# 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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### Why here “the simple group of order $168$ comes nearly as a counterexample”?

I’m reading an old paper by Saad Adnan and I don’t seem to understand a sentence at the very beginning of this paper. Conjecture.$~~$If the finite group $G$ has exactly $2$ conjugacy classes of ...
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### Direct products inside finite simple groups

For a finite group $G$, let $d(G)$ be the largest $k$ such that $G$ admits a subgroup isomorphic to a direct product of $k$ non-trivial groups. I am interested in families $G_n$ of finite simple ...
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### a group $G$ whose order has exactly two prime divisors is not simple, Burnside theorem?

I have to show that a group $G$ whose order has exactly two prime divisors is not simple. I was thinking to use the Burnside theorem which sais that if $\vert G \vert = p^aq^b$ where $p,q$ are primes ...
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### How do we compute the order of the Monster group?

How do we compute the order of the Monster group? The answer is quoted in many places, but when I trace back the references, I can't find any place where it's computed, or even a sketch of the ...
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### On the intersections of maximal subgroups of finite simple groups [closed]

I am looking for a proof or a counterexample for the following proposition: Let $G$ be a finite simple group and $M_{1}$ and $M_{2}$ be two maximal subgroups of $G$ with nontrivial intersection.Then ...
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### Let $G$ be a simple group. Show that if $H$ is a subgroup $G$ such that $[G:H]= 3$ then $H=\{1\}$ and $|G|=3$.

MY ANSWER: Since $G$ is simple the only normal subgroups of $G$ are $\{1 \}$ is $G$ itself. As $G / H$ is a quotient group it follows that $|H|\neq 3$ because otherwise the $H$ index in $G$ ...
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### Simple group of order $p^2 q r$, where $p, q, r$ are distinct primes, is isomorphic to $A_5$

As stated, I need to prove that, up to isomorphism, the only simple group of order $p^2 q r$, where $p, q, r$ are distinct primes, is $A_5$ (the alternating group of degree 5). Now I know the ...
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### Elementary (CFSG-avoiding) proof that the order of a group of composite order is bounded by the square of the order of its largest proper subgroup?

It seems that the following is true. For a finite group $G,$ define $u(G)$ to be the largest order of a proper subgroup of $G$. Then $\left|G\right|\leq u(G)^2$ provided that the order of $G$ is ...
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### Prove that there is no simple group of order $216=2^{3}\cdot 3^{3}$. [duplicate]

Prove that there is no simple group of order $216=2^{3}\cdot 3^{3}$. I feel like I am supposed to use the Index Theorem here but when I use Sylow's Third Theorem I have that $n_{3}\in\{1,4\}$. I am ...
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### Reference request: a list of (small) finite simple groups

I am currently in the midst of a project in which it would be useful to have a list of all (small) simple groups as a means to check calculations, not waste time, verify conjectures for small examples,...
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### Two questions about non-abelian finite simple groups

In OEIS, I found the positive integers $\ n\$ , for which there exists a non-abelian simple group with order $\ n\$ upto $\ 10^{10}\$. It can be found by entering the numbers $\ 60,168,360\$ Only ...
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### Can the fact that $\bigcap _{g\in G}gHg^{-1}\lhd G$ be used to prove that certain groups are not simple?

If $G$ is finite, and $H$ a proper subgroup (which existence is often easy to show by using Sylow Theory), it would be sufficient to prove $core_G(H)=\bigcap _{g\in G}gHg^{-1}\neq \{ 1\}$ to conclude ...
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### Rigorous proof of the Jordan-Hölder Theorem

As far as I am aware, the standard books on abstract algebra (Lang, Dummit, Rotman, Grillet, etc.) do not give a rigorous proof of the Jordan-Hölder Theorem. Here are two examples from Lang and ...
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### Proving that a group of order $p^nq$ for primes $p$ & $q$ is not simple.

Prove that a group of order $p^nq$ for primes $p$ & $q$ is not simple. I've been able to prove the theorem holds for $p=q$ and $p>q$. If $p<q$ the best I've been able to do is use Sylow to ...
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### Classification of finite simple groups $G$ with $\pi(G)=\{2,3,5\}$ [closed]

Let $G$ be a finite group. The set of prime divisors of $|G|$ is denoted by $\pi(G)$. I am looking for the classification of finite simple groups $G$ with $\pi(G)=\{2,3,5\}$.