Find the order of $\operatorname{Aut}(G)$. Let $G$ be a non-abelian group of order $20$. What will be the order of $\operatorname{Aut}(G)$?
$(a)$ $1$.
$(b)$ $10$.
$(c)$ $30$.
$(d)$ $40$.
If I take $G = D_{10}$ then I find that the order of $\operatorname{Aut}(G)$ is a multiple of $10$. In this case I use the fact that $G/Z(G) \cong \operatorname{Inn}(G)$. So I think $(d)$ is the correct option but I don't know the general result as to why it is true for any non-abelian group of order $20$. Please help me in this regard.
Thank you very much.
 A: This is not a well-formulated problem. There is a non-abelian group of order $20$ with an automorphism group of order $20$ and one with automorphism group of order $40$.
The group $G_1$ of shape $C_5 \rtimes C_4$, given by the presentation $\langle a,b \mid a^5 = b^4 = 1, bab^{-1} = a^2 \rangle$, is complete (i.e. it has trivial centre and all automorphisms are inner). Thus $\operatorname{Aut}(G_1) \cong G_1$ has order $20$.
On the other hand, $G_2 = D_{20}$ has centre of order $2$, so $\operatorname{Inn}(G_2) \cong D_{20}/Z(D_{20}) \cong D_{10}$ has size $10$, but $\operatorname{Out}(G_2)$ has size $4$ and is isomorphic to the Klein four-group, so $\lvert \operatorname{Aut}(G_2) \rvert =40$.
A: Are you allowed to assume that at least one of the options is correct?
Recall that $G/Z(G)$ cyclic $\Rightarrow$ G abelian. But $|G/Z(G)|$ must be a divisor of $|G|=20$. Hence the only non-cyclic options are $|G/Z(G)| = 4$, $10$ or $20$ (since prime order $\Rightarrow$ cyclic).
If $|G/Z(G)| = 4$ or $20$, then we are done. I do not know how to deal with the $|G/Z(G)| = 10$ case.
