Is a group always contained in a group that surjects onto its automorphism group?

Question

Let $G$ be a group. I am interested in embedding $G$ in a group $\tilde G$ such that there is a surjective map $\tilde G\rightarrow\operatorname{Aut}G$ whose restriction to $G$ yields the homomorphism $G\rightarrow\operatorname{Inn}G$ given by $g\mapsto \rho_g$ where $\rho_g$ is conjugation by $G$.

Does there exist a uniform construction of such a $\tilde G$?

I suppose another way to ask this question (edit 11/5/14 - this is actually a stronger question, see link below) is,

Is the restriction map $H^2(\operatorname{Aut}G, Z(G))\rightarrow H^2(\operatorname{Inn}G,Z(G))$ always injective, and does the class of $G$ (seen as an extension of $\operatorname{Inn}G$ by $Z(G)$) always lie in the image?

This feels like a long shot, so if the answer is no, are there some conditions on $G$ that make it true?

I am happy to suppose $G$ is finite if this is useful.

Thanks in advance.

EDIT 11/5/14: I realized after this question was fully answered that I actually wanted something stronger. I have followed up with another question here.

Answer 1

How about the holomorph $H=G \rtimes \operatorname{Aut}(G)$, where $\operatorname{Aut}(G)$ acts on $G$ naturally (ie. by $\phi \cdot g = \phi(g)$)?

Then $G$ is a normal subgroup of $H$, so for all $h \in H$ the conjugation map $\rho_h(g) = hgh^{-1}$ is an automorphism of $G$.

In this case the map $H \rightarrow \operatorname{Aut}(G)$ defined by $h \mapsto \rho_h$ is a surjective group homomorphism, and the restriction to $G$ is the natural map $G \rightarrow \operatorname{Inn}(G)$.

Answer 2

In fact, following Zen's comment, the holomorph $\tilde{G}=G \rtimes {\rm Aut}(G)$ always works. $G$ embeds into this in the obvious way, and the map $\tau:\tilde{G} \to {\rm Aut}(G)$ is defined by $\tau(g\alpha) = \rho_g \alpha$, where $\rho_G:x \mapsto gxg^{-1}$ is conjugation by $g$. Then $\tau$ is clearly surjective, and it has the required restriction to $G$, so we just need to check that it is a homomorphism.

$$\tau((g\alpha)(h\beta)) = \tau(g\alpha(h)\alpha\beta) = \rho_{g\alpha(h)}\alpha\beta$$ whereas $$\tau(g\alpha)\tau(h\beta) = \rho_g\alpha\rho_h\beta = \rho_g\alpha\rho_h\alpha^{-1}\alpha\beta,$$ so we have to check that $$\rho_{g\alpha(h)}(x) = \rho_g\alpha\rho_h\alpha^{-1}(x)$$ for all $x \in G$, which is routine, and I see that spin has beaten me too it anyway!

The free product, as suggested by user180040 also works,but that's seems like overkill!