Please check my solution.

Question

A group of order 48 must have a normal subgroup of order 8 or 16 .
Solution:Let G be a group of order n.
Let H be a normal subgroup of G.
Then G/H is a group.
Then by Lagrange's Theorem o(G/H)=o(G)/o(H)
So in this case order of G is 48 and divisors of 48 are 8 and 16.
so a group of order 48 must have a normal subgroup of order 8 or 16 .
Is may way correct?

Answer 1

Your method is incorrect. For example, the group $A_5$ of order $60$ has no non-trivial normal subgroups, but I could use your argument to show that it must have a normal subgroup of order $2$ or $4$.

Answer 2

Lagrange's theorem states if $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$. The converse is not true at all.

Have you learned about the Sylow theorems?