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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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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
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3 votes
1 answer
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Conjugacy classes of normal subgroup in group

Let $G$ be a finite group, and $N$ be a normal subgroup of index $p$. If every conjugacy class of $N$ is also a conjugacy class in $G$, what can we say about $G$ or $N$? Such instances occur if $G$ is ...
Maths Rahul's user avatar
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2 votes
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Suppose $H$ is a normal subgroup of $G$ with prime index $p$, does this implies that $H$ contains $G^p$?

The power subgroup is defined as $G^p =\left\{g^p \mid g \in G\right\}$. Question: If $H$ is a normal subgroup of $G$ with prime index $p$, does this imply that $H$ contains $G^p$? This statement is ...
ghc1997's user avatar
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3 votes
3 answers
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Does this condition imply normality for a subgroup?

(Note: This question emerged out of mere curiosity in the definition of a normal subgroup.) Let $(G, \cdot )$ be a group, and let $H$ be a subgroup. Now, for each coset in the set of (left ) cosets, $...
Vivaan Daga's user avatar
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Prove that $S_4/K \cong S_3$ using the fundamental theorem on homomorphism.

Let $A$ be the set formed by the elements of Klein group but identity, A= { (1,2)(3,4);(1,3)(2,4);(1,4)(2,3)}. Consider the set $Big(A)$ of bijection from A to itseld. With the operation of ...
user1335731's user avatar
4 votes
1 answer
133 views

An inequality about index in finite group

Some denotations Let G be a finite group. Let N be a normal subgroup of G.(Which means $\forall n \in N, \forall g \in G, gng^{-1}\in N$) Let M be an arbitary subgroup of N. Let H be the normalizer of ...
DTK's user avatar
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3 votes
2 answers
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Use of correspondence theorem for groups to prove that $o(N) = 2$

Let $G$ be a group and $H \triangleleft G$ simple such that $[G : H] = 2$. I have to prove that if $N \neq \{1\}$, $N \triangleleft G$ and $N \cap H = \{1\}$ then $o(N) = 2$. I know by third ...
Cyclotomic Manolo's user avatar
1 vote
0 answers
28 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
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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
3 votes
1 answer
234 views

Non-Simple, Centerless Group With Exactly One Non-Trivial Normal Subgroup

Question. (Is this statement true?) Given non-simple, centerless group $G$ such that there exists exactly one non-trivial normal subgroup $N\triangleleft G$, then $G/N$ must be isomorphic to some ...
JAG131's user avatar
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Conjugacy class size and normal subgroup

Let G be a finite group and let $x \in G$ and let $Cl_G(x)$ denote the conjugacy class of $ x \in G$ . It is known that $|G|=|Cl_G (x)||C_G (x)| $ where $C_G (x)$ is the centralizer of $x$ in G as ...
Math Student's user avatar
2 votes
1 answer
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Given normal subgroup $H\triangleleft G$ with prime index $p$, then $G\cong H\ \rtimes\langle\alpha\rangle.$

Update. Please see accepted answer for update to this question. It is in fact not true, and a very clear counter-example exists. Statement. Let $G$ denote some artbirary finite group. Given normal ...
JAG131's user avatar
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Finite index Lie subgroup in Lie correspondence

I am working with Lie groups and more specifically at finite index Lie subgroups. I was wondering how the fact that the index of the subgroup is finite would translate under the Lie group - Lie ...
noparadise's user avatar
2 votes
2 answers
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Can a non-Lie quotient of $GL_3(\mathbb{R})$ be $O_3(\mathbb{R}$)?

Recently a friend from my twitch stream asked me about the following question: Does there exist a normal subgroup $N\leq G=GL_3(\mathbb{R})$ such that $G/N\cong O_3(\mathbb{R})$? We know a lot about ...
Eric Nathan Stucky's user avatar
1 vote
0 answers
56 views

Virtual solvability of dense subgroups

Let $G$ be a (finitely generated) dense subgroup of $\mathsf{SL}(2;\mathbb{C})$. Is it possible that $G$ is virtually solvable? In other words, by Tit's alternative, does being dense necessitate the ...
cdkaram's user avatar
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2 votes
1 answer
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(Group)Homomorphism and Subgroups

Given a group $G$, a normal subgroup $H$ and a (surjective) homomorphism $\pi: G \to G/H$. Via Isomorphism Theorem, it is known that $ |G| = |Ker(\pi)||G/H|$. I understand this as grouping into ...
TheDuck's user avatar
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Group of order $2^kq$ has a normal Sylow $2$-subgroup [closed]

Let $G$ be a group of order $2^kq$ for some prime $q>2^k$ , denote $P$ and $Q$ one of the Sylow $2$-subgroup and Sylow $q$-subgroup of $G$ respectively. I have seen a comment that for such a group $...
Dian Wei's user avatar
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1 answer
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Can there be a subgroup $K$ of a group $G$ such that for some $a \in G$, $aK \subseteq Ka$ but $Ka \nsubseteq aK$? [duplicate]

I have a question, in a sense, about how asymmetric left and right cosets can be when dealing with an infinite, non-normal subgroup $K$ of a (non-abelian) group $G$. Specifically, my question is ...
MathNeophyte's user avatar
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0 answers
29 views

Groups of order $2^n p$ for $n\geq 1$ and $p$ prime with $2^n> (p-1)!$ are non-simple. Is my proof correct?

I'm doing my homework in Group Theory and as part of an exercise, I want to show the following Lemma: Let $n\geq 1$, $p$ a prime, s.t. $2^n > (p-1)!$ and $G$ a group of order $2^n p$. Then G has a ...
Joachim's user avatar
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The free product has the direct product as a factor group. What's the corresponding normal subgroup?

Let $G$ and $H$ be groups. Consider the free product $G * H$ and the direct product $G \times H$. There is a particular way of identifying $G \times H$ as a factor group of $G * H$. Namely, the ...
Dannyu NDos's user avatar
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-3 votes
1 answer
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All subgroups of a group of integers coprime with 7 [duplicate]

I am to list all the subgroups of $(\mathbb{Z_7^*}, \cdot)$. Recall that $\mathbb{Z_7^*}$ is a set of numbers s.t. $\gcd(a, 7) = 1$ I've listed all of them: $H_1 = \{ 1 \}, \\ H_2 = \{ 1, 2, 4 \}, \\ ...
Avgustine's user avatar
  • 149
0 votes
1 answer
46 views

Normal Subgroup criteria

Trying to prove the equivalence of the two following criteria for a subgroup H (of Group G) to be normal. $aha^{-1} \in H $ for all $a \in G, h \in H \implies aHa^{-1} = H $ for all $a \in G$ ...
TheDuck's user avatar
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1 vote
2 answers
90 views

Can I prove $G_s=H$ if $H$ is normal in $G$, where $S=G/H$

If $H$ is normal in $G$, where $S=G/H$, and $G$ acts on $S$, can I prove $G_s=H$ for any $s$ in $S$? (where $G_s$ is the stablizer of $s$) My try, if action is left multiply by $g$, to stablize $s=aH$,...
femto's user avatar
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Munkres' Topology theorem 68.7

Theorem 68.7 Let $ G = G_1 * G_2 $. Let $ N_i $ be a normal subgroup of $ G_i $, for $ i = 1, 2 $. If $ N $ is the least normal subgroup of $ G $ that contains $ N_1 $ and $ N_2 $, then $$ G/N \simeq \...
Davood Karimi's user avatar
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0 answers
15 views

Subgroup Lattice of $G = Z^*_{26}$ [duplicate]

I have been looking through my textbook and have noticed subgroup lattices. I understand the premise of them under addition, but this one example, in particular, I can't figure out. Let $G = Z^*_{26}$....
Refined's user avatar
-1 votes
1 answer
97 views

Semidirect product of groups

In recent days I am studying semidirect product of groups and I have come up with the following question which has already been answered here (From semidirect to direct product of groups), but I can't ...
Priya Sarkar's user avatar
0 votes
1 answer
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How to write a clear proof that if $N$ is a normal subgroup whose quotient group is abelian, the commutator subgroup is a subset of $N$?

I'm trying to find the right level of explicitness to use when writing algebra proofs. The challenge is that algebra proofs, once complete, can often devolve into simply a series of equations. But ...
SRobertJames's user avatar
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5 votes
1 answer
166 views

Group extension analogous to the symmetric group.

Recently, My Professor taught us about group extension. It is the following: A group $G$ is an extension of $Q$ by $N$ if we have the following short exact sequence: $1 \rightarrow N \rightarrow G \...
Pratina's user avatar
  • 149
1 vote
1 answer
86 views

Non-abelian groups with trivial socle.

Recently, my teacher told me about Socle of a group $G$. So, I know that the socle $Soc(G)$ of a group $G$ is the subgroup generated by all minimal normal subgroups of $G$. I was thinking then that ...
Priya Sarkar's user avatar
3 votes
2 answers
97 views

Normal Sylow Subgroups of Solvable Groups

If $G$ is square free such that $|G|=p_1\cdots p_n$, where $p_n>p_{n-1}>\cdots$. Then we can use the N/C Theorem, along with some induction, to show that the Sylow $p_n$-subgroup of $G$, $P_n$, ...
brandon72308's user avatar
1 vote
1 answer
89 views

Does there exist an element of order $4$ in $GL_2(\mathbb{Z})/GL_2(\mathbb{Z})'$?

For a group $G$, let $G'$ denote the commutator of the group $G$, and if $H \leq G$ the left cosets will be denoted as $gH$. Now, I understand the fact that $[SL_2(\mathbb{Z}):GL_2(\mathbb{Z})'] = 2.$ ...
Zen's user avatar
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1 vote
0 answers
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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
0 votes
1 answer
35 views

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
148 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
0 votes
1 answer
66 views

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
53 views

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
155 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
0 votes
0 answers
40 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
79 views

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
70 views

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
1 vote
1 answer
124 views

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
  • 386
1 vote
1 answer
63 views

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
  • 893
5 votes
1 answer
164 views

"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 ...
aleph2's user avatar
  • 914
1 vote
1 answer
84 views

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
0 votes
1 answer
127 views

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
  • 719
0 votes
1 answer
127 views

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
  • 386
3 votes
2 answers
119 views

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
66 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
110 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
0 votes
3 answers
115 views

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

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