# Finding Aut(Aut(Aut(C_73)))

1. Jan 22, 2015

### 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.

2. Jan 27, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Feb 5, 2015

### 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.

4. Feb 6, 2015

### 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.