# Finite simple groups.

• Apr 19th 2012, 07:59 PM
ENRIQUESTEFANINI
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?
• Apr 20th 2012, 04:13 PM
Sylvia104
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. (Wink)) 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.\$
• Apr 22nd 2012, 05:17 PM
ENRIQUESTEFANINI
Re: Finite simple groups.
Thank you very much Sylvia.