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5 votes
1 answer
364 views

Groups with no proper non-trivial fully invariant subgroup

Let $G$ be a finite group. A subgroup $H$ of $G$ is said to be characteristic if $\phi(H)\subseteq H$, $\forall \phi \in \operatorname{Aut}(G)$ and fully invariant if $\phi(H)\subseteq H$, $\forall \...
Nick Belane's user avatar
1 vote
0 answers
109 views

Center of factors of a finite $p$-group, obtained from a minimal normal subgroup

throughout a research problem about finite $p$-groups, I have a challenge as follows, Let $G$ be a finite non-abelian $p$-group, where $p$ is odd and $Z(G)$ is non-cyclic. ($Z(G)$ denotes the center ...
shankfei's user avatar
  • 111
12 votes
2 answers
698 views

Generators of a group and normal subgroups

Can we say anything about a minimal generating set of a finite group based on its normal subgroups? For example, can we bound their order, or say whether they come from the same conjugacy class? An ...
utx7563yu's user avatar
  • 175
2 votes
0 answers
174 views

Doubly transitive groups in which a point stabilizer has an abelian normal subgroup

Let $G$ be a finite doubly transitive group in its action on the set $X$, such that a point stabilizer $G_x$ ($x \in X$) has an abelian normal subgroup $N_x$. I have read that if $\vert N_x \vert$ is ...
THC's user avatar
  • 4,547
0 votes
1 answer
339 views

Commutator group and conjugacy classes

Let $G$ be a finite solvable group which is not nilpotent, and let $H=[G,G]$ be the commutator subgroup of $G$. Does the following hold for $G$ and $H$? "There exists $g \in G \setminus H$ and $h ...
User01's user avatar
  • 217
0 votes
1 answer
183 views

Finite subgroups of $\mathrm{O}_n(\mathbb R)$ from finite subgroups of $\mathrm{SO}_n(\mathbb R)$

Let $G$ be a finite subgroup of $\mathrm{SO}_n(\mathbb R)$. We also assume $G$ to be "maximal" in the sense that for every $g\in\mathrm{SO}_n(\mathbb R)\setminus G$, we have that $\overline{\...
Andrea Aveni's user avatar
1 vote
1 answer
317 views

Why $G/F(G)$ is isomorphic to a subgroup of ${\rm Out}(F(G))$?

I know two facts and I’ve managed to figure out how to prove one, but the other one is still a little confusing. Let $G$ be a finite solvable group and $F(G)$ is the Fitting subgroup of $G$. (1) $G/Z(...
user avatar
6 votes
1 answer
651 views

Why do we say the Fitting subgroup/generalized Fitting subgroup control the structure of a group?

I’m learning the Fitting subgroup these days. I’m interested in this topic and particularly in the role that it plays in the structure of groups. Many people on MSE mentioned that the Fitting subgroup/...
user avatar
2 votes
1 answer
112 views

Restriction of real irreducible 2-Brauer characters to subnormal subgroups

Question: Find a finite group $G$, a subnormal subgroup $H$ of $G$, a real-valued irreducible $2$-Brauer character $\chi$ of $G$ and a real-valued irreducible $2$-Brauer character $\mu$ of $H$ such ...
John Murray's user avatar
  • 1,090
6 votes
3 answers
428 views

Subgroup generated by a subgroup and a conjugate of it [closed]

Let $H\leq G$ be groups, and $a\in G$ so that $\langle H,a\rangle=G$. Does it follows that $\langle H\cup aHa^{-1}\rangle$ is a normal subgroup of $G$? My hope is that this is true, and my guess is ...
Iras's user avatar
  • 61
6 votes
3 answers
213 views

Groups whose poset of direct factors are lattices

Let $G$ be a finite group. Denote by $\mathcal{N}(G)$ the modular lattice of normal subgroups of $G$ and denote by $\mathcal{D}(G)$ the subposet of $\mathcal{N}(G)$ whose elements are the direct ...
Rajkarov's user avatar
  • 933
0 votes
2 answers
956 views

Existence of a cyclic non-normal subgroup in a $p$-group

Let $G$ be a finite non-abelian $p$-group, where $p$ is an odd prime, $N$ be a normal subgroup of $G$ of order $p$, where $\frac{G}{N}$ is non-abelian. Does there exist an element $g\in G$ such that ...
sebastian's user avatar
  • 487
12 votes
2 answers
488 views

Finite groups with lots of conjugacy classes, but only small abelian normal subgroups?

Denote the commuting probability (the probability that two randomly chosen elements commute) of a finite group $G$ by $\operatorname{cp}(G)$. By a result of Gustafson [2], $\operatorname{cp}(G)=\...
Alexander Bors's user avatar
2 votes
2 answers
527 views

Does the hyperoctahedral group have only 3 maximal normal subgroups?

An hyperoctahedral group $G$ is the wreath product of $S_2$ and $S_n$, where $S_{n}$ is the symmetric group on $n$ letters, or in other words the semi-direct product $G=S_2^n\rtimes S_n$, w.r.t. the ...
Lior Bary-Soroker's user avatar
8 votes
4 answers
659 views

Normal Covering of a Finite Group

Suppose $G$ is a finite group and $N_1, N_2, \cdots, N_k$ are proper normal subgroups of $G$. The set $\{ N_1, \cdots, N_k\}$ is called a normal cover for $G$, if $G = \cup_{i=1}^kN_i$. I need to the ...
Fatemeh Moftakhar's user avatar
12 votes
0 answers
775 views

$2$-group with two isomorphic normal subgroups of index $4$ with non-isomorphic quotients

Hypothesis: Let $P$ be a finite $2$-group with two isomorphic normal subgroups $M$ and $N$ such that $P/M\cong C_4$ (the cyclic group of order $4$) while $P/N\cong C_2^2$. By the lattice theorem, ...
verret's user avatar
  • 3,291
35 votes
4 answers
2k views

Being a subgroup: proof by character theory

Let me first cite a theorem due to Frobenius: Let $G$ be a finite group, with $H$ a proper subgroup ($H\ne (1)$ and $G$). Suppose that for every $g\not\in H$, we have $H\cap gHg^{-1}=(1)$. Then $...
Denis Serre's user avatar
  • 52.3k
7 votes
1 answer
958 views

Groups whose normal subgroups form a chain with respect to inclusion

Let G be a finite group. In general, given two normal subgroups N and K of G, we need not have N < K or K < N. The easiest example is the Klein 4-group V4 and its subgroups of order 2. So assume ...
Amin's user avatar
  • 307
16 votes
2 answers
2k views

Groups with all normal subgroups characteristic

Today in my research, I had to use fairly explicitly the rather tautological property of finite cyclic groups that every normal subgroup is characteristic, i.e. fixed by all automorphisms. This got me ...
Alex B.'s user avatar
  • 13k