Prove $\mathbb{Z}_4\ncong \mathbb{Z}_2\times\mathbb{Z}_2$

Question

Question: How can I show $\mathbb{Z}_4\ncong \mathbb{Z}_2\times\mathbb{Z}_2$ without considering index (order of elements), without using cyclic argument or using Sylow's theorem?

Is there a way to do it? I usually try to prove something is isomorphic to another thing, so I'm kinda lost trying to prove the opposite. And I thought about the two ways bellow but they didn't help me much.

Both of them are commutative and have the same cardinality ($\mathbb{Z}_4 =\{\overline{0},\overline{1},\overline{2},\overline{3}\} \implies \#\mathbb{Z}_4 =4$ and $\mathbb{Z}_2\times\mathbb{Z}_2 =\{(\overline{0},\overline{0}),(\overline{0},\overline{1}),(\overline{1},\overline{0}),(\overline{1},\overline{1})\} \implies \#(\mathbb{Z}_2\times\mathbb{Z}_2) =4$).

Thanks in advance. :)

Answer 1

Let $f:\mathbb Z_4 \to \mathbb Z_2 \times\mathbb Z_2$ be a group morphism, with $f(1)=x$. Then $f(2)=f(1)+f(1)=x+x=0=f(0)$, whence $f$ cannot be an isomorphism.