1. ## Finite simple groups.

Hi: Problem: Let U be a subgroup of G and 1 < |G:U| <5. Then |G| < 4 or G is not simple.

If |G:U| = 2 I can use a problem (problem A) that says: Let p be the smallest prime divisor of |G|. Then every subgroup of index p is normal in G. That is because in this case, 2 is a divisor of |G| and it is the smallest prime. For case |G:U| = 3 if |G| is not even then, by problem A I am done. If |G| is even, however, problem A is of no use. The same for |G:U| = 4. Any hint?

2. ## Re: Finite simple groups

There is a result in group theory that says that if $n=|G:U|$ then there is a normal subgroup $N$ of $G$ contained in $U$ such that $n$ divides $|G:N|$ and $|G:N|$ divides $n!.$ (I don't know what this result is called but I've found it useful. ) Thus for $n=3,$ $|G:N|=3$ or $6$ so $N\ne1;$ if $|G|>3,$ then $N\ne G$ either (since $N$ is contained in $U$ and $|G:U|=3),$ showing that $G$ is not simple.

If $|G:U|=4,$ then $4\mid|G:N|$ and $|G:N|\mid4!=24$ $\implies$ $G$ is not simple if $|G|>4.$

If $|G:U|=2$ then $U$ itself is a normal subgroup of $G$ and so $G$ is not simple if $|G|>2.$

3. ## Re: Finite simple groups.

Thank you very much Sylvia.