# Sylow Theorems

• May 19th 2008, 01:33 PM
alittletouched
Sylow Theorems
Hey all. I am just trying to figure out the following. It's the last thing I need to revise before my exam. Any help would be fantastic. Thank you! James

Take a group of order 117. This has prime factors 3 and 13. Prove that every group with order 117 has an index 3 cyclic subgroup (normal) and a centre which is not the identity

Take a group of order 118. This has prime factors 2 and 59. Prove that every group with order 118 has an index 2 cyclic subgroup (normal) and a centre which is not the identity

I think that these make use of the Sylow Theorems at some point but I am not sure exactly how to use them

Thank you
• May 19th 2008, 05:01 PM
ThePerfectHacker
For the group of order 117 you can take a subgroup H of order 3 and a subgroup K of order 13. The one with order 13 happens to be normal by showing the number of Sylow 13-subgroups is one. Since |H|,|K| are relatively prime it means |HK| = |H||K| = 39, furthermore HK is a subgroup since K is a normal subgroup. Thus, HK is a subgroup of index p=3, where p is the smallest prime dividing |G|, thus HK is a normal subgroup. Finally, HK is isomorphic to H x K which is a cyclic group since H and K are cyclic groups of relative prime orders.

For the group of order 118 you can take a subgroup H of order 59, it happens to be normal again by using Sylow theorems. It is cyclic since H it is prime order. And finally it is an index 2 subgroup.
• May 19th 2008, 09:23 PM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
Finally, HK is isomorphic to H x K which is a cyclic group since H and K are cyclic groups of relative prime orders.

unfortunately this is not correct, because HK in general doesn't have to be abelian. so, in general we'll have a semidirect

product rather than a direct one. i don't know how to prove the existence of such a cyclic subgroup right now but it's easy

to see that it will follow immediately if we can prove that the center of the group is non-trivial.

regarding the second problem, this claim in the problem that any group of order 118 has a non-trivial center is not true.

for example, the center of the dihedral group of order 118 is trivial. (in general, the center of the dihedral group of order $\displaystyle 2n$

for odd values of $\displaystyle n$ is always trivial.)
• May 20th 2008, 01:40 AM
alittletouched
Thank you very much. Is there any chance you could please explain why the number of Sylow p-groups being 1 makes it normal? Many thanks. James
• May 20th 2008, 01:51 AM
alittletouched
Am I right in thinking that the second Sylow theorem forces it to be normal as it must be conjugate to itself?
• May 20th 2008, 06:13 AM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
unfortunately this is not correct, because HK in general doesn't have to be abelian.

Of course, that was a mistake. (I was using the theorem that if H and K are disjoint normal and HK is the whole group then the group is isomorphic to H x K, the problem here is that I did not show H is normal).

Quote:

Thank you very much. Is there any chance you could please explain why the number of Sylow p-groups being 1 makes it normal? Many thanks. James
Yes. Write $\displaystyle |G| = p^am$ where $\displaystyle p\not | m$. Let $\displaystyle n$ be the number of Sylow p-subgroups then $\displaystyle n|m$ and $\displaystyle n\equiv 1(\bmod p)$. For example, $\displaystyle |G| = 2\cdot 59$ then the number of Sylow 59-subgroups satisfies $\displaystyle n|2$ and $\displaystyle n\equiv 1(\bmod 59)$, the only such number is $\displaystyle n=1$. Which means there is only one Sylow 59-subgroup. Now if $\displaystyle P$ is Sylow then $\displaystyle aPa^{-1}$ is also Sylow since $\displaystyle |P| = |aPa^{-1}|$. This means if if there is only one Sylow subgroup, like with the above example, then it is invariant under conjugation, i.e. $\displaystyle aPa^{-1} = P$ for all $\displaystyle a\in G$. This means that $\displaystyle P$ has to be normal, because normal subgroups are precisely the groups invariant under conjugation.
• May 20th 2008, 06:40 AM
alittletouched
Thank you again :) May I ask how you would show that H is normal? Thank you!