I apologize if the title is unclear, but I've just found this while reading Herstein's Topics in Algebra (p. 69) and I am not familiar with semidirect products. My query is this.
Herstein constructs a new group from $G$ by using an automorphism $T$ of $G$. If $T$ has order $r$, he adds the formal symbol $x$ with the following rules: $x^r=e \in G$; $x^{-1}gx=gT$ (his notation to indicate the image of $g$ through $T$). It follows that the set $\{x^ig \mid i \in \Bbb Z_r; g \in G\}$ has the structure of a group.
But consider the trivial case $T=I$, the identity automorphism. It seems to me that there is no restriction on the order of $x$ here. For example, we can create a group with the above-mentioned rules except $x^2=e, x \not = e$. If $G$ is finite, $\{ G, T\}$ will have twice as many elements as $G$. In general, it seems that we can let $o(x)$ be any multiple of $o(T)$. Why does Herstein not bother with this case?