# Thread: Cyclic group problem

1. ## Cyclic group problem

Prove the following: Let <a> be a cyclic group of order n. If n and m are relatively prime, then the function f(x)=x^m is an automorphism of <a>.

Hints:
1) If G=<a> and b is an element of G, the order of b is a factor of the order of a.
2) Suppose an element a in a group has order n. Then a^t = e iff t is a multiple of n (t=nq for some integer q).

2. ## Re: Cyclic group problem

The hints basically give you a proof of the question, just follow them and ask for more help if you get stuck on any step. It may be good for your intuition to see why this proposition is true for small cyclic groups (for example, take the cyclic group of order 2 and 3).

3. ## Re: Cyclic group problem

Gusbob,

Thanks for the reply. I'm not a math major and I was sick and missed the classes prior to this; so I'm a little lost in all of this. (The book doesn't help because it defines nothing for you, assumes you already know). Could I get a little hand holding Just kind of show me what you mean?

I know that x^m = x^gcd(m,n) = 1 : So then f(x) = 1 which shows that there is an inverse?

I also know that if b is an element of <a>, then b's order is a factor of a, and hint 2 gives that b^t = e iff t is a factor of n and from the above see that it is : This shows we have an identity?

Associativity is inherited and closure is inherited, then we have an automorphic group because this group would be isomorphic and by definition every isomorphism has an inverse which is also an isomorphism, and since the inverse is also an endomorphism of the same object it is an automorphism?

If the above is true, can you please explain how and why it is true? I just pulled what I could together from notes and stuff and I don't understand this cyclic group stuff because I wasn't in lecture on these days.

4. ## Re: Cyclic group problem

Originally Posted by spawn8214
Prove the following: Let <a> be a cyclic group of order n. If n and m are relatively prime, then the function f(x)=x^m is an automorphism of <a>.

Hints:
1) If G=<a> and b is an element of G, the order of b is a factor of the order of a.
2) Suppose an element a in a group has order n. Then a^t = e iff t is a multiple of n (t=nq for some integer q).
My recommendation is to catch up on your missing work - you may not be able to understand what I write down otherwise. But I will give you a sketch proof this time.

An automorphism on $G$ is an group isomorphism on $G$. To show that your function is a homomorphism, we observe that $f(a^{r+s})=a^{m(r+s)}=a^{mr}a^{ms}=f(a^{r})f(a^s)$. Because $G$ is cyclic generated by $a$, the image of this homomorphism is in $G$.

We deal with injectivity first. Take an arbitrary element $g$ of $\ker f$ and observe that $g^m=1$. That is, the order of $g$ is a divisor of $m$. Since the order $|g|$ of $g$ divides $n$ as well ,and $m,n$ are co-prime, we necessarily have $|g|=1$. There is only one such element in $G$, and so $\ker f = \{ 1 \}$.

For surjectivity, consider an element $g \in G$. The fact that $m,n$ are coprime implies $rn+sm=1$ for some integers $r,s$. As such $g=g^{rn+sm}=g^{sm}$ (using hint two for the second equality). In particular, $f(g^s)=g$

Of course, what I just said was complete gibberish if you don't know what a cyclic group is... I really do recommend you catch up on your work first, your edited post above seem to show a few gaps in basic group theory. Once you have done so, you should be able to write down a proper proof based on what I said.