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Questions tagged [sylow-theory]

For questions about Sylow theorems in the context of group theory. Not for use with questions regarding Sylow systems, which belong in solvable-groups.

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Show that $H$ is a normal subgroup of $G.$

This question was asked in my mock test of masters entrance test and I couldn't prove one of the question: Question $\to$ Let $G$ be a group of order $105$ and $H$ be it's subgroup of order $35$. ...
DTPW's user avatar
  • 149
1 vote
1 answer
32 views

Let $G$ be a group of order $p^nq$ where $p$ and $q$ are distinct primes and suppose $q \nmid p^i-1$ for $1 \leq i \leq n-1$. Prove $G$ is solvable

This is an extension of this post. Let $G$ be a group of order $p^nq$ where $p$ and $q$ are distinct primes and suppose $q \nmid p^i-1$ for $1 \leq i \leq n-1$. Prove that $G$ is solvable. This can be ...
Grigor Hakobyan's user avatar
2 votes
1 answer
42 views

If $|G| = p^a q^b$ for $p$, $q$ prime, and $G$ has exactly one Sylow-$p$ subgroup $P$ and one Sylow-$q$ subgroup $Q$, then $G \cong P \times Q$

I am hoping to show that if $|G| = p^a q^b$ for $p$, $q$ prime, and $G$ has exactly one Sylow-$p$ subgroup $P$ and one Sylow-$q$ subgroup $Q,$ then $G \cong P \times Q$. Here's where I've gotten so ...
Jake Khawaja's user avatar
4 votes
0 answers
91 views

If $G$ is a group of order $29 \cdot 30$, then $G$ has a normal Sylow-$29$ subgroup

Write $|G| = 29 \cdot 30 = 29 \cdot 2 \cdot 3 \cdot 5$. Of course, I must show $n_{29}=1$. Toward a contradiction, suppose $n_{29} \neq 1$. Then $n_{29}=30$. Counting nonidentity elements, $G$ has $30(...
Grigor Hakobyan's user avatar
1 vote
2 answers
137 views

If $G$ has automorphism $a \in Aut(G)$, that only fixes the identity and with $o(a)=2$, then $|G|$ is odd.

Let $G$ be finite group and $a \in Aut(G)$ with $o(a)=2$, if $a(g) \neq g$ for all $g \in G \setminus \{1\}$, then $|G|$ is odd. Hey Guys, I wanted to prove the Theorem above and was wondering if my ...
Stippinator's user avatar
10 votes
1 answer
356 views

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 ...
Panurge's user avatar
  • 1,827
1 vote
2 answers
90 views

Question regarding the properties of an automorphism group of a Sylow P subgroup

The context for this question has to do with proving: Groups of order $pq$ with $p < q$ have a normal subgroup of order $q$ and are cyclic iff $q$ is not congruent to $1$ mod $p$. I will leave out ...
froitmi's user avatar
  • 87
2 votes
0 answers
26 views

Quick question regarding a proof of the 1st Sylow Theorem [duplicate]

Here is the statement and proof: (Note: $N_G(H)$ denotes the normaliser of H in G) If $H \leq G $ and $| H | = p^k$ for some $k < n$, then there is some $P \leq G$ with $H \unlhd P$ and $|P| = p^{...
baslerbuenzli's user avatar
5 votes
2 answers
457 views

Generalized Sylow's theorem [closed]

I'm working through some exercises in Alperin and Bell's textbook "Groups and Representations." I came across a very interesting exercise which generalizes Sylow's theorem: Ex 7.4: If $|G|$ ...
Damalone's user avatar
  • 329
1 vote
1 answer
49 views

Is restriction to $p$ Sylow subgroup $\text{res} : H^1(H,M)\to H^1(H_p,M)$ injective?

'Galois cohomology of Algebraic number fields' written by K. Haberland reads the following lemma in page 66. Let $H$ be a finite group. Let $p$ be a prime number and $H_p$ be a fixed p Sylow subgroup ...
Poitou-Tate's user avatar
  • 6,351
1 vote
0 answers
28 views

Normalizer of an $A$-invariant Sylow $p$-subgroup

I was reading the Antionio Beltrán and Changguo Shao article On the number of invariant Sylow subgroups under coprime action and there is a part of the Lemma 2.5. which I do not undertand. First of ...
math_survivor's user avatar
1 vote
0 answers
27 views

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 ...
math_survivor's user avatar
2 votes
0 answers
31 views

Standard representation restricted to $S_5$ of $S_6$ acting on its set of sylow $5$-subgroups.

This is from a practice exam (paraphrasing): Let $\chi$ be a character of $S_6$, given by $\chi(g) = |\text{Fix}(g)|-1$ for the action of $S_6$ on the set of its sylow $5$-subgroups. Decompose the ...
Ben123's user avatar
  • 1,266
2 votes
0 answers
71 views

Show that there is no simple group of order 351

I want to show that there is no simple group of order $351 = 3^3 \cdot 13$. Let $G$ be a group. of order 351. Using Sylow III, I found that $n_3 \in \{1,13\}, \quad n_{13}\in \{1,27\}$. Let us suppose ...
idk31909310's user avatar
2 votes
2 answers
89 views

Let $P \in \text{Syl}_{5}(S_5)$. Show that the normalizer $N := N_{S_5}(P)$ is a monomial group.

Fix the field to $\mathbb{C}$. Let $P$ be a sylow $5$-subgroup of the symmetric group $S_5$. Let $N := N_{S_5}(P)$ be the normalizer of $P$. I want to show that $N$ is a monomial group, that is, that ...
Ben123's user avatar
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0 votes
0 answers
28 views

Show that every group of order $2 · 3 · 5 · 67$ has a normal subgroup of order $5$. [duplicate]

Show that if $G$ is a group of order $2010 = 2 · 3 · 5 · 67$, then $G$ has a normal subgroup of order $5$. I tried use Sylow's theorem to show that $k_5=1$. $k_5$ is equal to $1$, $6$, or $201$.
Ido's user avatar
  • 9
2 votes
1 answer
162 views

Proving that a group of order 360 has 10 Sylow 3 -subgroups ,36 Sylow 5 -subgroups,45 Sylow 2 -subgroups is a simple group

Edit: This is what the OP is trying to ask, I think. Let $G$ be a group of order $360$. Suppose that there are ten Sylow $3$-subgroups, 36 Sylow $5$-subgroups and 45 Sylow $2$-subgroups. Show that ...
Kirin's user avatar
  • 55
6 votes
1 answer
185 views

Classifying finite groups where order is multiplicative on elements with coprime orders

It's well known that $|g_1 g_2| = |g_1||g_2|$ whenever $g_1$ and $g_2$ are commuting elements of a group with $\gcd(|g_1|, |g_2|) = 1$. So, for example, $\gcd(|g_1|, |g_2|) = 1$ always implies that $|...
K_D's user avatar
  • 163
5 votes
1 answer
50 views

Can we conclude that an infinite bounded exponent periodic locally nilpotent group $G$ is the direct product of its Sylow $p$-subgroups?

When $G$ is a finite nilpotent group, we know that $G$ is the direct product of its Sylow subgroups. Now, assume that $G$ is an infinite locally nilpotent group. Also, consider that there exists a ...
Reza Fallah Moghaddam's user avatar
2 votes
1 answer
89 views

Sylow subgroups of semidirect products [closed]

Suppose that $G = A \times B$ is a direct product of finite groups $A$ and $B$. Let $P$ be a Sylow $p$-subgroup of $G$. We have an epimorphism from $G$ to $A$ so that the image of $P$ in $A$ is a ...
Greg's user avatar
  • 422
5 votes
2 answers
108 views

If $G$ is a finite group of order $|G|=936$, then there is a subgroup $H$ of $G$ with $|H|=117$

I'm very confuse with this exercise. The prime number decomposition of $|G|$ is $936=2^{3}3^{2}13$. That is, $G$ is not a $p$-group, has no order $pq$, $p^2q$, $p^mq$ or $p^mq^n$. And $|H|=117=3^{2}13$...
Artur Cruvinel Montibeller's user avatar
2 votes
0 answers
35 views

Normal $p$-complement exercise (Problem 5D.6 Isaacs' Finite Group Theory)

I have a question about the following problem [Finite Group Theory, Martin Isaacs, Chapter 5]: Suppose that $P \in Syl_{p}(G)$, and assume that $A \cap P=P^{'}$, where $A=\textbf{A}^{p}(G)$. If $P^{'}$...
Elianna 's user avatar
6 votes
1 answer
98 views

Can a group be fully characterised by its Sylow $p$-subgroups? [duplicate]

I was wondering if we have two groups, $G$, $H$, such that $|G| = |H|$ (and these orders are finite); and all their Sylow $p$-subgroups have the same structure (i.e. Syl$_p(G)$ isomorphic to Syl$_p(H)$...
AnarQ's user avatar
  • 71
5 votes
1 answer
237 views

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 ...
Elianna 's user avatar
0 votes
1 answer
61 views

Example of 3-solvable group of large 3-length

I would like to construct examples of $3$-solvable groups with large $3$-length. That means that the Sylow $3$-subgroups need to be large in a sense. Is there any general construction of such examples?...
primer's user avatar
  • 220
1 vote
0 answers
60 views

Group of order $2^kq$ has a normal Sylow $2$-subgroup [closed]

Let $G$ be a group of order $2^kq$ for some prime $q>2^k$ , denote $P$ and $Q$ one of the Sylow $2$-subgroup and Sylow $q$-subgroup of $G$ respectively. I have seen a comment that for such a group $...
Dian Wei's user avatar
  • 351
1 vote
0 answers
45 views

If a p-group is normal then can we say G is going to be a direct product of P

I recently came across a theorem that states that if G is a finite group and every p-group of G is normal then G is isomorphic to the direct product of its Sylow p subgroups. To prove this we use that ...
Bigalos's user avatar
  • 394
1 vote
2 answers
64 views

Deducing there exists exactly $5$ isomorphism classes of groups of order $12$.

There's a substantial amount that's been written about the semi-direct products of a group of order $12$ on this website. However, there's something that seems to be taken for granted each time the ...
Ty Perkins's user avatar
0 votes
0 answers
29 views

Groups of order $2^n p$ for $n\geq 1$ and $p$ prime with $2^n> (p-1)!$ are non-simple. Is my proof correct?

I'm doing my homework in Group Theory and as part of an exercise, I want to show the following Lemma: Let $n\geq 1$, $p$ a prime, s.t. $2^n > (p-1)!$ and $G$ a group of order $2^n p$. Then G has a ...
Joachim's user avatar
2 votes
1 answer
67 views

Do all the conjugates of $H\le G$ pairwise intersect in subgroups of the same order?

Let $H\le G$. Do all the conjugates of $H$ pairwise intersect in subgroups of the same order? If not in general, does it hold at least for Sylow $p$-subgroups? The question has a trivial answer when $...
Kan't's user avatar
  • 3,393
-4 votes
1 answer
78 views

Non-Abelian group of order $21$ subgroups [closed]

This is an exam problem from a couple of years ago... I)Let $G$ be a nonabelian group of order $21$. What are the possible orders of elements of $G$? II)Find the number of Sylow $p$–subgroups of G, ...
Aristarchus_'s user avatar
2 votes
1 answer
73 views

When does every element of Sylow $p$-subgroup is also a member of another Sylow $p$-subgroup?

Is it possible that every element of Sylow p-subgroup is also a member of another Sylow $p$-subgroup? Let $H_1$,$H_2$,....$H_n$ denotes different Sylow $p$-subgroup of group G, then for every $ x \in ...
femto's user avatar
  • 361
3 votes
2 answers
228 views

G has a element of order 2 not lying in center

Let G be order of 8,if G has a element of order 2 not lying center,how to prove G is isomorphic to $D_8$? A hint is consider the sylow-2 subgroup of $S_4$, I know it's $D_8$. So I want to construct a ...
ckx's user avatar
  • 416
4 votes
2 answers
547 views

Center of group of order 3773

What can we say about $|Z(G)|$ if $G$ is of order $3773 = 7^3 * 11$. Here $|Z(G)|$ means the size of the center. This is an exercise in book about Sylow theorem and I have no idea what Sylow theorem ...
27rabbit's user avatar
  • 101
2 votes
1 answer
125 views

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 ...
William Chuang's user avatar
3 votes
1 answer
172 views

Find a group of order 5784 that does not have a normal subgroup of index 12

This is a second part of a two-part question. In the first part I was asked to prove that any group of order $5784 = 2^3 \cdot 3 \cdot 241$ has subgroups of the following indexes: $3, 6, 8, 12,$ and $...
giorgio's user avatar
  • 583
1 vote
0 answers
57 views

Classifying Groups of Order 42 with an element of Order 6

Let $G$ be a group of Order 42 which has an element of order 6. I want to classify all groups like this up to isomorphism. Let $g\in G$ be the element of order 6. Then $C_6\cong\langle g\rangle\leq G$....
watertrainer's user avatar
3 votes
2 answers
99 views

Normal Sylow Subgroups of Solvable Groups

If $G$ is square free such that $|G|=p_1\cdots p_n$, where $p_n>p_{n-1}>\cdots$. Then we can use the N/C Theorem, along with some induction, to show that the Sylow $p_n$-subgroup of $G$, $P_n$, ...
brandon72308's user avatar
0 votes
1 answer
75 views

Some questions about the proof of the third Sylow theorem [closed]

Sylow's Theorem: Let $|G| = p^n · m$ with $p$ prime, m coprime to $p$, and $n ≥ 1$. The number $α(p)$ of $p$-Sylow groups of $G$ is a divisor of $m$ and of the form $α(p) = 1+kp$ for a $k ≥ 0$. Proof:...
Marco Di Giacomo's user avatar
2 votes
0 answers
133 views

Sylow $p$-Subgroups of $GL_n(\mathbb F_q)$

I know that one of the Sylow $p$-subgrouops of $GL_n(\mathbb F_q)(q=p^a)$ is the subgroup $P$ of upper triangular matrices with $1$ on diagonal, but is there a simple and direct way to see that every ...
Eric Ley's user avatar
  • 738
3 votes
2 answers
118 views

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 ...
paulina's user avatar
  • 695
0 votes
1 answer
65 views

Is there an infinite nilpotent group with one Sylow subgroup that is not normal? [closed]

It is known that a finite nilpotent group has every Sylow subgroup normal in it. Does this result generalize to infinite nilpotent groups or not ? If not, why, what is a counter example?
NotaChoice's user avatar
1 vote
1 answer
52 views

Group of order 60 and sylow subgroup counting

Im reading Dummit and foote page 145 where the book show $G$ is simple whenever $|G|=60$ and $n_5>1$ The book did it with a contradiction by assuming it is simple. Then deduced right away that $n_5=...
Remu X's user avatar
  • 1,071
1 vote
1 answer
63 views

Orbits of same size under conjugation action of Sylow $p$-subgroups by a normal subgroup $H$

Let $H$ be a normal subgroup of a finite group $G$ and $p$ a prime number, show that the orbits of the conjugation action induced by $H$ on $X = \{P \leq G \mid P \in Syl_p(G) \}$ all have the same ...
J P's user avatar
  • 893
-4 votes
2 answers
97 views

Order of a Sylow $p$-subgroup of the symmetric group of order $p!$ [closed]

I have been struggling with this question because many times I saw the statement "order of Sylow $p$-subgroup of symmetric group of order $p!$ is $p$" being used in exercises and texts ...
Luka's user avatar
  • 41
5 votes
1 answer
107 views

Are the character tables for $Z$-groups known?

A $Z$-group is a group whose Sylow subgroups are all cyclic groups. I know from here that if two $Z$-groups have the same character table, then they are in fact isomorphic groups. To my understanding, ...
NewViewsMath's user avatar
2 votes
1 answer
113 views

Groups of order $231$

In a certain exercise am I asked to show that a group of order $231$ has always an abelian subgroup of order $33$. It might be simple, but I still get struct at the same point. Here is my approach: ...
Emmy N.'s user avatar
  • 1,361
3 votes
1 answer
107 views

On p-Sylow subgroups of locally finite groups

I've come across this problem (I'm sorry if it's poorly written but it's a translation); for a $p$-Sylow I mean a $p$-subgroup that is not contained in any other $p$-subgroup: Let $G$ be a locally ...
temp's user avatar
  • 105
-1 votes
1 answer
158 views

Every group of order p^2 is abelian. [closed]

I am studying for my algebra qualifying exam in January, and I have a question regarding the proof I have for this question in Hungerford, problem 13 under the Sylow Theorems section. The way I ...
Sarah Hadaidi's user avatar
2 votes
1 answer
52 views

Supposed contradiction from Sylow theorems

Let $D_{506}$ denote the dihedral group with $1012 = 4 \cdot 23 \cdot 11$ elements. The Sylow theorems tell us that the number of 2-Sylow-subgroups (that is, subgroups of order 4) of $D_{506}$ divides ...
univalence's user avatar

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