Linked Questions

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?

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 ...

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 ...

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! ...

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 ...

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 $|G: N| \le n!$.

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?

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 $...

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 ...

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 $k!$.

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 ...

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 ...

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 ...

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 ...

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 ...

This is a 'Harder' problem 40 from Abstract Algebra(1996) by Herstein. I'm just not able to figure out how to do this. even though I found a very similar post. Following is a verbatim statement of the ...

Show that if $G$ is a simple group with a subgroup $H$ of index $n>1$, then $|G| \leq n!$. Hence show that a group of order $2^k \times 3$ can never be simple for $k>1$. So I have let $X$ be ...

Let G be a finite group, if $H\lneq G$ and $|G| \nmid ( \, [G:H]! \, ) $ then prove $G$ is not simple. I used contrapositive argument. Suppose $G$ is simple then we need to prove that $ |G| \mid ( \,...