1. ## 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

2. 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.

3. 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 $2n$

for odd values of $n$ is always trivial.)

4. 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

5. Am I right in thinking that the second Sylow theorem forces it to be normal as it must be conjugate to itself?

6. 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).

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 $|G| = p^am$ where $p\not | m$. Let $n$ be the number of Sylow p-subgroups then $n|m$ and $n\equiv 1(\bmod p)$. For example, $|G| = 2\cdot 59$ then the number of Sylow 59-subgroups satisfies $n|2$ and $n\equiv 1(\bmod 59)$, the only such number is $n=1$. Which means there is only one Sylow 59-subgroup. Now if $P$ is Sylow then $aPa^{-1}$ is also Sylow since $|P| = |aPa^{-1}|$. This means if if there is only one Sylow subgroup, like with the above example, then it is invariant under conjugation, i.e. $aPa^{-1} = P$ for all $a\in G$. This means that $P$ has to be normal, because normal subgroups are precisely the groups invariant under conjugation.

7. Thank you again May I ask how you would show that H is normal? Thank you!