# Thread: If every maximal subgr of G is simple and normal then |G| = 1, p, p^2 or pq.

1. ## If every maximal subgr of G is simple and normal then |G| = 1, p, p^2 or pq.

Hi:
In the book The Theory of Finite Groups: An Introduction, by Hans Kurzweil and Bernd Stellmacher, New York, 2004, in section 1.6, Products of Groups, there is Exercise 3:

Let $G$ be finite. Suppose that every maximal subgroup of $G$ is simple and normal in $G$. Then $G$ is an Abelian group and $|G| \in \{1, p, p^2, pq\}$, where $p, q$ are primes.

I proved this: Either $G$ has exactly one maximal subgroup or $G$ is a direct product of two maximal subgroups. And went on:

Case only one max: let it be $H$. Then $G$ cyclic, $H$ cyclic. And now I must use an elementary statement from another book, which bothers me most, and is this:

$T \unlhd G$, $T$ cyclic. Then $S \leq T \Rightarrow S \unlhd G$. (1)

Accordingly, $<1>$ and $H$ are the only subgroups in $H$. Then $|H| = p$ . And by means of $G/H$ I discover that $|G/H| = q$, and I'm done with this case.

Case $G = G_1 \times G_2$: for this case I tried to use commutators but I did not get anything. I think I could adapt the proof of (1), substituting the two conditions T maximal and T simple for T cyclic. But in proving (1) I directly manipulated elements of groups instead of groups, making an adaptation difficult.

Could you give me some hint?

EDIT:
I can now prove (1) using only "global properties": Every subgroup of a cyclic group G is a characteristic subr of G. Hence, if S subgr of T, S is characteristic in T. But T normal in G. Therefor, S normal in G.