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.


2 Answers 2


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.

  • 1
    $\begingroup$ What about $|G/Z(G)|=10$? $\endgroup$
    – the_fox
    Jan 27, 2019 at 14:37
  • $\begingroup$ Oh, thanks! Forgot that, don't know how to salvage the proof $\endgroup$
    – o.h.
    Jan 27, 2019 at 14:38
  • 3
    $\begingroup$ You can't show that $|G/Z(G)| \neq 10$ because $|Z(D_{20})|=2$. $\endgroup$
    – the_fox
    Jan 27, 2019 at 14:42

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