TITLE: Why does the automorphism used to construct the group have to be non-inner?
QUESTION [9 upvotes]: I have a question on why a particular assumption is made that the automorphism used to construct a certain group be non-inner.
In [Herstein, Topics in Algebra, p. 69], a construction of a nonabelian group of order 21 is given as follows:  Let $G=C_7 = \langle a \rangle$, $\phi: a \mapsto a^2$ be an automorphism of $G$ of order 3, and consider all formal symbols $x^i a^j$ subject to conditions $x^3=a^7=e, x^{-1}ax=a^2$, and we get a nonabelian group of order 21.  
The text continues that, more generally, if $G$ is a group, $T$ an automorphism of order $r$ of $G$ which is not an inner automorphism, pick a symbol $x$ and consider all elements $x^i g$, $i \in \mathbb{Z}, g \in G$, subject to $x^ig=x^{i'}g'$ iff $i \equiv i'(r), g =g'$, and $x^{-1} g^i x = gT^i$ for all $i$.  Then, we obtain a larger group $\{G,T\}$, with $G \trianglelefteq \{G,T\}, \{G,T\}/G \cong C_r$.  I think it is common to denote this group by the semidirect product $G \rtimes C_r$.  My question here is why is the assumption made that $T$ is ``not an inner automorphism.''

REPLY [5 votes]: Inner automorphisms are bijective homomorphisms $\theta_g:G\rightarrow G$ which have the form $$\theta_g(x)=g^{-1}xg$$
given a fixed $g\in G$.  If $G$ is abelian, then $xg=gx$ for every $x,g\in G$, which we can rewrite as as $g^{-1}xg=x$.  Thus every inner automorphism has the form $$\theta_g(x)=x\hspace{30pt}\text{for all }x\in G.$$
In other words, all inner automorphisms are the identity automorphism $\operatorname{id}_G$.  If one forms a semidirect product $G\rtimes_\phi H$ with the mapping $\phi(h)=\operatorname{id}$ for every $h\in H$, then $$G\rtimes_\phi H\cong G\times H.$$
In the context of semidirect products, this is a degenerate case.  To see a "non-trivial semidirect product," $\phi$ must map at least one $h$ to a non-identity automorphism of $G$, and thus when $G$ is abelian this automorphism cannot be inner.