Explicit construction of $(\mathbb{Z}_{3} \times \mathbb{Z}_{3}) \rtimes _{\varphi} \mathbb{Z}_{2}$

Question

I am new to group theory and I wanted to explore further the concept of external semidirect product by constructing $(\mathbb{Z}_{3} \times \mathbb{Z}_{3}) \rtimes _{\varphi} \mathbb{Z}_{2}$ for a particular homomorphism $\varphi : \mathbb{Z}_{2} \to Aut( \mathbb{Z}_3 \times \mathbb{Z}_3) $. Since any homomorphism sends identity to identity, we need only to determine $\varphi(1)$.
I've chosen $\alpha \in Aut( \mathbb{Z}_3 \times \mathbb{Z}_3) $ such that $\alpha(1,0) = (2,2)$ and $\alpha(0,1) = (2,1)$ and I've checked that this is indeed an automorphism. So $\varphi(1) := \alpha$
Next, I wanted to determine the order of an element in this new group according to the rule $$ (h_1,k_1)\circ (h_2,k_2) = (h_1 \phi(k_1)h_2,k_1 k_2)$$ for the general product $H \rtimes _{\phi} K$ with homomorphism $\phi$. I've got:
$((0,1),1)\circ((0,1),1)= ((0,1) + (2,1),0) = ((2,2),0)$
Next, since $\alpha(2,2) = \alpha( (1,0) + (1,0) + (0,1) + (0,1) ) = \alpha(1,0) + \alpha(1,0) + \alpha(0,1) + \alpha(0,1) = (2,2) + (2,2) + (2,1) + (2,1) = (2,0)$
we have $((0,1),1)\circ((2,2),0)= ((0,1) + (2,0),1)=((2,1),1)$
Finally, by the same argument
$((0,1),1)\circ((2,1),1) = ((0,1) + (0,2),0) = ((0,0),0)$
so somehow I've got an element of order 4 in a group of order 18, which is clearly impossible. I've checked each step several times and I can't see my mistake. Or am I wrong in assuming that any homomorphism will work?

Answer 1

Your chosen $\alpha$ has order $4$. Therefore it cannot be $\phi(1)$, since that must have order dividing $2$.