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.
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.