G a group of order 35, H a normal subgroup of order 7. Prove that if g in G has order 7, then g is in H.
What I'm doing is invoking Sylow's third theorem to show that there exist a single subgroup of order 5 and a single subgroup of order 7. And then the direct product of these has order 35, and then somehow… this means G has an element of order 35, which would mean G is cyclic, and if G is cyclic, each of its subgroups are cyclic. So H has an element of order 7 and if there's an element g in G of order 7, then it generates the subgroup H?
This is my train of logic? I could use some help on the somehow… part, or if that's even a valid way of proving that a group of order 35 is cyclic. Please help steer me back on track.
Best Answer
Since $|G|=35=7\cdot 5$, a subgroup $H$ of order $7$ is a Sylow-$7$-subgroup. Moreover, given that $H$ is normal, it must be the only Sylow-$7$-subgroup of $G$. Now, if $g\in G$ has order $7$, then $\langle g\rangle $ is a $7$-subgroup of $G$ and hence is contained in a Sylow-$7$-subgroup of $G$. But $H$ is the only Sylow-$7$-subgroup, so $\langle g\rangle\subseteq H$. Therefore, $g\in H$.