Finding Aut(Aut(Aut(C_73)))

[jimmycricket]

Im calculating $Aut(Aut(Aut(C_{73})))$ and have got as far as $Aut(Aut(Aut(C_{73})))\cong Aut(C_2\times C_2\times C_2) \times Aut(C_3)$
I thought this was the answer but I have been told that $Aut(C_2\times C_2\times C_2) \times Aut(C_3)\cong GL_3(\mathbb{F}_2)$
Can someone explain this to me please.

 

[Boorglar]

From what I get, $Aut(C_{73}) = U(73) = C_{72}$. $Aut(C_{72}) = U(72) = U(2^3) × U(3^2) = C_2 × C_2 × C_2 × C_2 × C_3$ (since $|U(3^2)| = phi(9) = 6$ and only abelian group of order 6 is $C_2 × C_3$. So I think you're missing a factor of $C_2$ in your second Aut( ) expression.

 

[Boorglar]

Sorry I just realized that I made a mistake. $U(2^3) = C_2 × C_2$ so your expression is actually correct. My apologies.