Cyclic group of order $4$

Question

I'm studying Artin's Algebra book and got stuck in the following problem:

13. (a) Let $G$ be a group of order $4$. Prove that every element of G has order $1, 2$ or $4$.

My (incorrect) reasoning was: if G has order $4$, then there exists an element $g \in G$ such that $g$ generates $G$: $G = \{1, g, g^2, g^3\}$, with $g^4 = 1$. Therefore $1$ has order $1$, $g$ has oder $4$, $g^2$ has order $2$ $((g^2)^2 = g^4 = 1)$, and g^3 has order $4$ $((g^3)^4 = (g^4)^3 = 1$, and it is the smallest such positive integer, because $(g^3)^2 = g^2 \neq 1$ and $(g^3)^3 = g \neq 1)$.

I know that is wrong because the following question is

13. (b) Classify groups of order $4$ by considering the following two cases: (i) $G$ contains an element of order $4$ (ii) Every element of $G$ has order < 4.

So I infer that I have used some wrong premise, but I'm really confused and can't identify which was it. Any help would be much appreciated.

Answer 1

Your mistake was in the first sentence

If $G$ has order 4, then there exists an element $g \in G$ such that $g$ generates $G$.

This is not true, $G$ having order 4 just means that it has 4 elements.