Questions tagged [normal-subgroups]

For questions concerning normal subgroups of groups. Consider using with the (group-theory) and/or the (abstract-algebra) tags too.

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Role of normality of the Klein group in solving a quartic using Lagrange resolvents

There are several ways how a useful resolvent for solving a quartic equation can be defined, e.g. (the roots are a, b, c, d) ...
David Kubecka's user avatar
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1 answer
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Show that the intersection of the distinct subgroups of a $p$-group of index $p$ is normal.

I've been struggling with this exercise (Exercise 8.2.28, Introduction to Abstract Algebra by Nicholson) for a few hours, and I haven't made much progress. Let $G$ be a group of order $p^n$ and let $...
iwjueph94rgytbhr's user avatar
2 votes
1 answer
133 views

On the finite minimal non-solvable groups

By minimal non-solvable group, I mean a non-solvable group whose all proper subgroups are solvable. Let $G$ be a minimal non-solvable group. If $G$ is not a simple group, then it can be easily ...
H.Shahsavari's user avatar
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1 answer
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How to find subgroups of direct product of cyclic groups [duplicate]

Find all subgroups of $\mathbb{Z_2} \times \mathbb{Z_5} \times \mathbb{Z_7}$. I know that this would be isomorphic to $\mathbb{Z_{70}}$, so that I would be looking for $8$ subgroups because $70$ has $...
MrMustache's user avatar
3 votes
1 answer
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The number of Hall $\pi$-subgroups of a $\pi$-separable group - Alexandre Turull article

This is an article which Alexandre Turull wrote. Lemma 2.1. states Lemma 2.1. Suppose $H$ is a finite group, acting on the finite group $F$, and assume that $|H|$ and $|F|$ are relatively prime. ...
math_survivor's user avatar
4 votes
1 answer
148 views

Let $K\lhd G$ be s.t. both $K$ and $G/K$ are simple. Show that either $K$ is the only proper normal subgroup of $G$, or $G \cong K \times (G / K)$.

Sorry about the title, I couldn't fit the whole exercise (Exercise 8.1.6, Nicholson Introduction to Abstract Algebra 4th edition): Let $K \triangleleft G$ be such that both $K$ and $G/K$ are simple. ...
iwjueph94rgytbhr's user avatar
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0 answers
32 views

Solvable radical and closed Lie subgroups

I have the following situation: let $\mathfrak{g}$ be an $n-$dimensional real Lie algebra of the connected Lie group $G$ and let $\mathfrak{r}(\mathfrak{g})$ be the solvable radical of $\mathfrak{g}$. ...
Tmath's user avatar
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-2 votes
1 answer
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Why is the direct product of finitely many nilpotent groups nilpotent? [duplicate]

I want to ask a question, and I found it here: Why is the direct product of a finite number of nilpotent groups nilpotent? But I am struggling to understand how can we take the product of two normal ...
NotaChoice's user avatar
1 vote
1 answer
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Assuming that $H \cap K = \{1_G\}$ and $G = \langle H, K \rangle$, prove that $G \cong H \times K$, why do we need both $H, K\mathrel{\unlhd}G$

The question is the same as here, and I understand the proof in it. But I don't know why do we need both $H, K\mathrel{\unlhd}G$? I think just one of them being a normal subgroup will be enough. W.L.O....
林敬珣's user avatar
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1 answer
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What are the different ways to claim that there does not exist any homomorphism between two groups

I have been working on the problems on homomorphism, the problem encountered is From the additive group $\mathbb{Q}$ to which one of the following groups does there exist a non-trivial group ...
Gajjze's user avatar
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1 vote
1 answer
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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
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5 votes
1 answer
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"Let $G$ be a finite group with a normal subgroup $N\cong S_3$. Show there is $H\le G$ s.t. $G=N\times H$." Does this $=$ really mean $\cong$?

I'm attempting Problem 60 on this problem set. Let $G$ be a finite group with a normal subgroup $N\cong S_3$. Show that there is a subgroup $H$ of $G$ such that $G = N \times H$. I'm guessing the ...
Luca T. Castrillón's user avatar
1 vote
1 answer
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Step in Butterfly Lemma by Zassenhaus - 1971: Allan Clark: Elements of Abstract Algebra

Theorem. (Zassenhaus) If $A$ and $C$ are subgroups of a group $G$ and if $B$ and $D$ are normal subgroups of $A$ and $C$, respectively, then there are isomorphisms $$ \frac{B\left(A \cap C\right)}{B\...
calculatormathematical's user avatar
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1 answer
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Prove that $[G;G]$ is a normal sub group of the group $G$ [closed]

I have mostly a question in my part "2-" . But any correction will be cool. Question: Let $G$ be a group and we define the commutator as follow $[g;h]=ghg^{-1}h^{-1}$ with $g, h \in G$. We ...
OffHakhol's user avatar
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Is there any general formula to calculate the order of elements in $U(n)$

Problem : For $g$ $\in$ $\mathbb{Z}$ , let $g' \in \mathbb{Z_{37}}$ denotes the residue class of $g$ mod $37$. Consider the group $U_{37}$ = {$g' \in \mathbb{Z_{37}} :1≤g≤37 \text{ with}, gcd(g, 37) = ...
Gajjze's user avatar
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3 votes
2 answers
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Abelian + normal $\implies$ central?

Here is the question, it is from a 2023 PhD qualifying exam at my university: Let $G$ be a finite group and $H$ a normal subgroup of $G$ of order 5. Prove that if $H$ contains an element not in the ...
Blue Ghost's user avatar
2 votes
0 answers
61 views

Finding the cosets of $S_3$

Struggling to understand where I'm going wrong here. I have $ G=S_3 $ and $ H = \{e, (1 2)\} $, and I want to list all the cosets with respect to H. This is what I have so far: Left cosets $eH=\{e, (...
ShashComandur's user avatar
1 vote
2 answers
73 views

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
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3 answers
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Proving $H\subset gHg^{-1}$ without the normality condition

In most textbooks a subgroup $H$ of group $G$ is defined as normal if $$i) \quad gHg^{-1} \subset H \quad \forall g \in G.$$ I understand that they don't write $gHg^{-1} = H$ as the definition of ...
Siddharth Prakash's user avatar
1 vote
1 answer
25 views

Finite index dense normal subgroups of completely metrizable groups

Is there some completely metrizable group $M$, which contains a normal subgroup $N\trianglelefteq M$ of finite index (at least $2$) that is dense in $M$?
user12345's user avatar
0 votes
2 answers
67 views

Showing a subgroup is a normal iff its left and right cosets (with distinct elements) are not disjoint and are equal

I am learning group theory and have come across the following problem statement Show that $K$ is a normal subgroup of a group $G$ if and only if $$\forall_{x,y\in G} xK\cap Ky \neq \emptyset \implies ...
Luk'yan Vilshansky's user avatar
1 vote
0 answers
31 views

Induced module is abelian

Let $G$ a profinite group and $H\subset G$ a normal open subgroup such that $G/H$ is abelian. Let $M$ be a discrete $H$-module. We say that the $H$-module is abelian if the image of the action map $\...
Mario's user avatar
  • 705
2 votes
0 answers
45 views

Precise meaning of H-admissible for a normal subgroup H of a group G

In the proof of Robinson's "A course in the theory of groups", 3.1.8 (theorem due to Wilson) he mentions H-admissible subgroups, but without, afaik, defining that. In this context, $H$ is a ...
Andrew Kay's user avatar
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2 answers
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$[G:H]=4$ where $H\neq 1$ means that $G$ is not simple.

The question is as follows: Suppose $G$ is a finite group with a nontrivial subgroup of index $4$. Prove that $G$ is not simple. I am on the hunt for a non-trivial normal subgroup. This proof ...
Chris Christopherson's user avatar
10 votes
1 answer
141 views

When are all normal subgroups of a direct product of finite groups a direct product of normal subgroups?

Let ${G_1}$ and ${G_2}$ be finite groups. When are all normal subgroups of their direct product ${{G_1}\times{G_2}}$ the direct product of normal subgroups in ${G_1}$ and ${G_2}$? So when is it true ...
ElectroSchOOp's user avatar
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1 answer
56 views

Typo? "If $G$ is a group, $H$ and $K$ are subgroups of $G$, and $K\unlhd G$. Is $H\cap K \unlhd H$ just if $K \subset H$?"

If $G$ is a group, $H$ and $K$ are subgroups of $G$, and $K\unlhd G$. Is $H\cap K \unlhd H$ just if $K \subset H$? I believe that my book may have a typo. Given $x \in H\cap K$, then $x \in H$ and $x ...
ends7's user avatar
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2 votes
1 answer
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Normal subgroup of $S_9$

Let's consider two subsets of the symmetric group $S_9$, $H=\{(123)^a(456)^b(789)^c: 0\le a,b,c \le 2 \}$ and $K=\{(147)^d(258)^d(369)^d:d=0,1,2\}$. In fact $H$ and $K$ are subgroups of $S_9$. ...
Fellow InstituteOfMathophile's user avatar
1 vote
2 answers
43 views

For a normal subgroup $H$ of $G$. how to prove $XH=HX$ for any part of $G$ ? Is union distributive over multiplication?

I am trying to prove that if $H$ is a normal subgroup of $G$ (i.e. $xH=Hx, \forall x\in G$) then for any subset $X$ of $G$ we have $XH= HX$. The hint given by the Wikipedia page is "take the ...
niobium's user avatar
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2 votes
0 answers
111 views

$n\ge 5$, prove $A_n$ is the only non-trivial proper normal subgroup of $S_n$.

Given than $A_n$ is simple for $n\ge 5$, prove $A_n$ is the only non-trivial proper normal subgroup of $S_n$. My work: Assume it is false, then there exists some non-trivial proper normal subgroup $K$ ...
GGplay's user avatar
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4 votes
2 answers
136 views

Let $p$ and $q$ be twin primes, both greater than or equal to $5$ , is every group of order $p^2q^2$ abelian?

I don't know if I made a mistake, but it seems that something is missing here. Let $p$ and $q$ be twin primes, i.e., $q = p + 2$. We want to prove that every group with order $p^2q^2$ is abelian, ...
pdrlmdd's user avatar
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1 answer
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The group $S_7$ has no subgroups of order $840$.

good afternoon everyone. I was studying group theory and came across the following exercise: Let $x=(3,4,8,6)(5,7)$, $y=(2,1)(8,6)(5,7)(3,4)$, and $G=\langle x,y\rangle$. Denote by $A,B,C$ and $D$ ...
Fqfqfqfqfqfq's user avatar
0 votes
1 answer
39 views

Clarification over subnormal series of a group

I am reading through a lecture note on group theory and it says the following: Let $H \leq G$ be a subgroup of group $G$. If $G_n \trianglelefteq \cdots \trianglelefteq G_1 = G$ is a subnormal tower ...
love and light's user avatar
3 votes
1 answer
232 views

Prove that any subgroup of $G$ containing a special subgroup of $G$ is also special.

Let $p$ be a prime number. Let $H\le G$ be a subgroup of finite group $G$. We say $H$ is special if there exists a normal series of form: $$ H=G_1 \unlhd G_2 \unlhd \dots \unlhd G_n=G $$ where $G_i$ ...
Hermi's user avatar
  • 1,506
2 votes
1 answer
230 views

Let $S\subseteq G$ for a group $G$. What is the relationship between $\langle S\rangle_G$ and ${\rm ncl}_G(S)$?

This is a big-list question that I suppose could be a community wiki post. If this is too broad or otherwise a bad question, I'm sorry. The Question: Let $S\subseteq G$ for a group $G$. What is the ...
Shaun's user avatar
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10 votes
1 answer
376 views

Is it possible for a normal subgroup of a finite group have greater number of elements in the minimal generating set?

Let $G$ be a finite group, and $1 \lhd N \lhd G$. With $G = \langle A \rangle$ and $N = \langle B \rangle$ be minimal. Is it possible for $|B|>|A| $? Main motivation behind this question was ...
Leon Kim's user avatar
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5 votes
4 answers
245 views

Show that a subgroup of order $3$ is normal in a group of order $15$

I've been trying to show that any group of order $15$ is cyclic and the only missing part in my proof is to show that the subgroup with $3$ elements (which we know exists by Cauchy's theorem) is ...
ephe's user avatar
  • 346
-1 votes
1 answer
52 views

Are commutator subgroups simple?

I am interested in the following setting: $$ 0\rightarrow [G,G]\rightarrow G \rightarrow \mathbb{Z}^r\rightarrow 0, $$ in particular in the case $r=1$. Is $[G,G]$ simple in this case? I am aware of ...
MathBug's user avatar
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6 votes
2 answers
291 views

Exercise 9, Section 2.5 of Hungerford’s Abstract Algebra

If $|G| = p^nq$, with $p\gt q$ primes, then $G$ contains a unique normal subgroup of index $q$. Why does author mention both unique and normal? Subgroup of index $q$ is Sylow $p$-subgroup of $G$. So ...
user264745's user avatar
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1 vote
1 answer
103 views

If there is subgroup of finite index $n$ and $G$ is finitely generated then there is a subgroup of $G$ that has index $n$ and is characteristic in $G$

Prove that if a group $G$ contains a subgroup $H$ of finite index, then $G$ contains a normal subgroup $N$ of finite index such that $N \leq H$. Show furthermore that if $G$ is finitely generated with ...
love and light's user avatar
2 votes
0 answers
28 views

Normal closure of closed subgroup is closed in a f.g. profinite group?

Is there an example of a topologically finitely generated profinite group $G$ and a closed subgroup $H$ such that we simultaneously have: The profinite normal closure $L$ of $H$ is open in $G$. The ...
Henrique Augusto Souza's user avatar
1 vote
1 answer
57 views

Series of nested normal subgroup (composition series) induces a sequence of quotient groups

In Group theory in a nutshell by A. Zee on pg. 66 he introduces sequences of nested normal subgroups: $G \rhd H_1 \rhd H_2 \rhd \dots \rhd H_k \rhd I$. Then he says that this induces a sequence of ...
Jens Wagemaker's user avatar
0 votes
1 answer
90 views

Sylow $p$-subgroups are all cyclic then $G$ has normal subgroup $N$ and such that $G/N$ and $N$ are both cyclic

Problem: If $G$ is a finite group whose Sylow $p$-subgroups are all cyclic then $G$ has normal subgroup $N$ and such that $G/N$ and $N$ are both cyclic. Whenever I need to find normal subgroup, I ...
N00BMaster's user avatar
0 votes
1 answer
61 views

Similarities Between Rings and Groups: Ideals and Normal Subgroups

I've been pondering the parallels between rings and groups and the commonalities shared by ideals and normal subgroups. There's a theorem stating that if $R$ is a ring and it happens to be a field, ...
hadadrefael's user avatar
0 votes
1 answer
34 views

Internal direct product $\mathrm{H= \phi(N_1) \times \phi(N_2)}$

Let $\mathrm{G=N_1 \times N_2}$ be an internal direct product of groups and let $\phi:G \to H$ be a surjective homorphism. Is it true that we also have the internal direct product $\mathrm{H= \phi(N_1)...
J P's user avatar
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0 votes
1 answer
63 views

If every quotient group of G by non-trivial normal subgroups are finite, then G is finite

If every quotient group of $G$ by a non-trivial normal subgroup is finite, then $G$ is finite. This a statement that I'm supposed to prove if it's true or not. If $G = (\mathbb{Z}, +)$, then all non-...
J P's user avatar
  • 333
1 vote
2 answers
46 views

If every quotient group of a group G by non-trivial normal subgroups is abelian, G is abelian

If every quotient group of a group G by non-trivial normal subgroups is abelian, G is abelian. I'm asked about the veracity of this statement. I thought about the group $\mathrm{D_3}$ and the only ...
J P's user avatar
  • 333
1 vote
1 answer
113 views

Restriction of irreducible representation to subgroup of index 2

Let $G$ be a finite group, $H \unlhd G$, $[G:H] = 2$, $\pi$ - irreducible complex representation of $G$, $\epsilon$ is also a representation of $G$ such that $\epsilon(g) = id_{\mathbb{C}}$ if $g \in ...
Jayden's user avatar
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0 votes
1 answer
78 views

Does this alternative argument prove that if $H,K\triangleleft G$ and $H\cap K=\{e\}$, then $hk=kh$ for all $h\in H$, $k\in K$?

From Printers book.Ex. H2 chapter 14 Theres a hint ($hk=kh \iff hkh^{-1}k^{-1}=e$), and using the fact H,K are normal. I did it using the hint and it was ok. My question is why cant I just say that by ...
letsgetthismsc's user avatar
4 votes
1 answer
160 views

Must two groups with all subgroups of the same order be isomorphic? [duplicate]

Let $G,H$ be finite groups such that there's bijection $\varphi$ from the subgroups of $G$ into the subgroups of $H$ satisfying: $|g_1|=|\varphi(g_1)|$; $g_1 < g_2$ if and only if $\varphi(g_1)&...
Derivative's user avatar
  • 1,506
2 votes
2 answers
59 views

Quotient group isomophic to $\mathbb{Z}$

In a certain document I am asked to show that if, given a group $G$, there is a normal subgroup $N$ such that $G/N \simeq \mathbb{Z}$, $\forall n \in \mathbb{N}$ there exists a subgroup $H$ such that $...
Daniel C.'s user avatar
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