In a previous question, I was told that
$$Z_2 \rtimes_\varphi Z_2= G_2$$
$G_2$
$$\begin{array}{c|c|c|c|c}
\cdot & e
& a & b& c\\\hline
e & e & a & b & c \\\hline
a &a & e & c& b\\\hline
b & b & c & a& e \\\hline
c & c & b & e&a
\end{array}$$
By the Wikipedia page I know that $\varphi: Z_2 \to \text{Aut}(Z_2)$
I’ve reasoned that Aut$(Z_2)$ is the trivial group.
Then, there is only one option for $\varphi$ which creates $G_1$
$$\begin{array}{c|c|c|c|c}
\cdot & e
& a & b& c\\\hline
e & e & a & b & c \\\hline
a &a & e & c& b\\\hline
b & b & c & e & a \\\hline
c & c & b & a & e
\end{array}$$
Where am I wrong?
2026-03-27 22:18:55.1774649935
Semidirect product of $Z_2$ with itself
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