# 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.

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 $$\lvert \operatorname{Aut}(G_2) \rvert =40$$.

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.

• What about $|G/Z(G)|=10$? Jan 27, 2019 at 14:37
• Oh, thanks! Forgot that, don't know how to salvage the proof
– o.h.
Jan 27, 2019 at 14:38
• You can't show that $|G/Z(G)| \neq 10$ because $|Z(D_{20})|=2$. Jan 27, 2019 at 14:42