# An isomorphism and functions

#### Artusartos

I was a bit confused the last paragraph before "Corollary 4.6.4". It says that we have the isomorphism $\alpha : Z_k \rightarrow Aut(Z_n)$ but then says that $\alpha(a^j)(b^i)=b^{m^ji}$.

In a regular function $f: X \rightarrow Y$, we take one element from X and end up with an element in Y, right? But in this isomorphism, we take two elements $a^j$ and $b^i$ (and b^i is not even necessarily in Z_k). and end up with an element in $Z^x_n$. So how can that happen with a function?

#### Attachments

• 59.4 KB Views: 285
• 58.5 KB Views: 287
Last edited:
Related Linear and Abstract Algebra News on Phys.org

#### conquest

Hello,

Your alpha function sends elements from Z_k to the automorphisms of Z_n. This means that for every element a of Z_k alpha applied to a is a function from Z_n to Z_n. So alpha is an isomorphism and the formula gives you a description of the image of a^j under this isomorphism namely exactly that function that sends b^i to B^(bladiebla).

The only obstacle here is to keep track of where elements are going alpha is a perfectly fine map and so is alpha(a).

#### Artusartos

Hello,

Your alpha function sends elements from Z_k to the automorphisms of Z_n. This means that for every element a of Z_k alpha applied to a is a function from Z_n to Z_n. So alpha is an isomorphism and the formula gives you a description of the image of a^j under this isomorphism namely exactly that function that sends b^i to B^(bladiebla).

The only obstacle here is to keep track of where elements are going alpha is a perfectly fine map and so is alpha(a).
Thank you.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving