Linked Questions

Let $G$ be a finite group of order $n$ and $p$ be the minimal prime number dividing $n$. Assume that $H \subset G$ is a subgroup of index $p$. Prove that $H$ is normal: $H \trianglelefteq G$ To do ...

Assume that $G$ is finite with $p$ the smallest prime dividing its order. Suppose $H < G$ with $[G:H]=p$. Prove that $H \lhd G$. I've seen this question a few times on here but all the proofs I ...

Let $G$ be a finite group, $H \leq G$ such that $[G : H]$ is the least prime which divides $|G|$. Show that $H$ is normal in $G$. Can someone explain what information we get from knowing that $[G : H]...

Show that if a group $G$ has odd order, any subgroup $H$ of index 3 in $G$ is normal in $G$. I think this is equivalent to the following: Let $H$ and $K$ be subgroups of a group $G$, with $K \leq H$...

If $|G|=p^{n}$ Then Why is it that every subgroup of order $p^{n-1}$ is normal?

Basically what the title says. If $|G|=27$ for example, how would one prove that $N$ is normal in $G$ if $|N|=9$

This is a problem from Herstein that I have been stuck upon for ages. I am becoming increasingly disappointed and disillusioned about my abilities due to this problem. Let G be finite and let $p$ ...

Let $G$ be a finite group, and let $H$ be a subgroup of $G$ such that the index $(G\colon H)$ of $H$ in $G$ is the smallest prime that divides the order of $G$. Can we say anything about whether or ...

It is known that any subgroup of a group having index 2 is normal. Note that here 2 is a prime. Is such a result true for some other odd prime ?. That is, "does there exist an odd prime $p$ such ...

Let $G$ be a group and $H$ a non-normal subgroup of $G$. Please prove that the order of $G$ is divided by a prime number which is less than $[G:H]$.

Given that $H\leq G$ of index $p$ in $G$, where $p$ is the smallest prime integer such that $p \mid |G|$, then $H \trianglelefteq G$. I would appreciate some hints, as I don't even know where to begin....

If $G$ is a finite group of order $n$ and $p$ is the smallest prime dividing $|G|$, then prove that any subgroup of index $p$ is normal. This falls under the category of Cayley's Theorem, but i have ...

I am trying to prove that a group of order $p^2q^2$ where $p$ and $q$ are primes is solvable, without using Burnside's theorem. Here's what I have for the moment: If $p = q$, then $G$ is a $p$-group ...

Could you please give an example of a subgroup of index $3$ which is not normal ? I know every subgroup of index $2$ is normal but if index is $3$ , I have no idea whether all of the subgroups are ...

I had seen this proof at many places, but everywhere sylows theorem is used. So is their any way to solve it without using sylows theorem?