The proposition is this: Let $G$ be a group of order 77. Show that $\operatorname{Aut}(G)$ is abelian but not cyclic.

There's also a hint to use Sylow theorems for both $G$ and $\operatorname{Aut}(G)$.

I know that since we have $11\not\equiv 1\mod 7$, and $77=7\times 11$, then every group of order 77 is cyclic, this its automorphism group is abelian.

Furthermore, we know that $\operatorname{Aut}(G)$ is isomorphic to the group of elements of $Z_{77}$ which are invertible under multiplication, thus $\operatorname{Aut}(G)\simeq Z_{60}$.

Now there are only two abelian groups of order 60 (up to isomorphism) and one way to show that $\operatorname{Aut}(G)$ isn't cyclic is to show that it doesn't have an element of order 4, which is equivalent to showing that there's no integer $m$ such that $m^4 \equiv 1 \mod 77$.

I calculated all the squares $\mod 77$ and then managed to show what I wanted, but this is obviously not a fast or smart way of proving what I want to prove. I can't really think of a better way to show this, so any ideas are appreciated!

Thank you for all the help in advance.

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    $\begingroup$ If you look at $G \cong \mathbb{Z}_{7} \times \mathbb{Z}_{11}$, can you immediately see two (or three) automorphisms of order $2$? $\endgroup$ Dec 12 '20 at 22:44
  • $\begingroup$ I think they're the automorphisms sending $(1,1)$ to $(6,1), (1,10), (6,10)$ respectively. I didn't really think of them immediately but I think I realized how to find them now. And since there are these three automorphisms, there cannot exist an automorphism of order 4, because that would mean that we could only have one automorphism of order 2 (since that cyclic group of order 4 would be the only 2-Sylow subgroup, since it would be normal), which isn't true. Nice one! Thank you! $\endgroup$ Dec 12 '20 at 23:23
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    $\begingroup$ More abstractly, you can take the inversion in one factor and the identity in the other, or inversion in both factors. That gives three automorphisms of order $2$. And if we can write an abelian group as a product of $k$ nontrivial factors, that gives $2^k - 1$ automorphisms of order $2$. But a cyclic group has at most one element of order $2$. Thus the automorphism group of an abelian group that can be written as a nontrivial product never is cyclic. $\endgroup$ Dec 12 '20 at 23:30

Write $C_n$ for the cyclic group of order $n$. The automorphism group of $C_{77}$ is $C_{77}^\times$. (This is a group of order $60$, but it's better not to write $Z_{60}$ for it.)

By the Chinese remainder theorem $$C_{77}^\times\cong C_{7}^\times\times C_{11}^\times\cong C_6\times C_{10}\cong C_2^2\times C_3\times C_5.$$ Can you convince yourself of the fact that there is no element of order $60$ in there?

Overkill: use the classification of finite abelian groups.

  • $\begingroup$ Thanks for the remarks on my notation, only now I realized that I used notation $Z_{60}$ to say group of order 60, and that's just bad. About the solution I see the idea and I understand that I have to remember the Chinese remainder theorem more often in this setting. Thank you very much! $\endgroup$ Dec 12 '20 at 23:24

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