $-1$. So $\chi(g) = m-n$ and $\chi(1) = m+n$. If $\chi(g) \equiv ...
Derek Holt's user avatar
  • 90.2k
8 votes
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Is every finitely generated simple group $2$-generated?

I haven't read it thoroughly enough to understand the construction, but this 1986 paper by Guba constructs a finitely generated simple group all of whose $2$-generated subgroups are free, and which is ...
Jeremy Rickard's user avatar
8 votes
Accepted

Prove that the given simple group can be generated by two elements.

First of all, by Cauchy's theorem we know there are some elements $a,b$ with orders $5$ and $3$ respectively. By Lagrange's theorem it follows that $|\langle a,b\rangle|$ is divisible by both $3$ and $...
Mark's user avatar
  • 40.3k
8 votes
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On a minimal nonsolvable group

No such group exists. Then assumed conditions imply that $N$ is a minimal normal subgroup of $G$ or, equivalently, the conjugation action of $G$ on $N$ gives $N$ the structure of an irreducible ${\...
Derek Holt's user avatar
  • 90.2k

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