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We know that

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

I am kindly asking to know the possible presented references including the following point, if anybody is aware of them.:

The classification all finite groups G whose possess a single proper non-trivial normal subgroup.

Thanks for your time.

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What you're asking for is essentially one step further than the classification of all finite simple groups. I doubt much is known past a few examples. The true answer is probably at least an order of magnitude more complicated then the enormous theorem. –  Jim Mar 10 '13 at 17:40
@m.k. actually, the normal subgroup need not be simple, it is possible to construct examples where it is just characteristically simple (even abelian). I will elaborate with some examples when I am on a proper computer. –  Tobias Kildetoft Mar 10 '13 at 17:48
As @TobiasKildetoft said, all such groups are (prime) cyclic extensions of characteristically simple groups, and those are just direct products of (the same) simple group. –  user641 Mar 10 '13 at 17:55
@Jim: You are right in saying that this is orders of magnitudes more difficult that classifying finite simple groups. The types of examples are described in Andreas Caranti's post. You could regard most of these as fully classified, but the exception is groups in which the unique normal subgroup is a noncentral elementary abelian $p$-group, and the quotient group is a nonabelian simple group. Classifying these is equivalent to describing all irreducible representations of all finite simple groups in all characteristics (and also the associated second cohomology groups). –  Derek Holt Mar 10 '13 at 22:43
@BabakS.: You don't need to apologize for asking the question. I upvoted it because I think it's a good one, I just wanted to point out that it's a very hard question and so we shouldn't expect much more then some families of examples. –  Jim Mar 10 '13 at 23:14

1 Answer 1

up vote 7 down vote accepted

You will get similar examples by taking a finite, nonabelian simple group $S$, and extending it by an outer automorphism of prime order to a group $G$.

Somewhat dually, you can take a quotient $P$ of prime order of the Schur multiplier of $S$, and extend $P$ by $S$ to a group $G$.

Another class of soluble examples can be obtained by starting with two distinct primes $p, q$. Consider the period $n$ of $p$ modulo $q$. Then in the multiplicative finite field $\mathbf{F}_{p^{n}}$ there is a subgroup $Q$ of order $q$ that acts irreducibly on the additive group $P$ of $\mathbf{F}_{p^{n}}$. The semidirect product $G = PQ$ will have the property, with $P$ the only nontrivial, proper normal subgroup.

Coming back to insoluble examples, one can take a direct power $S^{p}$ of a nonabelian, finite simple group $S$, with $p$ prime. If you let a cyclic group $C_p$ of order $p$ permute cyclically the factors in $S^{p}$, you should get another example, with $S^{p}$ as the distinguished normal subgroup.

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Thanks for your time. I think I could construct such that group. –  Babak S. Mar 10 '13 at 17:54
Interestingly (at least to me), the example with the two distinct primes gives precisely those finite groups whose order is not a power of a prime, but such that the order of any proper subgroup is a power of a prime. –  Tobias Kildetoft Mar 11 '13 at 1:09
@TobiasKildetoft: I am very thankful if you make me the group touchable. Thanks both of you, Andreas and you. –  Babak S. Mar 11 '13 at 9:22
@BabakS. I am not sure what you mean by touchable. –  Tobias Kildetoft Mar 11 '13 at 12:53
@BabakS. A very concrete set of examples are the dihedral groups of odd prime degree (so of order $2p$ for an odd prime $p$). –  Tobias Kildetoft Mar 11 '13 at 13:29

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