$\forall H, \exists G\unrhd H$ with automorphisms of $H$ obtained by restricting inner automorphisms of $G$ to $H$.

Question

18. Show that if $H$ is any group then there is a group $G$ that contains $H$ as a normal subgroup with the property that for every automorphism $\sigma$ of $H$ there is an element $g\in G$ such that conjugation by $g$ when restricted to $H$ is the given automorphism $\sigma$, i.e., every automorphism of $H$ is obtained as an inner automorphism of $G$ restricted to $H$.

For $G=H\rtimes_\varphi K$, view $G$ as the set of ordered pairs $\{(h,k)\}$ s.t. $(h_1,k_1)(h_2,k_2)=(h_1(\varphi(k_1)(h_2)),k_1k_2)$

My guess is that $G=\operatorname{Hol}(H)=H\rtimes \operatorname{Aut}(H)$. For example, let $H=\Bbb Z_2\times\Bbb Z_2=\{1,x,y,xy\}$, then $\operatorname{Aut}(H)\cong S_3$, where each symmetry permutes the $3$ non-identity elements of $H$. $G=\operatorname{Hol}(H)=(\Bbb Z_2\times\Bbb Z_2)\rtimes_\varphi S_3$:$$\varphi:S_3\rightarrow\operatorname{Aut}(\Bbb Z_2\times\Bbb Z_2)\cong S_3$$$$\varphi(12)=\sigma_{(12)}:x\mapsto y,y\mapsto x,xy\mapsto xy$$$$\varphi(123)=\sigma_{(123)}:x\mapsto y,y\mapsto xy,xy\mapsto x$$$$\vdots$$

We kind of embed both $H$ and $\operatorname{Aut}(H)$ in a larger group, in this case $G=\operatorname{Hol}(H)\cong S_4$, and $H\unlhd G$. Every automorphism $\sigma$ of $H=\Bbb Z_2\times\Bbb Z_2$ appears in the codomain of $\varphi$ and it corresponds to exactly one element in $S_3$ ($\varphi$ is a bijection). If we see $S_3$ as a subgroup of $S_4$, then $\sigma$ corresponds to $g=(1,\sigma)\in G$ s.t. conjugation by $g=(1,\sigma)$ when restricted to $H$ is $\sigma$, because this is exactly how we define a semidirect product: $K$ acts on $H$ by conjugation. e.g. $\sigma_{(12)}:x\mapsto y,y\mapsto x,xy\mapsto xy$ corresponds to the ordered pair $g=(1,(12))\in G$ s.t. conjugation by $g$ when restricted to $H$ (used during construction of the semidirect product) is $\sigma_{(12)}$.

Is my guess correct? I know the example looks clumsy but I want to know if there is any problem with the above. Thanks!

Answer 1

Your guess is correct.

Generally if you take a groups $H$ and $K$, and a group homomorphism $f: K \rightarrow \operatorname{Aut}(H)$, you can use $(H,K,f)$ to form the semidirect product $G = H \rtimes_f K$.

Then some basic facts you can check using the definitions:

• $H' = \{(h,1) : h \in H \}$ is a normal subgroup of $G$, isomorphic to $H$.
• $K' = \{(1,k) : k \in K \}$ is a subgroup of $G$, isomorphic to $K$.
• Every element $g \in G$ induces a automorphism of $H'$ via conjugation, hence an automorphism of $H$.
• For $k \in K$, conjugation by $(1,k)$ induces an automorphism of $H'$ which corresponds to the automorphism $f(k)$ of $H$.

In particular if you take $K = \operatorname{Aut}(H)$ and $f$ to be the identity map, the semidirect product $G$ is (by definition) the holomorph. It should be immediate that in this semidirect product $G$, conjugation by elements of $K'$ gives you every automorphism of $H$.