Why is the mapping φ:a^i→a^{2i} an automorphism of G with order 3?

[Lee33]

Homework Statement

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 $\phi$?

 

[Dick]

Homework Statement

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 $\phi$?

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

 

[pasmith]

Homework Statement

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 $\phi$?

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 $\phi^3(a^i)$?

 

[Lee33]

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

 

[Dick]

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

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 $\phi^3$.

 

[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?

 

[Dick]

Ah, 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?

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?

 

[Lee33]

Dick - No I can't see that. Why is that?

 

[Dick]

Dick - No I can't see that. Why is that?

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.