To be use for both group theory and semigroup theory.
Questions tagged [maximal-subgroup]
157 questions
35
votes
4 answers
$(\mathbb{Q},+)$ has no maximal subgroups
I have a problem that I don't have any idea.
Show that group $(\mathbb{Q},+)$ has no maximal subgroups.
Muniain
- 1,543
11
votes
0 answers
Any non-trivial finitely-generated group admits maximal subgroups
I want to solve the following problem from Dummit & Foote's Abstract Algebra:
This is exercise involving Zorn's Lemma (see Appendix I) to prove that every nontrivial finitely generated group possesses maximal subgroups. Let $G$ be a finitely…
user1337
- 24,859
11
votes
2 answers
$H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.
I need to prove that $H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.
My proof to the $(\Rightarrow)$ direction seems too much trivial:
Let us assume there exist $A$ so that $A/H\lhd G/H$. Then by definion, $H$ must be normal…
catch22
- 3,145
8
votes
0 answers
When does a finite group $G$ only have a single maximal subgroup containing given $A
Let $G$ be a finite group. Let $A=G$ and the order of…
Michał Zapała
- 1,962
8
votes
1 answer
$(\mathbb{Q},+)$ has no maximal subgroup.
Definition. A maximal subgroup of a group G is a proper subgroup $M$ of $G$ such that there are no subgroup $H$ of $G$ such that $M
sigma
- 3,278
8
votes
1 answer
Intersection between two conjugates of a maximal subgroup.
Let $G$ be a non-abelian finite group such that every proper subgroup of $G$ is abelian. Suppose $M$ is a maximal subgroup of $G$ which is not normal in $G$. I was asked to show that
$\bigcup_\limits{g \in G} gMg^{-1}$
has at least $ 1 + |G|/2 $…
E.E.
- 364
7
votes
3 answers
Maximal subgroup contains either the center or the commutator subgroup
Here's the problem:
Suppose $G$ is a group and $H$ is a maximal subgroup of $G$. Show that either $Z(G) \leq H$ or $G' \leq H$.
I know that if $G$ is abelian, then $G = Z(G)$, $G' = \{e\}$, and $G' \leq H$.
Furthermore, if $G$ is not abelian but…
Ali Mirzaei
- 113
5
votes
0 answers
What is called the "maximal normal subgroup" contained in a subgroup?
Let $G$ be a group. We know that the subgroup generated by a family of normal subgroups is again normal, hence for each subgroup $H$ of $G$, there is a maximal subgroup of $G$ contained in $H$ that is normal in $G$. I was interested in knowing if…
Jianing Song
- 2,545
- 5
- 25
5
votes
1 answer
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 subgroups of $ SU(6) $. My guess so far is that the…
Ian Gershon Teixeira
- 10,176
5
votes
2 answers
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 $G$ is not nilpotent. (In particular, $G$ is…
Guest
- 1,657
5
votes
0 answers
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{Z}_p\times \mathbb{Z}_p \times \cdots \times…
PARVEEN
- 51
5
votes
2 answers
What if an automorphism fixes every maximal subgroup pointwise. Is it then the identity?
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 merely mapped back to itself.
user1007655
5
votes
1 answer
Let $p$ be a prime number and let $G$ be a finite $p\text{-group}$. Let $M$ be a maximal subgroup of $G$.
QUESTION: Let $p$ be a prime number and let $G$ be a finite $p\text{-group}$. Let $M$ be a maximal subgroup of $G$. Show that $M$ is a normal subgroup of $G$ and that $| G: M | = p$.
THE HINT GIVEN IS: By strong induction on $n$, where $| G | = p…
Silvinha
- 361
5
votes
2 answers
Characterizing maximal subgroups of cyclic groups.
Suppose $G = \langle x \rangle $ is a cyclic group of order $n\geq 1$. Prove that a subgroup $H \leq G$ is maximal if and only if $H = \langle x^p \rangle$ for some prime $p$ dividing $n$.
Source question: Dummit and Foote, "Abstract algebra", ex…
user661541
5
votes
1 answer
Minimal order of a counterexample to Wall’s conjecture
There used to be once a rather well known and interesting conjecture, that was formulated by Gordon E. Wall:
The number of maximal subgroups of a finite group $G$ does not exceed $|G|$
That conjecture appeared to be false. However, the only…
Chain Markov
- 16,012