If $G$ is non-abelian, then $\operatorname{Inn}(G)$ is not a normal subgroup of the group of all bijective mappings $G \to G$

Question

Let $(G,\cdot)$ be a group and let $\mathfrak{S}(G)$ be the set of all bijective mappings from $G$ to $G$.

Show that: If $G$ is non-abelian, then $\operatorname{Inn}(G):=\{\kappa_a \mid a\in G\}$ is not a normal subgroup of $(\mathfrak{S}(G),\circ)$

First of all, this is not homework, but we had it in university as an exercise and didn't get a solution from our lecturer. So I am just looking for a correct proof, since we have easter holidays and it is not possible to ask him.

Thanks in advance!

Answer 1

Since $G$ is not abelian, there exists $g \in G \setminus Z(G)$, and hence there exists $h \in G$ with $h^{-1}gh \ne g$. So the inner automorphism $\kappa_h$ does not fix $g$. Let $x$ be the transposition $(1,g) \in {\rm Sym}(G)$. Then $(1,g)^{-1}\kappa_h(1,g)$ does not fix $1$ and so cannot be lie in ${\rm Aut}(G)$.

Answer 2

If $G$ is finite,

Let $G$ be a group of order $n$ sice $G$ is nonabelian $6\leq n$.

$Inn(G)\cong G/Z(G)$ so $|Inn(G)|\leq n < n!/2$

Since $S_n$ has uniqe normal subgroup of order $n!/2$, $Inn(G)$ is not normal in $S_G$. (notice that $S_n\cong S_G$).

Answer 3

$\operatorname{Inn}(G)\unlhd$ $\mathfrak{S}(G)$ if and only if, for all $a\in G$ and $\sigma\in\mathfrak{S}(G)$, there is $b\in G$ such that $\sigma^{-1}\kappa_a\sigma=\kappa_b$, where $\kappa_x$ is the conjugation by the element $x$. In turn, $\sigma^{-1}\kappa_a\sigma=\kappa_b$ if and only if (definition), for every $g\in G$: \begin{alignat}{1} &\sigma^{-1}(\kappa_a(\sigma(g)))=\kappa_b(g) &&\iff \\ &\sigma^{-1}(a^{-1}\sigma(g)a)=b^{-1}gb &&\iff \\ &a^{-1}\sigma(g)a=\sigma(b^{-1}gb) &&\color\red{\stackrel{g=1}{\Longrightarrow}} \\ &\sigma(1)a=a\sigma(1) &&\Longrightarrow \\ &G\text{ is abelian} \\ \end{alignat} where the last implication holds because $\{\sigma(1),\sigma\in\mathfrak{S}(G)\}=G$. The claim follows by taking the contrapositive.