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Questions tagged [maximal-subgroup]

To be use for both group theory and semigroup theory.

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Counting maximal subgroups of $\mathbb{Z}_m^n$

Let  $$ \mathbb{Z}_m=\mathbb{Z} / m\mathbb{Z} $$ How many maximal subgroups does  $$ \mathbb{Z}_m^n=\underbrace{\mathbb{Z}_m \times \mathbb{Z}_m \times \cdots \times \mathbb{Z}_m}_n $$  have? (m need ...
tys's user avatar
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1 answer
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Maximal subgroups of $SL(n, 2)$

According to Aschbacher's theorem, if $H$ is a maximal subgroup of $SL(n,q)$, then either it belongs to one of the classes $C_1 - C_8$, or $H$ is absolutely irreducible and $H/(H\cap Z(SL(n,q)))$ is ...
Mr. Nobody's user avatar
8 votes
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188 views

When does a finite group $G$ only have a single maximal subgroup containing given $A<G$?

Let $G$ be a finite group. Let $A<G$. When is it true that $G$ has only a single maximal subgroup containing (or equal to) $A$? The necessary condition is that there must be $x \in G$ such that $\...
Michał Zapała's user avatar
1 vote
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29 views

Reference for Cardinality of Parabolic Subgroup of Symplectic Group over Finite Fields

I am looking for a source to reference that gives the cardinality of parabolic subgroups of the Symplectic group $Sp$ over a finite field $\mathbb F_q$. What I want is essentially exactly what is in ...
Ryan L's user avatar
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5 votes
1 answer
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Is the irreducible $ SU(3) $ subgroup of $ SU(6) $ maximal?

Is the 6 dimensional $ (2,0) $ irrep of $ SU(3) $ maximal in $ SU(6) $? For those of you who are interested in context, I started wondering this the other day when I tried to write down the maximal ...
Ian Gershon Teixeira's user avatar
1 vote
2 answers
69 views

Generalisation of notion of maximal subgroup / maximal subalgebra

I'm trying to trace a definition. For this question fix some group $G$ (we generalise to an arbitrary algebra later). Definitions: Write $G'\leq G$ when $G'$ is a subgroup of $G$. Call $G'$ a ...
Jim's user avatar
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2 answers
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If $M$ is maximal subgroup of a finite group $G$, if $M$ is normal in $G$, then $|G:M|$ is prime. [duplicate]

I want to show that if $M$ is maximal subgroup of a finite group $G$, if $M$ is normal in $G$, then $|G:M|$ is prime. I feel pretty close to getting this one, but I'm getting stuck Suppose $M \leq G$ ...
Grigor Hakobyan's user avatar
1 vote
2 answers
100 views

Does any infinitely generated group have no maximal subgroups?

I've known that every finite group has maximal subgroups. And, use Zorn's Lemma, every finitely generated group has maximal subgroups. Also, there are examples that have no maximal subgroups, like $(\...
BlowingWind's user avatar
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1 answer
68 views

maximal subgroup of the general linear group

Maybe I'm being silly. As algebraic groups over an algebraically closed field $K$ of characteristic not $2$, is the conformal orthogonal group $CO_{2n}$ a maximal subgroup of $GL_{2n}$(or $CO_{2n+1}$ ...
scsnm's user avatar
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The maximal free abelian subgroup that can be embedded in $GL(n,\mathbb{Z})$

I am stuck on this problem and cannot seem to find a good reason for drawing the required conclusion. The problem is as follows: Given $SL(n, \mathbb{Z})$ a subroup in $GL(n, \mathbb{Z})$. How can ...
Yushi MuGiwara's user avatar
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35 views

Primitive maximal subgroups of $S_{n}$

I notice the following result in GTM163. Consider $S_{n}$ act on $\lbrace 1,2,\cdots,n \rbrace$ in a natural way. Then the maximal subgroups $M$ of $S_{n}$ fall into three classes: $(i)$ (intransitive)...
Dehai's user avatar
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2 votes
1 answer
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G2 as a subgroup of SO(7) from extended Dynkin diagrams?

As it is shown in the post Subgroups of $E_8$ by using extended Dynkin diagrams it is possible to find subgroups of a group from the extended Dynkin diagram by simply cutting nodes. However, the ...
p6majo's user avatar
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1 answer
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Is there an explicit maximal simple group?

It is not hard to prove the following lemma: LEMMA: Let $(G_i)_{i \in I}$ be a chain of simple groups. Then $G = \bigcup_{i \in I} G_i$ is a simple group. Let $N$ be a normal subgroup of $G$ and ...
Jan Matula's user avatar
3 votes
3 answers
107 views

Let $H\le G,G$ finite. If $M_H$ is a maximal subgroup of $H$, is it the case that $M_H = H \cap M$ for some $M$, a maximal subgroup of $G$?

I have a question regarding finite groups and their subgroups. Specifically, let $G$ be a finite group, and let $H$ be a subgroup of $G$. I am interested in understanding whether the following ...
Khaled's user avatar
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Maximal subgroup of a finite group contains all other subgroups.

A maximal subgroup $M$ of a finite group $G$ is a proper subgroup of $G$ such that no proper subgroup $H$ exists such that $H$ contains $M$. I want to prove that all other proper subgroups of $G$ ...
Md Iqbal Kotha's user avatar
2 votes
1 answer
42 views

Do all unitary simple groups $U_{2n+1}(2)$ have maximal subgroups of the form $3^{2n}:S_{2n+1}$?

In the ATLAS, the unitary simple groups $U_5(2)$ and $U_7(2)$ have maximal subgroups of structures $3^4:S_5$ and $3^6:S_7$, respectively. It seems that they are subgroups of the generalized symmetric ...
Isaac 's user avatar
  • 79
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0 answers
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Size of maximal subgroups in the direct product of finite groups

Let $\pi(G)$ be the set of all prime divisors of the order of a finite group $G$. Prove that: if $M$ is a maximal subgroup in $D=G \times G$ then $\pi(M)=\pi(G)$. My attempt: 1) If $G$ is $p$-group ...
Khaled's user avatar
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$G$ a group with center $\{e\}$ and $A$ a maximal subgroup of $G$ that is also abelian and not normal. How to show that $A$ is a Frobenius complement?

I have been sitting on this homework problem for days now: As the title says, I have a group (which doesn't have to be finite. Even Frobenius groups aren't defined as finite in our course) which only ...
joern's user avatar
  • 33
3 votes
1 answer
201 views

All maximal subgroups are Sylow subgroups

Let $G$ be a group in which all maximal non-trivial subgroups are Sylow subgroups. Then $G$ isn't a simple non-abelian group. I know how to prove this by relying on the theorem that if all maximal ...
Mr. Nobody's user avatar
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is there are a method to build subgroups of the multiplicative group $\left( \mathbb{Z}/n\mathbb{Z}\right)^*$

The multiplicative group $\left( \mathbb{Z}/n\mathbb{Z}\right)^*$ is defined by : $ \left( \mathbb{Z}/n\mathbb{Z}\right)^*= \{\, \bar{x} \in \mathbb{Z}/n\mathbb{Z}\;\;:\;\;gcd(x,n)=1 \,\} $ then we ...
Es-said En-naoui's user avatar
3 votes
2 answers
138 views

Algorithm that list all Maximal subgroups (up to conjugacy) of a finite group

Question: Is a polynomial time algorithm known to find all maximal subgroups (up to conjugacy) of a given group? (by polynomial time, I mean polynomial in the input size). If there is, please share a ...
Jins's user avatar
  • 564
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0 answers
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Why can’t Zorn’s lemma be used to show any module has maximal submodule [duplicate]

So you can use Zorn’s lemma to show that if $R$ is a ring then $R $ has a maximal left ideal. Show you would let $X$ be the set of all proper left ideals of $R$. Then can show any chain has an upper ...
Anonmath101's user avatar
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Aschbacher Class $2$ subgroup structure

In $PGL(12,3)$, there should be an Aschbacher Class $2$ subgroup the image of $GL(2,3)^6 \wr{\rm Sym}(6)$. I am trying to locate the image of $GL(2,3)^6$ in Magma using derived subgroup but it doesn't ...
scsnm's user avatar
  • 1,283
1 vote
2 answers
78 views

Example of a group such that the perfect maximal subgroup is $1$.

I an looking for an example of a group say $G$ such that the perfect maximal subgroup is $1$. I cannot find any references or come up with an example of my own. Could someone please cite a reference?
Sam Hoffman's user avatar
4 votes
1 answer
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construct the normaliser of a subgroup and then construct the subgroup

There is a maximal nontoral (not contained in a conjugate of a fixed maximal torus) elementary abelian $2$-subgroup of rank 6 in $G = PGL(8,7)$. I denote this group by $A$. Its normaliser $N_{G}(A)$ ...
scsnm's user avatar
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3 votes
1 answer
54 views

inclusion among finite symplectic groups

This might be a silly question, but do we have the following? $Sp(2,q) < Sp(4,q) < Sp(6,q) < Sp(8,q) <....$ I checked the list of maximal subgroups of $Sp(n,q)$ for $n = 4,6,8,10,12$ and ...
scsnm's user avatar
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2 votes
0 answers
93 views

A question on a 2-group with an elementary Abelian maximal subgroup

Let $G$ be 2-group of order $2^{n+1}$($n\geqslant2$) which has a maximal subgroup $N\cong\mathbb{Z}_{2}^{n}$. It is straightforward to check that if $G$ is an Abelian group, then $G$ is isomorphic to $...
H.Shahsavari's user avatar
2 votes
0 answers
344 views

Every maximal subgroup has prime index.

Let $G$ be finite group, and assume that every maximal subgroup of $G$ has prime index. Show that $G$ is solvable. I tried to prove it myself, but I'm afraid there may be mistakes in my proof. $\...
Mr. Nobody's user avatar
5 votes
2 answers
244 views

Let $H$ be a maximal subgroup of a finite group $G$ such that $|G:H|=4$. Then there exists $K\leq H$ such that $|H:K|=3$

Let $H$ be a maximal subgroup of a finite group $G$ such that $|G:H|=4$. Then there exists $K\leq H$ such that $|H:K|=3$. My attempt: Since maximal subgroups of nilpotent groups have prime index, so $...
Guest's user avatar
  • 1,587
3 votes
1 answer
240 views

Maximal compact subgroups of $\mathrm{GL}(2,\mathbb Q_p)$

In Bump, Daniel, Automorphic forms and representations, Cambridge Studies in Advanced Mathematics. 55. Cambridge: Cambridge University Press. xiv, 574 p. (1997). ZBL0868.11022., Exercise 4.5.1 on p. ...
Nightgap's user avatar
  • 1,259
0 votes
1 answer
33 views

Let $S$ be any non-empty subset of a group $G$. Does there exists a UNIQUE MAXIMAL subgroup $H$ such that $S$ is union of (some) left cosets of $H$?

It can be shown easily that there exists subgroup(s) such that $S$ is union of its left cosets. Trivial example: Take the trivial subgroup $H_0=\{e\}$ and then $S=\bigsqcup\limits_{g\in S}gH_0$. Then ...
MathBS's user avatar
  • 3,124
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0 answers
193 views

Let $G$ be a finite group and $M$ be a maximal subgroup of $G$. If $G = Z(G)M$, then $M$ is normal in $G$

I need to prove that if $G$ is a group, $M$ is a maximal subgroup of $G$ and $Z(G) \nsubseteq M$,then $M \unlhd G$. Is true that $G = Z(G)M$, right? Is this enough?
avsdcl's user avatar
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2 votes
0 answers
46 views

If $H$ is a maximal subgroup of $G$, then the action of $G$ on the cosets of $H$ is primitive.

If $H$ is a maximal subgroup of $G$, then the action of $G$ on the cosets of $H$ is primitive. This is a corollary in my group theory lecture notes with no proof. My thought is the following (I don't ...
SunshineAndRain's user avatar
0 votes
1 answer
299 views

Show $S_n$ is a primitive permutation group.

This is Exercise 7.2.1 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search for "symmetric group primitive" in the group-theory tag, this is ...
Shaun's user avatar
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5 votes
0 answers
205 views

Maximal cyclic subgroups

Can we classify the groups for which every maximal cyclic subgroup is of same order and intersection of any two maximal cyclic subgroups is identity? For example in case of abelian groups $$G=\mathbb{...
PARVEEN's user avatar
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0 votes
1 answer
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Zorn's lemma and maximal elements

This continues a discussion begun at checking Zorn's lemma on an example where an example was offered to help understand maximal elements and Zorn's lemma. That example used the set {1,...,100} ...
crabtree's user avatar
2 votes
1 answer
237 views

Understanding the definition of Sylow $p$-subgroups

Here is the definition of Sylow $p$-group (source: wikipedia) For a prime number $p$, a Sylow $p$-subgroup of a group $G$ is a maximal $p$-subgroup of $G$, i.e. a subgroup of $G$ that is a $p$-group (...
Esha's user avatar
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1 vote
0 answers
109 views

The Frattini subgroup of the standard Wreath product of two quasicyclic groups is the group itself.

This is part of Exercise 5.2.5 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE. Here are some previous questions ...
Shaun's user avatar
  • 45.5k
5 votes
2 answers
264 views

What if an automorphism fixes every maximal subgroup pointwise. Is it then the identity? [closed]

This question came up in the discussion over here My first thought was that then it fixes the Frattini subgroup. Any help? For reference we found that the answer is no when each maximal subgroup is ...
user avatar
2 votes
2 answers
605 views

Let $G$ be a finite $p$-group, where $p$ is prime. Show that if $G$ is not cyclic, then $G$ has at least $p+1$ maximal subgroups.

Question: Let $G$ be a finite $p$-group, where $p$ is prime. Show that if $G$ is not cyclic, then $G$ has at least $p+1$ maximal subgroups. I've seen this thread: Let $G$ be a finite $p$-group with ...
User7238's user avatar
  • 2,494
-1 votes
2 answers
138 views

Non-trivial automorphisms that fix maximal subgroups?

I thought this question up myself. It's somewhat similar to a question I asked previously Progress on a conjecture of Burnside... For the relevant definitions consult basic texts on group theory. (...
user avatar
2 votes
1 answer
150 views

About non existence of maximal subgroup in additive group $\mathbb{Q}$

It's known that the rationals $\mathbb{Q}$ as an additive abelian group have no maximal subgroup. I think I understand the proof provided here, which is elementary and IMO elegant. I'll re-write the ...
usb_naming_is_confusing's user avatar
1 vote
0 answers
207 views

Frattini subgroup of $p$-group

Suppose $G$ is a finite, non-trivial $p$-group and $\Phi(G)$ is the Frattini subgroup, defined as the intersection of all maximal subgroups of $G$. Since $G$ is finite, there are finitely many maximal ...
Luke's user avatar
  • 765
0 votes
3 answers
403 views

Why must we use Zorn's Lemma to prove every subgroup of $G$ is contained in a maximal subgroup of $G$?

Why must Zorn's Lemma be used to prove every subgroup is contained in a maximal subgroup? I understand the proof of it, but I feel like it's such an obvious statement that can be proved by common ...
user62783's user avatar
  • 127
1 vote
1 answer
105 views

Let $G$ be a group and $M$ be a maximal abelian subgroup of $G$. Prove $C_M(g)\lneq M$ for every $g\in G-M$.

Question: Let $G$ be a group and $M$ be a maximal abelian subgroup of $G$. Prove $C_M(g)\lneq M$ for every $g\in G-M$. Thoughts: I figured I could appeal to Zorn's Lemma here, but I was really trying ...
User7238's user avatar
  • 2,494
1 vote
3 answers
502 views

Show that $S_n$ is a maximal subgroup of $S_{n+1}$.

Question: Show that $S_n$ is a maximal subgroup of $S_{n+1}$. Thoughts: This question is answered here: Maximal subgroup of $S_n$, by adjusting the notation a bit. I was wondering though if there ...
User7238's user avatar
  • 2,494
2 votes
1 answer
88 views

Let $H\leq G$ where $H$ is max simple. Prove either $G$ is simple or there exists a minimal normal subgroup $N$ of $G$ such that $G/N$ is simple. [duplicate]

Question: Let $G$ be a finite group and $H$ a maximal, simple subgroup of $G$. Prove that either $G$ is simple or there exists a minimal normal subgroup $N$ of $G$ such that $G/N$ is simple. Thoughts:...
User7238's user avatar
  • 2,494
0 votes
0 answers
91 views

Frattini subgroup in abelian group

Let $G$ a abelian group. I recall that $G^p=\{g^p:g\in G\}$ ($p$ is primes) and $\Phi(G)$ is Frattini supgroup of $G$ i.e. intersection of maximal subgroup of $G$. I want proof that $\Phi(G)=\...
user910194's user avatar
3 votes
1 answer
158 views

Isaacs Algebra : A Graduate Course definition for diagonal subgroup

I actually have a number of related questions. In Problem 7.2, Isaacs defines a subgroup $D$ of $G= M \mathop{\dot{\times}} N$ to be diagonal if $$ D \cap M = 1 = D \cap N \text{ and } DM = G = DN. $$ ...
gregg's user avatar
  • 31
2 votes
1 answer
102 views

Finite group with prime index maximal subgroups

Let $G$ be a finite group and $p$ the largest prime that divides the order of $G$. Suppose that all the maximal subgroups of $G$ have a prime index. Then, the Sylow $p$-subgroup is normal in $G$. ...
Zed's user avatar
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