Linked Questions
49 questions linked to/from Normal subgroup of prime index
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4answers
3k views
$G$ is a finite group such that $|G| = n$ and $p$ is minimal prime dividing $n$. $H \subset G, [G:H] = p$. Prove $H$ is normal in $G$ [duplicate]
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 ...
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2answers
2k views
If a subgroup has smallest prime index, then it is normal [duplicate]
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 ...
3
votes
1answer
678 views
$H \leq G$ s.t. $|G : H|$ is the least prime dividing $|G|$. Show $H$ is normal in $G$. [duplicate]
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]...
-1
votes
1answer
954 views
Show that if a group $G$ has odd order, any subgroup $H$ of index 3 in $G$ is normal in $G$? [duplicate]
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 $K \leq H$...
-4
votes
1answer
632 views
Proving that for $|G|= p^n$, that some subgroup $N$ is normal in G, where $|N|=p^{n-1}$ [duplicate]
Basically what the title says. If $|G|=27$ for example, how would one prove that $N$ is normal in $G$ if $|N|=9$
0
votes
1answer
794 views
Why is every subgroup of order $p^{n-1}$ normal? [duplicate]
If $|G|=p^{n}$
Then
Why is it that every subgroup of order $p^{n-1}$ is normal?
3
votes
0answers
458 views
Let G be finite and let $p$ be the smallest prime dividing $|G|$. Let $H \le G$ be of index $p$. Prove that $H$ is a normal subgroup of $G$. [duplicate]
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 $p$ ...
0
votes
2answers
280 views
normality of subgroup of odd prime index [duplicate]
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 ...
0
votes
1answer
206 views
Non-normal subgroup [duplicate]
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 $[G:H]$.
1
vote
1answer
121 views
Is this type of a subgroup always normal? [duplicate]
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 ...
0
votes
1answer
127 views
$|G|<\infty$. $p$ the smallest prime dividing $|G|$. $|G:H|=p$. then H is normal. [duplicate]
Since $p$ is the smallest prime dividing $|G|$, and $|G:H|=p$, we know that $H$ is the largest proper subgroup of $G$. So the normalizer $N_G(H)$ must be equal to either $G$ or $H$.
If $N_G(H)=G$, ...
1
vote
0answers
96 views
If $H\leq G$ of index $p$ in $G$ and $p\mid |G|$ then $H\trianglelefteq G$ [duplicate]
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....
0
votes
2answers
62 views
How do i show any subgroup of index $p$ is normal? [duplicate]
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 ...
0
votes
1answer
27 views
$p$ smallest prime dividing $G$ subgroup with index $p$ normal in $G$ [duplicate]
I am struggling to understand this proof.
Suppose that $p$ is the smallest prime that divides $|G|$ show that any subgroup of index $p$ is normal in $G$.
Proof
Let $\phi \rightarrow S_p$ be the ...
0
votes
0answers
22 views
About coset and normal subgroup [duplicate]
Let $H\le G$ and suppose that $[G:H]$ is the least positive prime factor of $|G|$.
How to show that $H\triangleleft G$?
I am confused in how to establish relation between the number of cosets of $H$ ...