-4

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

user3491700
  • 153
  • I think you need to state your question more clearly. What are the conditions and what is the desired conclusion? Are you claiming that there always exists a normal subgroup with this property? Are you claiming that every sugroup with $p^{n-1}$ elements is normal? I added some LaTeX formatting, you'll see how it works immediately. – Jesko Hüttenhain Mar 02 '17 at 08:05
  • The writing in the title and the content of the question seem to differ. According to the title $N$ should be of order $3^2 = 9$. And also what's the point of introducing $B$? – Stefan4024 Mar 02 '17 at 08:06
  • http://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index – Prahlad Vaidyanathan Mar 02 '17 at 08:07
  • Sorry for the confusion, I messed up the question. I essentially am asking the title question, allow me to edit – user3491700 Mar 02 '17 at 08:11
  • I know that obviously normality is gNg^(-1)=N and this is necessary to prove. However I feel I am missing an obvious piece of information because I am blanking out on where to start – user3491700 Mar 02 '17 at 08:22

1 Answers1

2

In the link from Prahlad's comment you can see how if $H$ is a subgroup of $G$, s.t. $[G:H] = p$ and $p$ is the smallest prime divisor of $H$, then $H\unlhd G$

Here's one different (if I dare to say a simplier) approach that works for $p-$groups. In fact we'll prove that for each $1 \le i \le n$ there exists a normal subgroup of order $p^i$. As $G$ is a $p-$ group we have that $Z(G) \not = 1$, so by Cauchy Theorem there exists an element $x$ of order $p$, s.t. $x \in Z(G)$. Now it's fairly trivial to prove that $\langle x \rangle \unlhd G$. Now that the base case is proven assume that this is true for all $i$ s.t. $i \le k$. By induction hypothesis $G$ has a normal subgroup $K$ of order $p^k$. Now we can consider the group $G/K$, which as a $p-$ group contains a normal subgroup of order $p$. Using the Correspondence Theorem we have that this group corresponds to a normal subgroup of $K$ of order $k+1$. Hence by induction the claim is proven.

Stefan4024
  • 35,843
  • 1
    But what you prove is a different claim (though it is not entirely clear which one the OP asks about). The existence of a normal subgroup of a given order does not imply that all subgroups of that order are normal. – Tobias Kildetoft Mar 02 '17 at 08:53
  • @TobiasKildetoft I might be wrong, but the title of the question says that "some $N$ is a normal, where $|N| = p^{n-1}$. I understand this as prove that there exists a normal group of order $p^{n-1}$. – Stefan4024 Mar 02 '17 at 08:55
  • @TobiasKildetoft To clarify it never says prove that ALL group of that order are normal. (which is true too). – Stefan4024 Mar 02 '17 at 08:56
  • 1
    As I said, the title is not very clear. It could also mean that the only assumption on $N$ is the order (this fits better with the text in the body). In either case, the question is a duplicate (the target might just change), but this is a nice answer anyway. – Tobias Kildetoft Mar 02 '17 at 08:57