Ok, so the normal subgroup looks like H = {e, x} since it's a subgroup of order 2.

The definition of a normal group tells us ghg^-1 ∈ H for every g∈ G and every h ∈ H. Since H = {e,x}, only one thing can happen: we must have ghg^-1 = h for h ∈ {e,x}. In particular, this means the left and right cosets of H in G are the same, and H can be used to form a factor group, G/H.

By Lagrange's theorem, we have |G| / |H| = [G:H], or the number of cosets of H in G. So we have 60/2 =30, so G/H is a group of order 30.

It is known that the quotient group of a cyclic group is cyclic, but the converse need not hold. So I'm guessing you maybe left that condition out? The order isn't prime, so we can't conclude it's cyclic from that. You're kinda on your own, there.