Linked Questions

0
votes
1answer
2k views

H a subgroup of index n, then G has a normal subgroup K with [G : K] ≤ n!. [duplicate]

Prove that if G is a group and H a subgroup of index n, then G has a normal subgroup K with [G : K] ≤ n! I'm having trouble proving this because frankly I have no idea where to start. Any tips?
0
votes
3answers
2k views

Show there is a normal subgroup $K$ of $G$ with $K\subset H$ and such that the order of $K$ divides $n!$ [duplicate]

Let $G$ be a finite group and suppose $H$ is a subgroup of $G$ having index $n$. Show there is a normal subgroup $K$ of $G$ with $K\subset H$ and such that the order of $K$ divides $n!$ . any ...
2
votes
1answer
749 views

If $(G : H) = r$ and there exists $K \triangleleft G$ contained in $H$ such that $(G : K) = r!$ [duplicate]

Possible Duplicate: How to prove that if $G$ is a group with a subgroup $H$ of index $n$, then $G$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$ I'm trying to prove the ...
1
vote
1answer
511 views

Quotient group with normal subgroup dividing the order of another group [duplicate]

Let G be a group with subgroup H and let $\Omega$ be the set of right cosets of H in G. Show that if G is a group with a subgroup of index n then G has a normal subgroup with index dividing n! ...
1
vote
2answers
79 views

Normal subgroup of index a divisor of n! [duplicate]

I’ve found this exercise on “basic algebra”: Show that if a finite group $G$ has a subgroup $H$ of index $n$ then $H$ contains a normal subgroup of $G$ of index a divisor of n!. My attempt: I’ve ...
-2
votes
2answers
89 views

Given a subgroup $H $, find a normal subgroup of $G$ with index less than or equal to $n!$ [duplicate]

Let $G$ be a group and $H$ a subgroup with finite index $n$. Prove that $G$ has a normal subgroup $N$ such that $N\subseteq H$and $|G: N| \le n!$.
0
votes
0answers
54 views

Prove that there is a subgroup $K$ : $k! | [H:K]$ [duplicate]

Consider a group $G$, and $H \in G$ - subgroup with $[G:H] = k$, then prove that there is exists normal subgroup $K$ in $H$ such that $k! | [G:K]$? Actually I have no ideas. Any hints?
22
votes
7answers
14k views

Does the intersection of two finite index subgroups have finite index?

Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $...
16
votes
2answers
7k views

For $G$ group and $H$ subgroup of finite index, prove that $N \subset H$ normal subgroup of $G$ of finite index exists

Let $G$ be a group and $H$ be a subgroup of $G$ with finite index. I want to show that there exists a normal subgroup $N$ of $G$ with finite index and $N \subset H$. The hint for this exercise is to ...
0
votes
1answer
1k views

Let $G$ be a simple group of order $n$. Let $H$ be a subgroup of $G$ of index $k$. Show that $n$ divides $k!$

Let $G$ be a simple group of order $n$. Let $H$ be a subgroup of $G$ of index $k$ with $H\ne G$. Show that $n$ divides $k!$.
0
votes
1answer
1k views

$G$ simple group and there exists a subgroup of index $n$. Show that: $|G|$ divides $n!/2$

Let $G$ be a simple group and there exists a subgroup $H$ of index $n \in \mathbb{N}_{\ge 3}$. Show that: $|G|$ divides $n!/2.$ I saw a similar question here. There was written: "The proof is really ...
0
votes
2answers
199 views

Every simple subgroup of $S_4$ is abelian

Q1 Prove that every simple subgroup of $S_4$ is abelian. Q2 Using the above result, show that if $G$ is a nonabelian simple group then every proper subgroup of $G$ has index at least $5$. My attempt ...
4
votes
1answer
205 views

On proving that a finitely generated group has a finite number of subgroups with index $n$.

Theorem: If $G$ is a finitely generated group, then it has a finite (maybe zero) number of subgroups of index $n$ for any $n\in \mathbb{N}$. Here is a sketch of a proof. It consists of assuming such ...
1
vote
1answer
67 views

$G$ can't be a simple group if $|G| > n!$ and $|G:H| <n$.

Prove that if the order of the group $G$ is bigger than $n!$ and $H < G$ is a subgroup with $|G:H| <n$, then $G$ cannot be a simple group. We got the hint that we should represent $G$ on the ...
0
votes
1answer
81 views

Questions regarding the existence of normal subgroup with finite index

If $G$ is a group with a subgroup $H$ of finite index $n$, then $G$ has a normal subgroup $N$ whose index in $G$ is finite. I found a proof of the question here: How to prove that if $G$ is a group ...

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