Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a famous Theorem telling that:

For $n≥5$, $A_n$ is the only proper nontrivial normal subgroup of $S_n$.

For the proof, we firstly start with assuming a subgroup of $S_n$ which $1≠N⊲S_n$. We proceed until at the last part of proof's body, we assume $N∩A_n=\{1\}$. This assumption should be meet a contradiction with normality of $N$ in $S_n$. There; we get $N=\{1,\pi $} in which $\pi$ is an odd permutation of order $2$. Now for meeting desire inconsistency, I have two approaches:

(a) Since every normal subgroup, having two elements, lies in the center of $G$ so, our $N⊆ Z(S_n)=\{1\}$ for $n≥5$ and then $N=\{1\}$. enter image description here

(b) Clearly, $1≠N$ acting on set $\Omega=\{1,2,...,n\}$ is intransitive wherein $|\Omega|≥5$ and according to the following Proposition $S_n$ would be imprimitive. enter image description here

Proposition 7.1: If the transitive group $G$ contains an intransitive normal subgroup different from $1$, then $G$ is imprimitive (Finite Permutation Groups by H.Wielandt).

May I ask if the second approach is valid? I am fond of knowing new approach if exists. Thanks.

share|cite|improve this question
Yes it is valid, but by the time you know |N|=2, you know N is central. Perhaps use the primitivity result earlier to conclude that N is transitive, and so N does intersect An. – Jack Schmidt Jun 18 '12 at 18:31
@JackSchmidt: Thanks Jack. Honestly, I did the second one and wanted to give it to my Prof. :-) – Babak S. Jun 18 '12 at 18:39
Make sure you can easily prove the result in Wielandt (it is easy, orbits of a normal subgroup are blocks), and it sounds fine. (a) is even easier though :-) – Jack Schmidt Jun 18 '12 at 18:41
up vote 5 down vote accepted

You are almost there. Try to prove that $Z(S_n)= 1$ for all $n \geq 3$. Then if $N$ is non-trivial and normal, you assume $N \cap A_n = 1$. This implies $N \subseteq Z(S_n)$. Why? Because in general, if $N \unlhd G$ and $N \cap [G,G] = 1$ then $N \subseteq Z(G)$.
We conclude that the normal subgroup $N \cap A_n \neq 1$. At this point I assume that you know that $A_n$ is a simple group for $n \geq 5$. Hence $N \cap A_n = A_n$, so $A_n \subseteq N \subseteq S_n$. Since $index[S_n:A_n] = 2$, it follows that $N=A_n$ or $N=S_n$.

share|cite|improve this answer
Thanks Nicky for the answer. – Babak S. Jun 18 '12 at 19:16
@Babak, you are welcome! Your question got me thinking about the general case: can one classify all finite groups $G$ possessing a single proper non-trivial normal subgroup $N$? Here $N$ must be characteristic and characteristically simple and hence a direct product of isomorphic simple groups. Can more be said here about the structure of $N$ and $G$? Have to think about this. – Nicky Hekster Jun 18 '12 at 21:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.