# Thread: Finite simple groups.

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

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

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

3. ## Re: Finite simple groups.

Thank you very much Sylvia.