On groups with $\operatorname{Inn}(G) \cong D_8$ or $A_4$.

Question

Let $G/Z(G) \cong D_8$, where $D_8$ is the dihedral group of order $8$ and $Z(G)$ is the center of $G$. What are the possibilities for the group $G$ and does there always exist a non-inner automorphism group $G$?

Same question with $A_4$ in place of $D_8$.

Thanks.

Answer 1

There are infinitely many such groups. For example, $G / Z(G) \cong D_8$ when $G \cong A \times D_{16}$ with $A$ any abelian group. Similarly, $G / Z(G) \cong A_4$ when $G \cong B \times \operatorname{SL}_2(3)$ with $B$ any abelian group.

The "right question" to ask here is for those groups $G$ such that $G / Z(G) \cong D_8$ (or $A_4$), where $G$ has no abelian direct factors. I do not know the answer to that question. You should have a look at capable groups.

Answer 2

For the dihedral group of order $8$ with presentation $\langle x, y| x^2=y^2=(xy)^4=1\rangle$, consider the group $G=\langle x, y| x^2=y^2=(xy)^4\rangle$. $G$ is of order $48$ with center of order $6$, generated by $x^2$.

For the group $A_4$ with presentation $\langle x, y| x^2=y^3=(xy)^3=1\rangle$, consider the group $G=\langle x, y| x^2=y^3=(xy)^3\rangle $ of order $72$ with center of order $6$, generated by $x^2$.