Let $N$ be a finite, normal subgroup of $G$, and $H$ be a subgroup of $G$ such that $[G:H]$ is finite. We were tasked to show that if $[G:H]$ and $|N|$ are relatively prime, then $N\leq H$.
Initially, I thought of just using the Lagrange Theorem for each group-subgroup relationships that I have above, but realized that $|G|$ and $|H|$ might be groups of infinite order. I am having trouble and I don't know if what I thought initially is correct. Just give me hints on how to start and I'll do the rest. I still want to prove it on my own. Thank you very much.
Best Answer
Since $N$ is normal, $HN$ is a subgroup of $G$, and since $HN/N \cong H/(H \cap N)$, we have $|HN:H|=|N:H \cap N|$. These last indices are finite and divide $|G:H|$ and $|N|$. Since these last numbers are relatively prime, it follows that $|HN:H|=1=|N: H \cap N|$. Hence, $HN=H$, which is equivalent to $N \subseteq H$.