All Questions
109 questions
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Choice of group action on Sylow subgroups
This question is in relation to exercises of the form "show that a group of order $n$ is not simple."
Suppose $G$ is a finite group, and $p$ is a prime dividing the order of $G$. Generally, ...
- 13
2
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1
answer
108
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Prove no group order 96 is simple
Does this proof hold?
$96=2^53$
Let $H$ be a 2-sylow subgroup in $G$.
Then $|H|=32$
I know by a prior lemma that $[N[H]:H]$ is congruent to $[G:H]$ (mod p) so therefore $|N[H]|/32$ is congruent to 1 (...
1
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0
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63
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About Hall subgroups. Given $\pi$, with at least 2 prime numbers, prove that there exists a finite group G such that G has not a Hall $\pi$-subgroup
Given $\pi$, a set of prime numbers with at least 2 prime numbers, prove that there exists a finite group G such that G has not a Hall $\pi$-subgroup.
Given a set of prime numbers, $\pi$, I want to ...
- 59
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1
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78
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Simple group of order $168$ embeds into $A_8$
Show that a simple group of order $168$ must be isomorphic to a subgroup of $A_8$.
I read the link here and I didn't find direct information about this particular problem. But here is my attempt at ...
- 3,023
8
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2
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173
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When the number of $p$-Sylows in a Simple group is odd and $p$ is odd
From a paper by Hall, if $n$ is the number of $p$-Sylow subgroups in some finite group $G$ for some prime $p$ dividing $|G|$ then $n$ is a product of numbers of two kinds
Those numbers which are the ...
- 3,178
10
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1
answer
401
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No simple group of order 756 : Burnside's proof
I'm interested in a proof of the non-simplicity of groups of order 756. W.R. Scott, Group Theory, p. 392, exerc. 13.4.9, gives it as an easy exercise, but depending on rather advanced results.
I have ...
- 1,847
1
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0
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29
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On the number of invariant Sylow subgroups under coprime action - Antonio Beltrán and Changguo Shao article
This is an article that Antonio Beltrán and Changguo Shao wrote. Lemma 2.5. states: [All groups are supposed to be finite (this is mentioned before)]
Lemma 2.5. Let $A$ be a group acting coprimely on ...
4
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1
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248
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Problem 5C.3 Isaacs' Finite Group Theory
I have a question about the following problem [Finite Group Theory, Martin Isaacs, Chapter 5]:
Let $G$ be simple and have an abelian Sylow 2-subgroup $P$ of order $2^{5}$. Deduce that $P$ is ...
- 171
2
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1
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132
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Prove the index of a proper subgroup of a simple group of order 17971200 is at least 14.
I didn't find a solution for this problem or other usual approaches that could directly work. So, here is my attempt. I am self-studying and reviewing group theory recently, and would like to know if ...
3
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2
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135
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No simple group of order $p^nq^m$, with barely invoking Sylow theorems
It is a well known fact that for two distinct primes $p$ and $q$, and natural numbers $m, n \geq 1$, there can be no simple group of order $p^nq^m$. Most proofs I have seen of this statement either ...
- 713
2
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1
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405
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"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 ...
- 425
2
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1
answer
176
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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. (...
- 105k
2
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1
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642
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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....
- 105k
3
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1
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306
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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....
- 105k
1
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1
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177
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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 $...
- 1,031
7
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1
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318
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Isaacs Character Theory - exercise 4.11
I am trying to prove the following statement from Isaac's "Character theory of finite groups":
Let $G$ be simple and let $S \in \operatorname{Syl_{2}}(G)$ be elementary abelian; $|S| = q.$ ...
- 583
4
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3
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375
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Sylow-$2$ Subgroups of a simple group of order 168
I'm stuck somewhere in the following claim, I would appreciate if you could help:
Claim: Let $G$ be a simple group of order $168(=2^3\cdot 3\cdot 7).$ Then all Sylow $2$-subgroups of $G$ are dihedral.
...
- 457
4
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3
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572
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Is a group of order $2^kp$ not simple, where $p$ is a prime and $k$ is an positive integer?
Is a group of order $2^kp$ not simple, where $p$ is a prime and $k$ is an positive integer?
I did this for the groups of order $2^k 3$. Here the intersection of two distinct Sylow $2$-subgroups (if ...
- 370
0
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1
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68
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Let $G$ be a simple group which acts on $\Omega$. Let $\alpha \in \Omega$ such that $|O(\alpha)|=p$. Prove the order of $p$-sylow subgroup is $p$.
Let $G$ be a finite simple group which acts on $\Omega$.
Let $\alpha \in \Omega$ such that $|O(\alpha)|=p$, ($O$ is the orbit of $\alpha$, $p$ is a prime number).
Prove the order of $p$-sylow subgroup ...
- 2,322
0
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0
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54
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How to show certain subgroups of $S_p$ are simple?
I met the following exercise: Let $G$ be a transitive subgroup of $S_p$, where $p$ is a prime and the notion of transitive is defined in the usual sense (action on $\{1,2,...,p\}$). Let $H$ be the ...
- 1,511
3
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1
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526
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Let $G$ be a group of order $120$ has a normal subgroup of order $2$ or index $2$.
Let $G$ be a group of order $120$ with no normal subgroup of order $5$. Then $G$ has a normal subgroup of order $2$ or index $2$.
My attempt: Let $n_5$ denote the number of Sylow $5$-subgroups of $G$....
- 1,637
3
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2
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178
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A finite nonabelian group is not simple if any two of its elements who are conjugate to each other commute.
Problem: Let $G$ be a finite nonabelian group. Assume any two elements $x,y \in G$ conjugate to each other also commute, show that $G$ is not simple.
I write down a proof of this problem based on its ...
- 1,103
7
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0
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132
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Proof there are no perfect groups of order 3024
How can I prove that there are no perfect groups of order $3024$?
My attempt is the following:
Each non-trivial finite perfect group admits a non-abelian simple quotient.
This holds because if the ...
- 9,694
1
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0
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41
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No group of order 160 is simple [duplicate]
I'm trying to prove that no group of order 160 is simple. The following is my approach.
Let $G$ be a group of order $160$. Note $160 = 2^55$. I can easily get that $n_2 = 1$ or $5$ and $n_5 = 1$ or $...
- 63
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1
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851
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There is no simple group of order $36$.
I tried to do this as an exercise and wanted to ask if my proof is correct or if it is missing something. Thank you so much.
Let $G$ be a group such that $\lvert G \rvert = 36 = 2^2 \cdot 3^2.$ Show ...
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0
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113
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If $|G| = n < 60$, and $n$ is composite, then $G$ is not a simple group. Why? [duplicate]
If $|G| = n < 60$, and $n$ is composite, then $G$ is not a simple group. I am not totally sure how to solve this.
So far, I have tried thinking of every possible theorem I can think of.
A simple ...
user856333
4
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0
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767
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No simple group of order $2025$
There is no simple group of order $2025$.
If $G$ is simple then by the Sylow theorem, there must be $81$ Sylow $5$-subgroups and $25$ Sylow $3$-subgroups. Also, I can see that if Sylow $5$ subgroups ...
- 101
1
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1
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175
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If $G$ is a simple group of order 168, $n_3=28$, then a Sylow 3-subgroup $T$ of $N_G (P_3)$ acts transitively on the set of Sylow 7-subgroups of $G $
I am looking at Abstract Algebra, 3rd ed., by Dummit and Foote, page 208. In classfying groups of order $168$, we first assume that there is a simple group of order $168$ and prove that
(1) $G $ has ...
- 3,830
3
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1
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454
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Number of conjugacy classes of elements of order $7$ in a group of order 168
Let $G$ be a simple group of order $168$ (Here, we don't assume we know there is a unique such group). Compute the number of conjugacy classes of elements of order $7$ in $G$ [Hint: Consider Sylow $3$-...
- 5,157
1
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0
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93
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A detail in the proof that a group of order 168 is simple
I am doing an excercise from Gallian's Contemporary Abstract Algebra that wants me to prove that the group $PSL(2,Z_7)$ is simple. I have reached the point $n_2=7$, $n_3=7$ or $28$ and $n_7=8$.
I ...
2
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0
answers
211
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Prove that a group of order $540$ is not simple
Prove that a group of order $540$ is not simple
In this problem I have done the following steps.
Sylow's Theorem gives $n_3=10$ and $n_5=36$. Let $P$ be a Sylow $3$ subgroup and $Q$ a Sylow $5$ ...
4
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0
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143
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No Simple Group of Order 1080
I came up with what I think is a nice elementary proof, and I just want to double-check there are no mistakes:
Proof that there is no simple group $G$ of order $1080=2^33^35$.
Suppose $G$ is simple.
...
- 8,406
3
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2
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68
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Any group of order $5^2q^m$ is not simple if both $q$ is a prime and $m$ a positive integer such that $5^2\nmid q^m!$
Let $G$ be a group of order $5^2 q^m$ where $q$ is a prime and $m$ a positive integer such that $5^2\nmid q^m!$. Show that $G$ is not simple.
Attempt: Assume that $G$ is simple. The fact that $5^2\...
- 6,653
0
votes
3
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159
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Understanding a proof that $|G|=pq,(p<q) \Rightarrow G$ is not simple.
Proposition
Let $G$ be a finite group and $|G|=pq$ where $p<q$ and $p,q$ are prime. Then, $G$ is not simple.
The questioneer of this page A finite group of order $pq$ cannot be simple. writes the ...
- 3,354
1
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1
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75
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Permutation Group of Order $7920$ and Degree $11$ (Isaacs 8C.2)
Suppose that $G$ is a permutation group of degree $11$ and order $7920=11 \cdot 10 \cdot 9 \cdot 8$. Prove that $G$ is simple.
A hint to the problem is to show that $|N_G(P)|=55$ where $P \in{\rm Syl}...
2
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1
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150
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Simple group of order $1092 =2^2\cdot 3\cdot 7\cdot 13$ has a single conjugacy class of subgroups of index $14$, but no subgroup of index $13$
Let $G$ be a simple group of order $1092 =2^2\cdot 3\cdot7\cdot13.$ Prove that $G$ has a single conjugacy class of subgroups of index $14,$ but no subgroup of index $13.$
Extending from this post: ...
- 428
0
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2
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476
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Show that a group of order $175$ is not simple.
$|G| = 175 = 5^2 \times 7$
After small calculation I found that only possible value of $n_5 = 1$ and $n_7 = 1$.
How to prove that $G$ is not simple group?
0
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1
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92
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There is no simple group of order 400 [duplicate]
There is no simple group of order 400.
My attempt:
Firstly, I started supposing that $G$ was not simple, and using the Sylow theorems I got that $G$ must have $25$ $2$-Sylow and $16$ $5$-Sylow.
The ...
- 17
0
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1
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65
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$|G| = pqr^2$, $G$ not simple [closed]
If $G$ is a group with cardinality $pqr^2$ and $p,q,r$ different primes $\ge 2$, and given $1+np$ doesn't divide $qr^2$ for any $n \in \mathbb{Z}^+$,
How can you show $G$ is not simple?
I feel like ...
user863652
1
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0
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70
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Show that $G$ is not simple. [duplicate]
If $G$ is a group of order $|G|=10^6$, then $G$ cannot be simple.
I hame hoping to get feedback on my proof attempt; I am using:
Lemma: Let $|G|=p^am$, where $a>1$ and $m>1$. Assume $G$ is ...
- 374
2
votes
1
answer
180
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Show that G is not simple where $|G| = 2^2 \cdot 5^3 \cdot 7^9$
Show that if a group $G$ has order $|G|=2^2\cdot5^3\cdot7^9$, then $G$ is not simple.
$\textbf{Def:}$ A group $G$ is $\textbf{simple}$ if and only if $|G|>1$ and $\lbrace 1_G\rbrace$ as well as $G$ ...
- 374
1
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1
answer
660
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Every simple group of order $60$ is isomorphic to $A_5$ - proof by contradiction
I'm trying to prove the theorem that states that
Every simple group of order $60$ is isomorphic to $A_5$.
I'm trying to do it by assuming by contradiction that it doesn't hold and reach
a ...
- 5,138
2
votes
1
answer
207
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Struggling to understand proof, there is no simple group of order $525$
Prove there is no simple group of order $525.$
$N(H)$ is the normalizer of a set $H$.
Proof:Let $L_7$ be a Sylow $7$ subgroup of $G$. It follows that $|N(L_7)|=35$ by the sylow theorems. Let $L$ be a ...
0
votes
1
answer
289
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Calculate the number of order $p$ elements in a group (using Sylow)
I'm having trouble understanding how to calculate the number of elements of order $p$ (being $p$ prime number) in a group $G$ so that
$$|G|=p^a\cdot m,$$
being $a\in\mathbb{N}$, $a\geq 1$, and $p\nmid ...
- 3,609
2
votes
1
answer
177
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Simple group of order 10000
Prove that no group of order 10,000 is simple.
Here is my attempt.
For contradiction, suppose $G$ of order $10000$ is simple. Notice that
$$10,000 = 5^4 \cdot 16.$$
By the Sylow theorems, $n_5 = 1 + ...
- 109
4
votes
0
answers
235
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No simple group of order 720, again
In his Notes on Group Theory, 2019 edition (http://pdvpmtasgaon.edu.in/uploads/dptmaths/AnotesofGroupTheoryByMarkReeder.pdf p. 83 and ff.)
Mark Reeder gives a proof of the non-existence of simple ...
- 1,847
1
vote
1
answer
56
views
Order of elements not in any Sylow $2$-subgroup
Let $G$ be a finite simple group with $P\in{\rm Syl}_2(G)$ being an elementary abelian group. Suppose $ P=C_G(x) $ for all $x\in P\setminus \{1\}$. Show that every element in $G$ is either an ...
- 5,826
5
votes
1
answer
176
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Number of $5$-Sylows of a simple group of order $660$.
Let $G$ be a simple group of order $660$. I am trying to find $n_5$ - the number of Sylow $5$-subgroups of $G$.
I have easily proved that $n_5 \in \{1, 6, 11, 66\}$ using Sylow Theorems. Besides, I ...
- 396
2
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1
answer
90
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$G$ finite simple group with more than $n/p^2$ conjugacy classes then every $p$-Sylow are abelian
Suppose to have $G$ a simple group of order n. Let p a prime dividing n. If G has more than $\frac{n}{p^2}$ conjugacy classes then prove that all p-Sylow subgroups are abelian.
I've no idea how to ...
- 545
2
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1
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1k
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Let $G$ be a group of order $1365$. Prove $G$ is not simple.
I'm having troubles with this.
Let G be a group of order $1365$. Is $G$ simple?
Normally, we aim to find a single Sylow p-subgroup and since its normal, we get the results.
However, factoring $1365=3 \...
- 597