Suppose that a group G has a subgroup of order n. Prove that the intersection of all subgroups of G of order n is a normal subgroup of G. I have no idea where to go on this besides letting A= the intersection of all subgroups of order n.
Suppose that a group G has a subgroup of order n. Prove that the intersection of all subgroups of G of order n is a normal subgroup of G. I have no idea where to go on this besides letting A= the intersection of all subgroups of order n.
This is the sketch of the proof.
Let H be a subgroup of group G of order n and M be the intersection of all subgroups of G of order n. It follows immediately that M is a subgroup of G. We remain to show that M is normal in G. We see that all subgroups of order n in G are either conjugate subgroups of H or subgroups of order n in G that are not conjugate to H but are conjugate to other subgroups of order n in G.
Let K be the intersection of all conjugates of H in G. Since the intersection of all conjugates of H is normal in G (link 1)(link 2) and $\displaystyle M \subset K$, we see that $\displaystyle gMg^{-1}=M$ for every $\displaystyle g \in G$. Thus M is a normal subgroup of G.
Edit : The last line holds because M is the intersection of all conjugate subgroups of subgroups of order n in G, where $\displaystyle M = \bigcap_{i \in I}\bigcap_{g \in G}gH_ig^{-1}$ and each $\displaystyle H_i$ for $\displaystyle i \in I$ denotes a subgroup of a group G of order n.