1
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1answer
15 views

Group of order $|G|=pqr$, $p,q,r$ primes has a normal subgroup of order

Let $p,q,r$ be positive primes, $p<q<r$, and let $G$ be a group with $|G|=pqr$. Show that there exists a normal subgroup $H$ of $G$ of order $qr$. I've seen this post Groups of order $pqr$ and ...
1
vote
1answer
30 views

Sylow subgroups of a non abelian group $G$ with $|G|=21$ and $|G|=39$

I am trying to solve the following exercise: ¿How many Sylow subgroups has a non abelian group $G$ of order $21$ and $39$ respectively. I could do the following: a) $|G|=21=3\cdot 7$. I'll call ...
3
votes
1answer
43 views

There are no simple groups of order $27p$

Statement Show that there are no simple groups of order $27p$ for any $p$ prime number. I got stuck with this problem, I'll write what I've done so far: Suppose $G$ is a group with $|G|=3^3p$, $p$ ...
1
vote
1answer
45 views

Sylow subgroups problem

Let $G$ be a finite group and $p<q$ such that $p^2$ doesn't divide $|G|$. Let $H_p$ and $H_q$ be Sylow subgroups of $G$ with $H_p \lhd G$. Show $H_pH_q \lhd G \space \implies H_q \lhd G$. From the ...
1
vote
1answer
36 views

Sylow p-subgroups and Sylow theorem

Find all Sylow 3-subgroups of $S_3\times S_3$? This is what I already found: Since $O(S_3\times S_3)=36=2^2 3^2$ Sylow- $3$ subgroups have order $9$. If $n_3$ is the no. of Sylow- $3$ subgroups, Then ...
1
vote
1answer
43 views

If $G$ is a finite group whose $p$-Sylow subgroup $P$ lies in its center, then there is a normal subgroup $N$ of $G$ with $P\cap N=\{e\}$ and $PN=G$

If $G$ is a finite group and its $p$-sylow subgroup $P$ lies in the center of $G$, prove that there exists a normal subgroup $N$ of $G$ with $P\cap N=\{e\}$ and $PN=G$
1
vote
0answers
33 views

Two Open Ended Questions in Sylow Theory

Sylow Theorems are very powerful in finite group theory. Two natural questions come to mind: 1) Given a finite group $G$ and a $p$-subgroup $H$ of $G$, how many Sylow $p$-subgroups of $G$ contain ...
3
votes
1answer
39 views

A Group Having a Cyclic Sylow 2-Subgroup Has a Normal Subgroup.

I want to prove the following: Let $G$ be a group of order $2^nm$, where $m$ is odd, having a cyclic Sylow $2$-subgroup. Then $G$ has a normal subgroup of order $m$. ATTEMPT: We ...
9
votes
1answer
105 views

A group of order $120$ has a subgroup of index $3$ or $5$ (or both)

What I have tried that number of $2$-sylow subgroup can be $1,3,5$ or $15$.I have solved the problem when the number of $2$-sylow subgroup is $1,3,5$. But I am not able to solve it for $15$. Any help ...
0
votes
2answers
50 views

Number of non isomorphic groups of order $122$, My attempt through Sylow theory.

$|G|=122 = 2 . 61$ No. of sylow $2$ subgroups $= 1$ or $61 = n_2$ No. of sylow $61$ subgroups $= 1 = n_{61}$ Let the group of order $61$ be $H_{61}$ and the group of order $2$ be $H_2$ Then : ...
2
votes
2answers
36 views

If $H$ is a normal subgroup of a finite group $G$ and $|H|=p^k$ for some prime $p$. show that $H$ is contained in every sylow $p$ subgroup of $G$

If $H$ is a normal subgroup of a finite group $G$ and $|H|=p^k$ for some prime $p$. show that $H$ is contained in every sylow $p$ subgroup of $G$ Attempt: $|H|=p^k \implies |G|=p^{n_1} q^{n_2} ...
1
vote
2answers
40 views

Show that if $G$ is a group of order $168$ that has a normal subgroup of order $4$ , then $G$ has a normal subgroup of order $28$

Show that if $G$ is a group of order $168$ that has a normal subgroup of order $4$ , then $G$ has a normal subgroup of order $28$. Attempt: $|G|=168=2^3.3.7$ Then number of sylow $7$ subgroups in $G ...
2
votes
1answer
51 views

Smallest possible odd integer that can be the order of a non-Abelian group.

Smallest possible odd integer that can be the order of a non-Abelian group. Attempt: A non abelian group means $Z(G) \subset G$ . Hence, it suffices to find the smallest odd integer $n$ such that ...
2
votes
1answer
51 views

The smallest composite integer $n$ such that there is a unique group of order $n$?

We need to find the smallest composite integer $n$ such that there is a unique group of order $n$. Attempt: Let us suppose that $n= ab$ is a composite integer where $a,b$ are integers such that ...
2
votes
1answer
43 views

The only group of order $255$ is $\mathbb Z_{255}$ ( Using Sylow and the $N/C$ Theorem)

Let $G$ be a group such that $G =255 = 3.5.17$. Let $H$ be a sylow $17$ sub group of $G$. Then, by Sylow's theorem : the number of sylow $17$ sub groups can be $n=1,18,35,...$ . Since, $n$ should ...
1
vote
1answer
32 views

Let $G$ be a finite group and let $a$ be an element of $G$. Then, $|cl ~(a)| = |G:C(a)|$

Let $G$ be a finite group and let $a$ be an element of $G$. Then, $|cl ~(a)| = |G:C(a)|$ where $cl ~(a)$ refers to the conjugacy class of $a$. The proof in the book which I am reading asks to prove ...
0
votes
2answers
68 views

Use Sylow's theorem to show that $G = HN_G(P)$

I am stumped on this question. Does anyone have some helpful hints or a solution to this question? Thanks! Let $G$ be a group of finite order. Let $H$ be a normal subgroup of $G.$ Let $P$ be a ...
3
votes
1answer
63 views

Proving that the intersection of a Sylow p-group with a normal subgroup is also a Sylow p-group

An exercise from Dummit and Foote pg. $101$ ex. $9$ asks to show the following: Let $G$ be a group of order $p^{a}n$ where $p$ does not divide $n$ and let $N\unlhd G$ so that $|N|=p^{b}m$ where ...
0
votes
1answer
49 views

Inconsistent definition of Sylow p-subgroup

Here is the definition of a Sylow $p$-subgroup from Wikipedia: For a prime number $p$, a Sylow $p$-subgroup (sometimes $p$-Sylow subgroup) of a group $G$ is a maximal $p$-subgroup of $G$, i.e., a ...
-1
votes
1answer
87 views

Elements whose orders are multiple of $p$ [closed]

Let $G$ be a non-solvable group, $N$ an abelian minimal normal $p$-subgroup of order $p^r$ with $p\notin \pi(G/N)$, $N=C_G(N)$ and $K=G/N\cong A_5$. By these assumption we can conclude that $G$ has ...
3
votes
1answer
104 views

Computing values of centralizers in a non-solvable group with a given property

A finite group G satisfies property $P_n$ if for every prime integer $p$, $G$ has at most $(n−1)$ non-central conjugacy classes the order of the representative element of which is a multiple of $p$. ...
7
votes
1answer
140 views

A question about Sylow subgroups

Let $G$ be a finite group and $P\neq\{e\}$ be a Sylow $p$-subgroup of $G$ and $P^g\neq P$ be its conjugate in $G$. If we know that $P\cap P^g\neq \{e\}$, can we conclude that $Z(P)\cap Z(P^g)\neq ...
4
votes
2answers
147 views

Is a finite group always a element-wise product of Sylow subgroups?

Let $G $ be a finite group, and let $ p_1, \ldots, p_n $ be the distinct primes dividing $|G|$. For each $i $, let $ P_i $ be a Sylow $ p_i $-subgroup of $ G $. I seem to recall a theorem saying $ ...
1
vote
3answers
41 views

How many Sylow-$ 3$ subgroup does $G$ have?

Let $G$ be a noncyclic group of order $21$. How many sylow-$3$ subgroup does G have? The possible orders of Sylow $3$ subgroups is $1, 7$. But how to check the exact number?
2
votes
0answers
37 views

Neccessary Condition involving Sylow-Subgroups for $p$-Solvability

Let $G = AB$ and let $P \unlhd A$, where $P \in Syl_p(G)$. If $P$ permutes with all Sylow $q$-subgroups of $B$ with $q \ne p$, then $G$ is $p$-solvable. Any suggestions on how to proof?
0
votes
1answer
25 views

Sylow-Subgroups and arbitrary groups where their order contains the same prime-power.

Let $|G| = p^k m$ with $p$ and $m$ being coprime. Then it is well known that there exists a subgroup $S$ of $G$ with $|S| = p^k$, the so called Sylow-$p$-subgroups. Now let $U \le G$ be some subgroup ...
3
votes
0answers
35 views

Simple question on Sylow subgroups.

Given a finite group $G$, $|G|=p^am$, $P,Q\in\operatorname{Syl}_p(G)$, then we know that $Q=P^x$ for some $x\in G$. Thus every element of $Q$ is the image of an element of $P$ under the isomorphism ...
0
votes
0answers
33 views

On the possible subgroups of a Sylow subgroup

Let $G$ be a finite group, say $|G|=p^am$, with $(p,m)=1$, Let $n$ be the number of $p$-Sylow subgroups of $G$. Call them $P_1,\dots,P_n$. Is true that every subgroup of $G$ of order $p^b$ with $b\le ...
2
votes
1answer
48 views

Question about Sylow Theorem and normalizer

I'm dealing with the following problem. Let $G$ be a finite group, $H$ and $K$ Sylow 3- 5- subgroups respectively of $G$. Suppose that 3 divides $|N(K)|$, show that 5 divides $|N(H)|$. I've ...
0
votes
1answer
54 views

Subgroups of a group of order 60 with a normal subgroup of order 2 (Sylow)

This is the problem 38 of the chapter 24 in the Gallian's Algebra. Suppose that $G$ is a group of order $60$ and $G$ has a normal subgroup $N$ of order $2$. Show that: $G$ has normal ...
1
vote
3answers
27 views

$P\in Syl_p(G)\Rightarrow|G:N_G(P)|=|Syl_p(G)|$

Let $G$ be a finite group s.t. $|G|=p^rm$, where $(p,m)=1$. Let then $P$ be a $p$-Sylow subgroup of $G$, i.e. $P\le G$ with $|P|=p^r$. We want to show that $|G:N_G(P)|$ is the number of $p$-Sylow ...
2
votes
1answer
76 views

Question about $p$-Sylow subgroups of the quotient group

I have been working on the following problem. Let $G$ be a finite group, $N\trianglelefteq G$ and $p$ a prime, then $n_{p}(G/N)\leq n_{p}(G)$. I have beeen trying to solve it, but it seems I ...
0
votes
1answer
47 views

Sylow questions on $GL_2(\mathbb F_3)$.

Consider $G:=GL_2(\mathbb F_3)$. I have to extrapolate as much information about it as I can. Without computations. First of all: I think someone else has already done this before me, hence if you ...
0
votes
3answers
69 views

If Q is a p-Sylow-Group of H there is a p-Sylow-Group P of G with $\phi(P)=Q$ while $\phi:G\rightarrow H$ epimorphism

Let G be a finite group and $\phi: G \rightarrow H$ a group-epimorphism. Proof: If $Q\in Syl_p(H)$ there is a $P\in Syl_p(G)$ with $Q=\phi(P)$.
4
votes
1answer
55 views

Question about Sylow $p$-subgroups

If a group $H$ has order $255$ then the Sylow theorems tell us that it must have a Sylow $p$-subgroup of order $5$ and there are either $1$ or $51$ of them, also there is either $1$ Sylow $p$-subgroup ...
5
votes
2answers
94 views

How many nonabelian groups of order 2009? (Check work)

I just need someone to check this argument. Let $G$ be a nonabelian group of order $2009$. The prime factorization of $2009$ is $7^2 \cdot 41$. Let $n$ be the number of Sylow 7-subgroups. Then $n ...
3
votes
1answer
99 views

A Group of Order $540$ is not simple

Why is a group of order $540$ not simple? The hints I have been given are not helpful. Here's what I have been told. Let $G$ be such a group. Then there are $36$ Sylow $5$-subgroups; let $H$ be ...
2
votes
2answers
101 views

This is about Sylow subgroups of Alternating group $A_n$ (Multiple choice)

This is a question from a competitive exam. For a positive integer $n\ge 4$ and a prime number $p\le n$ denote $U_{p,n}$ to be the union of all $p$-sylow subgroups of alterbating group $A_n$. Also ...
1
vote
0answers
38 views

Fixed Points and Sylow-Subgroups of Subgroups who are also Sylow-Subgroups of whole Group

Let a finite group $G$ act on a set $\Omega$. For some $\alpha \in \Omega$, denote by $G_{\alpha} = \{ g \in G : \alpha^g = \alpha \}$ the stabiliser of $\alpha$ in $G$. I want to show that: (i) if ...
4
votes
2answers
89 views

What do Sylow 2-subgroups of finite simple groups look like?

What do Sylow 2-subgroups of finite simple groups look like? It'd be nice to have explanations of the Sylow 2-subgroups of finite simple groups. There are many aspects to the question, so I envision ...
2
votes
1answer
25 views

Question about a non-abelian group of order $p^2q$

Suppose $p<q$, where $p,q$ are primes and we have a non-abelian group $G$ of order $p^2q$. Is it true that it has a subgroup which is not normal? I try to use Sylow's theorems. We take Sylow ...
2
votes
1answer
61 views

A question about the involution in simple groups.

Let $G$ be a finite simple group of Lie type over a finite field ($F_q$) of order $q$ with characteristic $p\neq 2$. Suppose $S$ is $2$-sylow subgroup of $G$ and is not abelian. I have two question ...
1
vote
2answers
118 views

Sylow $7$-subgroup of a group of order $4\cdot3\cdot5\cdot7$ is normal

How to show that the sylow $7$-subgroup of a group of order $420$ is normal. I Know that it is true using GAP. But how to show it without using GAP. I don't know how to start this. Thanks for any ...
0
votes
1answer
33 views

Ruling out orders when applying Sylow's theorems

Going through examples of applications of the Sylow theorems in Fraleigh's book, when proving that no group of order 36 is simple, after concluding that $| H \cap K | = 3$ for two $3$-Sylows $H$,$K$, ...
7
votes
1answer
106 views

Show that if the prime $p$ divides $|G|$, then $|X|$ is divisible by $p$.

Question : Let $p$ be a prime number that divides the order of the finite group $G$. Let $X$ = $\bigcup_{P \in Syl_p(G)}P$. Show that $|X|$ is divisible by $p$.
1
vote
2answers
35 views

A Criterion for being Sylow p-group

Show that if $H$ is a $p$-group of finite group $G$ and $N_G(H)=H$ then $H$ is a Sylow $p$-group of $G$? Or prove the following more general property,$$[G:H]\equiv1\ (\mod\ p)$$
2
votes
2answers
96 views

An application of Sylow theorems in p-groups!

If $G$ is a finite group of order $p^{n}$ (which $p$ is a prime number) and have only one subgroup of order $p^{n-1}$ ,namely $H$ ,then $G$ is cyclic ! My "proof" is as follows: suppose $$x\in G-H$$ ...
0
votes
1answer
36 views

Normal Sylow $p$-subgroup of a normal subgroup

Any hints for the following question - I am sure that I am missing something very simple here. $K$ is a normal subgroup of $G$ and $P$ is a Sylow $p$-subgroup of $K$. If $P$ is a normal subgroup of ...
0
votes
0answers
23 views

Sylow subgroups and normalizers [duplicate]

Let $H$ be a Sylow 3-subgroup and $K$ a Sylow 5-subgroup of a finite group $G$. Suppose $|H|=3$ and $|K|=5$ and $N_G(K)$, has an element of order 3. Show that $N_G(H)$ has an element of order 5.
1
vote
1answer
33 views

$2$-Sylow subgroups of $S_{14}$

I need to describe what are the $2$-Sylow subgroups of $S_{14}$. I don't know how to start thinking about this since is it too large. I know that all that I must do is to find one of them since all ...