(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Assume that G is a group of order 63 that has two Sylow subgroups whose intersecion is non-trivial. Show that G has an element of order 21.

2. Relevant equations

3. The attempt at a solution

So by Sylow's theroem, I know that 63=3^2*7, that the sylow 7 subgroup is a normal subgroup of G with 6 order 7 elements. My sylow 3 subgroup is of order 9 and there can be one or 7 sylow three subgroups. Since I know that s7 (sylow 7 subgroup) is normal.

So since we have two sylow subgroups whose intersection is non-trival, ths means that we have 7 sylow 3 subgroups. This means we have 14 elements of order 3 in are group G. Since we have a unique sylow 7 subgroup, we have 6 elements of order 7 in G.

Let H = sylow 7 subgroup.

So for every a in G, aha^-1 is a subgroup of order 7, so we must have aHa^-1=H (since H is normal). So N(H) = G (N(H)= normalizer of H). SInce H has prime order, it is cyclic and abelian. So C(H) contains H (C(H)= centralizer of H in G). So 7 divides |C(H)| and |C(H)| divides 63. So C(H)= H or C(H)=G, or |C(H)|= 21.

Is this enough to get that xy=yx wth |x,y|=21?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Centralizer question

**Physics Forums | Science Articles, Homework Help, Discussion**