Apologies if this question has been asked, I cannot seem to find it and I would greatly appreciate an explanation before my midterm tomorrow. My question reads
Let p be a prime, and let H ≤ Dp be a proper subgroup (i.e., not all
of Dp). Show that H is cyclic.
My only thought was that there are only two generating elements of Dp, reflection and rotation, and so a proper subgroup we could make is only using one generating element - thus subgroup is cyclic. But this is obviously not a legitimate proof (or even legitimate logic), it ignores the possibility of using both generators and placing more restrictions on them to form a proper subgroup, and I never use the fact that Dp is an n-gon with prime n. So yea, I'm missing a lot and I can't figure out why or what it is. If I could further request a more colloquial/intuitive explanation I would greatly appreciate it. Thanks.
By Lagrange, any (proper) subgroup $H$ of $D_p$ must have either $2$ or $p$ elements. If $|H| = 2$ then it is obviously cyclic. Assume therefore that $|H| = p$.
Any group of prime order has, again by Lagrange, no non-trivial subgroups, so it is cyclic.