Sylows Theorem |

This example shows clearly that the consideration of suitable group actions can be very helpful, at least in group theory. Applications to other fields of mathematics will follow soon.Example:The regular representation ofGyields, in accordance with formula, theG-sets[G choose k], for1 <= k <= | G |. IfGis finite andpa prime dividing| G |, say| G | = p, where^{r}·q, r ³1, q=p^{s}tpdoes not dividet, then we can putk:=pand consider the particular^{r}G-set[G choose p, as H. Wielandt did in his famous proof of^{r}]Sylow's Theoremin order to show thatGpossesses subgroups of orderp. His argument runs as follows:^{r}pis the exact power of^{s}pdividing the order of[G choose p. This is clear from^{r}]as each power of|[G choose p^{r}] | =(p^{r}q)/(p^{r})·(p^{r}q-1)/(1)...(p^{r}q-(p^{r}-1))/(p^{r}-1),pcontained in the denominator cancels. Thusp-subsets^{r}Mexist, the orbit length of which is not divisible byp. We consider such an^{s+1}Mand show that its stabilizerGis of order_{M}pby proving that^{r}pis both an upper and lower bound: For each^{r}m ÎMandg ÎGwe have that_{M}gm ÎM, henceOn the other hand, the fact that| G_{M}| <= | M | = p^{r}.pdoes not divide the orbit length^{s+1}| G(M) | = | G | / | Gyields_{M}|This proves the first item of| G_{M}| >= p^{r}.Sylow's TheoremAssumeGto be a finite group andpto be a prime divisor of its order. ThenThe subgroups

Gcontains subgroups of orderp, for each power^{r}pdividing its order^{r}| G |.S £Gof the maximalp-power order are called theSylowofp-subgroupsG.They have the following properties:

- Each
p-subgroupUofGis contained in a suitable Sylowp-subgroupS.- Any two Sylow
p-subgroups ofGare conjugate subgroups.The proof of the second and third item follows from a consideration of double cosets. Assume a

p-subgroupUofGand a Sylowp-subgroupS.Then we derive from the corollary thatwhere(| G |)/(| S |)= å_{g Î D}(| U |)/(| U ÇgSg^{-1}|),Ddenotes a transversal ofU \G/S.If all the intersections in the denominator on the right hand side were proper subgroups ofU,then the right hand side were divisible byp,which contradicts the left hand side. Hence there must exist ag ÎD,such thatU £gSgSince^{-1}.gSgis a Sylow^{-1}p-subgroup, too,Uis contained in a Sylow subgroup, which proves the second item.The third item follows by taking for

Ua Sylowp-subgroupS':shows that(| G |)/(| S |)= å_{g Î D'}(| S' |)/(| S' ÇgSg^{-1}|)S'=gSgfor a suitable^{-1},g Î D',whereD'denotes a transversal ofS' \G/S.

harald.fripertinger@kfunigraz.ac.at,

last changed: August 28, 2001

Sylows Theorem |