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Showing that a group of order $231$ is a semi direct product

I am having a real hard time with this problem. My understanding is that to show that $G$ is a semi direct product, I must find 2 subgroups of $G$ such that one is normal, they have trivial ...
baslerbuenzli's user avatar
2 votes
0 answers
41 views

Understanding the case when $P_3$ is normal and $P_2 \cong \mathbb Z _2 \times \mathbb Z_2$ is not normal to classify a group of order $12$

I am studying the classification of groups, when the Klein 4-group arises as a subgroup. The first example comes from Artin about groups of order $12$. He uses proof by cases, and the only case that ...
Talmsmen's user avatar
  • 1,314
2 votes
0 answers
61 views

Suppose $P$ is a $p$-subgroup of the $p$-solvable group $X$. Prove $O_{p'}(N_X(P)) \subseteq O_{p'}(X)$.

I want to prove the following lemma Suppose $P$ is a $p$-subgroup of the $p$-solvable group $X$. Then $O_{p'}(N_X(P)) \subseteq O_{p'}(X)$. $O_{p'}(X)$ means the $p'$-core of $X$. That is, $O_{p'}(X)...
Lee's user avatar
  • 77
1 vote
1 answer
79 views

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, ...
fatlips22's user avatar
4 votes
1 answer
137 views

Must a nonabelian group of odd order always have a normal Sylow subgroup?

Let $G$ be a nonabelian group of odd order $n = |G|$ that's not a prime power, so that all the Sylow subgroups are proper. Must it have at least one normal Sylow subgroup? Of course this $G$ is ...
Richard_Loves_Music's user avatar
0 votes
2 answers
78 views

How to prove the action of a group on a set of its Sylow p subgroups faithful in these cases (and why?)?

Firstly, I was learning material that included this. Let $S$ be the set of Sylow 5 subgroups of $S_5$. Then, we can consider the action of $S_5$ on $S$ by conjugation. From this we get the associated ...
baslerbuenzli's user avatar
2 votes
1 answer
77 views

Inducing a homomorphism with a nontrivial kernel.

I am in the middle of reading a proof. So far I have a group $G$ of order $q^2p^2$, both prime. I have two Sylow $q$-subgroups $Q_1$ and $Q_2$ both of order $q^2$. I know $Q_1\cap Q_2\neq \{e\}$, $|\...
MCMath's user avatar
  • 135
2 votes
1 answer
108 views

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 (...
Hannah Taylor's user avatar
2 votes
1 answer
53 views

$G/P$ is isomorphic to $C_4 \times C_4$

$G$ is the group of matrices \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} where $a, b, c \in \mathbb{Z}/5$ with $a, c \neq 0$. This is a group of order $80$ with one (normal) $5$-Sylow ...
ScintillatingWolves's user avatar
3 votes
2 answers
165 views

Sylow subgroups of $\mathrm{SL}_2(\mathbb{Z}_p)$ and $\mathrm{SL}_2(\mathbb{F}_p)$

Let $p,l$ be odd primes and $p\neq l$. I am interested in determining the $l$-Sylow subgroups of the groups $\mathrm{SL}_2(\mathbb{Z}_p)$ and $\mathrm{SL}_2(\mathbb{F}_p)$. I have read that for $\...
Ben's user avatar
  • 1,270
0 votes
0 answers
47 views

No simple groups of order of order 4095, 4389, 5313 or 6669 using normalizers

I am working on problem 6.2.10 from Dummit and Foote. It asked us to prove there are no simple groups of order 4095, 4389, 5313 or 6669 using the method it calls “Playing $p$-subgroups off against ...
Mattan Feldman's user avatar
4 votes
2 answers
101 views

Correspondence of Sylow p-Subgroups between Group $G$ and Quotient Group $G/Q$

Let $G$ be a finite group, $p$ a prime factor of $|G|$, $Q$ a $p$-subgroup of $G$, and $Q$ a normal subgroup of $G$. Let $\pi: G \to G/Q$ be the quotient homomorphism. Prove that $\pi$ induces a one-...
ZhouYang's user avatar
  • 147
1 vote
1 answer
107 views

Question about relations of Sylow $p$-subgroups in $G$ and $G/H$

I want to know if there's any theorem that relates how many sylow p-subgroups are in $G$ and $G/H$, where $H$ is a Sylow group. I "invented" a lemma that is somewhat useful, I leave it below ...
lostMathematician's user avatar
1 vote
1 answer
65 views

Group of order $2^3 \cdot 7^2$ has a normal subgroup of order $7$ or $49$

Let $G$ be a group of order $392 = 2^3 \cdot 7^2$, then $G$ has a normal subgroup of order $7$ or $49$. Attempt. Suppose $G$ does not have a normal subgroup of order $49$. Then in particular, $n_7 \...
Aram Nazaryan's user avatar
5 votes
1 answer
90 views

Isomorphism between simple group of order 168 and PSL(2,7)

In our group theory classes at the university, we devoted several seminars to derive the uniqueness (up to isomorphism) of a simple group of order $168$. However, i am stuck on last steps, that prove ...
King Crimson's user avatar
0 votes
1 answer
65 views

List, up to isomorphism, all abelian groups of order at most 50, containing at least 3 elements of order 2 and at least 1 element of order 3.

I have this question I want to do. I think that I have a rough understanding in how to tackle it, but at some point I get lost. My idea is sort of like this: I first discard all orders which are not ...
enfield's user avatar
  • 395
1 vote
0 answers
63 views

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 ...
S.Lara's user avatar
  • 59
0 votes
1 answer
78 views

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 ...
Aram Nazaryan's user avatar
1 vote
1 answer
93 views

Intersection of commutator subgroup and normalizer of a $p$-Sylow subgroup is normal in $G$

I've been trying to solve the following question without much success and I'm kind of lost on what else I could try to do: Assume that the commutator subgroup $G^{'} = [G,G]$ of a finite group $G$ is ...
user594756's user avatar
2 votes
0 answers
80 views

Show that a group of order $10^6$ surjects onto a group of order at most $10^4.$

The Problem is: Let $G$ be a group of order $10^6.$ Show that $G$ admits a nontrivial homomorphism onto a group $H$ of order at most $10^4.$ My approach Here, $|G|=2^65^6.$ One canonical way to ...
Rabi Kumar Chakraborty's user avatar
0 votes
0 answers
54 views

Order of the group of upper triangular matrices over $\mathbb{Z_p}$ [duplicate]

I was learning about Frobenius proof of Sylow's first theorem, and in the proof he uses Cayley's theorem to make a map from the group $G$ to $S_n$ and then uses the permutation matrix to map from $S_n ...
Victor Hugo's user avatar
1 vote
2 answers
73 views

Determine the number of subgroups of order $9$ in the alternating group $𝐴6$ [closed]

Using Sylow's theorem, I found that all possible cases are $1, 4, 10,$ and $40,$ but I can only conclude that $1$ is not possible. How can I determine that $4, 10,$ and $40$ are not possible?
user1417084's user avatar
8 votes
2 answers
173 views

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 ...
J. Linne's user avatar
  • 3,178
1 vote
1 answer
90 views

$G$ solvable group of order $pm$ has subgroup of order $m$ when $\mathrm{gcd}(p,m) = 1$ and $P$ a Sylow $p$-subgroup with $N_G(P) = P$.

As mentioned in the title, the problem I'm looking to solve is: Let $G$ be a solvable group of order $pm$ and $\mathrm{gcd}(p,m) = 1$. Let $P$ a Sylow $p$-subgroup with $N_G(P) = P$. Prove $G$ has a ...
MathLearner12's user avatar
0 votes
2 answers
121 views

If G is the INTERNAL direct product of its Sylow subgroups, then every Sylow subgroup is normal in G.

I am working on an old qualifying exam problem, which is the following: Prove that a finite group $G$ is the internal direct product of its Sylow subgroups if and only if every Sylow subgroup is ...
qualsqualsquals's user avatar
2 votes
2 answers
97 views

$G$ finite group, $H$ normal subgroup, $P$ Sylow $p$-subgroup of $H$, then $G = HN_G(P)$

This is related to: This question Consider the same premise. $G$ a finite group, $H$ normal in $G$, and $P$ a Sylow $p$-subgroup of $H$. Then the claim is $G = HN_G(P)$. I want to ask if my attempt ...
HorribleATMath's user avatar
1 vote
1 answer
87 views

A group of order 2015 has a normal subgroup of order 155.

I am reviewing an old qual problem (see title). I know that if $|G|=2015$, there are unique and hence normal subgroups of order 13 and 31. I also know that if $H,K$ are subgroups of $G$ and $K$ is ...
qualsqualsquals's user avatar
2 votes
0 answers
71 views

All groups of order $2010$ are solvable

Show that every group of order $2010 = 2 \cdot 3 \cdot 5 \cdot 67$ is solvable. This is easy. $n_{67} =1$ since $n_{67} \mid 30$ and $n_{67} \equiv 1 \mod 67$. Now, $P \in \mathrm{Syl}_{67}(G)$ is ...
Aram Nazaryan's user avatar
1 vote
1 answer
74 views

Subgroup of $2$-Sylow subgroup of $S_{17}$

Let $G$ be a $2$-Sylow subgroup of $S_{17}$. We have to prove that $G$ has a subgroup isomorphic to $\mathbb Z_8\times \mathbb Z_8$. My progress: It is easy to calculate that $|G|=2^{15}$. Now, any $p$...
Tabludif's user avatar
  • 163
2 votes
1 answer
126 views

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
  • 178
1 vote
1 answer
65 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 ...
Aram Nazaryan's user avatar
2 votes
1 answer
65 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
110 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(...
Aram Nazaryan's user avatar
1 vote
2 answers
150 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 ...
user avatar
10 votes
1 answer
401 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
  • 1,847
1 vote
2 answers
105 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
5 votes
2 answers
547 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
  • 321
1 vote
0 answers
34 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
29 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
81 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
0 votes
0 answers
32 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 $201$.
Steve's user avatar
  • 19
2 votes
1 answer
172 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
216 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
57 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
105 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
  • 434
5 votes
2 answers
116 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
45 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
102 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
4 votes
1 answer
248 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
63 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

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