# Group Theory

1. Oct 30, 2013

### Lee33

1. The problem statement, all variables and given/known data

Let $G$ be a cyclic group of order $7,$ that is, $G$ consists of all $a^i$, where $a^7 = e.$ Why is the mapping $\phi:a^i\to a^{2i}$ an automorphism of $G$ of order $3$?

The attempt at a solution

I know the group $G$ is formed by the elements $\{ e, a, a^2, a^3, a^4, a^5, a^6 \}$. Now under the mapping of $\phi$, the order of elements is changed to: $e, a^2, a^4, a^6, a, a^3, a^5$. Thus, $\phi$ is both one-one and onto. What I don't understand is why $G$ is of order $3$ by the map $\phi$?

2. Oct 30, 2013

### Dick

Play around with it. Order 3 means $\phi^3$ is the identity. Is it? Is $\phi^3(a)=\phi(\phi(\phi(a)))=a$?

3. Oct 30, 2013

### pasmith

The automorphisms of $G$ are themselves a group, $\mathrm{Aut}(G)$, under the operation of composition of functions. An automorphism $\phi \in \mathrm{Aut}(G)$ being of order 3 means that $\phi^3$ is the identity function on $G$, $\mathrm{Id}_G \in \mathrm{Aut}(G)$.

So what's $\phi^3(a^i)$?

4. Oct 30, 2013

### Lee33

But how did they get $\phi^3$? How did they know $\phi$ has order 3? For example, does it have order $8$ or $9$?

5. Oct 30, 2013

### Dick

The order of the automorphism, n, is the smallest number such that $\phi^n$ is the identity automorphism. They didn't just know that it is 3, they figured it out. And they are asking you to do the same. You figured out the mapping for $\phi$ pretty easily, now do $\phi^3$.

6. Oct 30, 2013

### Lee33

Ohh, thank you! I understand now. I thought they came up with it out of the blue.

But is there an easy way to verify the order of the automorphism?

7. Oct 30, 2013

### Dick

Sure, there's an easier way than working out the whole group mapping. I think you should work that out by playing with it. Once you've done that and given it some thought, do you see that if $\phi^n(a)=a$ where a is a generator then $\phi^n(x)=x$ for all elements of the group?

8. Oct 31, 2013

### Lee33

Dick - No I cant see that. Why is that?

9. Oct 31, 2013

### Dick

Think about it. Every element of the group is a power of a and $\phi^n(a)=a$ and $\phi$ is a homomorphism. Put those things together.