Mathematics Stack Exchange
Mathematics Stack Exchange

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.

A group is simple if it has no proper, non-trivial normal subgroups (a subgroup $H\leq G$ is proper if $G\neq H$, and is non-trivial is $H\neq 1$). 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.

Simple groups can be seen as the "building blocks" of groups. This is explained in this question.

652 questions
12
votes
4 answers

Why is $A_5$ a simple group?

I am currently trying to understand the following proof that $A_5$ is a simple group. I only have the proof in German, so the following is only my translation. I really want to understand this proof. I have seen other proof (e.g. in this question or…
Martin Thoma
  • 9,821
6
votes
2 answers

Showing group of order 9555 is not simple by using following Hint

There is already another proof of this theorem, but I'm curious about solving this problem as written in Dummit&Foote(Ex.6.2 12) Show there is no simple group $G$ with $|G|=9555$. No simple groups of order 9555: proof Hint: Let $Q \in Syl_{13}(G)$…
Maddy
  • 1,875
3
votes
2 answers

$|G| = pqr$ with $p$, $q$ and $r$ distinct primes. Show G is not simple.

$|G| = pqr$ with $p$, $q$ and $r$ distinct primes. Show G is not simple. I know this might have been asked and answered before. I just wanted someone to tell me if my argument is OK: Let $|G| = pqr$, and assume $p < q < r$. We have at least one…
Wulfgang
  • 617