Let p be a prime and H be a subgroup of a group G of order 2p. If H is not normal in G, then by Lagrange's Theorem, H must have order 2. This is because the only possible orders for H are 2, p, or 2p. If the order is 2p, then H = G and is normal. If the order is p, then the index of H in G is 2, meaning H is normal. Therefore, if H is not normal, its order must be 2.

1. Let p be a prime. Show that if H is a subgroup of a group with order 2p that is not normal, then H
has an order of 2.
I feel that I need to use Lagrange's Theorem but I am not sure how to get started. Anything will
help, thanks!
Solution
You are correct you need Lagrange's Thm. Notice the only divisors of 2p are 2 and p. This
implies that the order of H is either 2,p, or 2p. If order of H = 2p then H = G and H is normal in
G If order of H = p then the index of H in G (usually denoted [G:H]) is 2. This means there are
only two distinct cosets of H in G. This guarantees H is normal. It is fairly straight forward to
prove any subgroup of index 2 is normal. If you would like a proof write back. Thus if H is not
normal it must have order 2.