# Automorphism

• Jan 21st 2009, 11:12 AM
knguyen2005
Automorphism
1) What is Aut(G) and how do you show that the automorphism is isomorphic to a group i.e. Aut(C5) =~ C4, and what is Aut(C9) is isomorphic to which cyclic group, is it group C8?

2) Describe Aut(Cn) explicitly for n = 24
How do you show that Aut(C12) =~ C2 * C2 and Aut(C14) =~ C6 ?. Find the correspoding isomorphism for Aut(C24)

Another question is: Does it mean that, C2 * C2 = C4
C4 * C4 = C16
C5 * C5 = C25
and so on ... so in general Cn * Cn = C(n*n). I dont know

Many thanks
• Jan 21st 2009, 12:48 PM
aliceinwonderland
Quote:

Originally Posted by knguyen2005
1) What is Aut(G) and how do you show that the automorphism is isomorphic to a group i.e. Aut(C5) =~ C4, and what is Aut(C9) is isomorphic to which cyclic group, is it group C8?

2) Describe Aut(Cn) explicitly for n = 24
How do you show that Aut(C12) =~ C2 * C2 and Aut(C14) =~ C6 ?. Find the correspoding isomorphism for Aut(C24)

Another question is: Does it mean that, C2 * C2 = C4
C4 * C4 = C16
C5 * C5 = C25
and so on ... so in general Cn * Cn = C(n*n). I dont know

Many thanks

The automorphism of the cyclic group Z/nZ is $(Z/nZ)^{\times}$, which is of order $\phi(n)$ (link)

$\phi(9)=6$, thus Aut(C9) is not isomorphic to C(8).
$\phi(mn) = \phi(m)\phi(n)$, when gcd(m,n)=1.
$\phi(3*8)=\phi(3)\phi(8) = 2 *4 =8$, so the order of Aut(C24) should be 8.

A cyclic group is an abelian group by definition.

$C_{m} \times C_{n} = C_{mn}$, when gcd(m,n)=1. Thus $C_{4} \times C_{4} \neq C_{16}$

Note: Someone might give you a more thorough answer for this. This is my attempt to this problem.