Find map of a non-generator element, under Isomorphism.

Question

The group $\langle (\mathbb Z/9)^\times,\times\rangle$ has elements given by: $\{1, 2, 4, 5, 7, 8\},$ and their orders are:

$o(1)= 1;$
$o(2)=|\{2,4,8, 7, 5, 1\}| = 6;$
$o(4)=|\{4,7,1\}|= 3;$
$o(5)=|\{5, 7, 8, 4, 2, 1\}| = 6;$
$o(7)= |\{7, 4, 1\}|= 3; $
$o(8)=|\{8, 1\}|= 2.$

The cyclic group $\left<\mathbb Z/6\mathbb Z, +\right>$ has the set of elements $=\{0,1,2,3,4,5\},$ with their order as:

$o(0)= 1;$
$ o(1)= |\{1,2,3,4,5,6\}|= 6;$
$ o(2)= |\{2,4,0\}|=3;$
$ o(3)=|\{3,0\}|= 2;$
$ o(4)=|\{4, 2, 0\}|= 3;$
$ o(5)=|\{5,4,3,2,1,0\}|=6.$

Hence, both have two generators, two elements of order $3,$ one element each of order $2,$ and $1.$

The number of Isomorphisms possible are: $2,$ given by :

1.$\varphi(2)=1; \varphi(5)=5; \varphi(1)= 0; \varphi(8)=3; $

2.$\xi(2)=5; \xi(5)=1; \xi(1)= 0; \xi(8)=3; $

Consider the case of $\varphi$:

As, given two cyclic groups of order six; so, using any generator's map to the co-domain, should be able to find the map for $\varphi(4), \varphi(7).$

But, am unable to do this for finding the map of $\varphi(7),$ as shown below:

$\varphi(4)= \varphi(2\cdot 2)= 1+1= 2.$

Had, chosen $\xi$, with $\xi(2)=5$ then $\xi(4)=\xi(2\cdot 2)= 5+5=10\equiv 4\pmod 6.$

But, how to find $\varphi(7),$ is unclear as the prime term $'7'$ inside the brackets should be expressed as product of smaller terms, and have to use $\varphi(2).$

Answer 1

You noted (sort of), that $7\equiv 2^4\pmod9 $, so knowing $\varphi (2)$ makes it easy to get $\varphi (7)$. Namely it's $\varphi (7)=\varphi (2)^4=4\varphi (2)$. So you get the values under each isomorphism. That's the point of the statement "the (iso)morphism is determined by generators". It's actually true even when you have more than one generator. But the case of a cyclic group is particularly simple: one generator.

Similarly, since $4\equiv 2^2\pmod9 $, we have $\varphi (4)=\varphi (2)^2=2\varphi (2)$. You can use either generator to do this. There's $2$ because Euler's totient function evaluated at $6$ is $2$. In general for a cyclic group there are $\phi(n)$ generators. It's also the number of isomorphisms, and (of course, automorphisms) because generators go to generators.