# Another problem involving cyclic groups.

1. Jun 14, 2012

### jmjlt88

Show that in a finite cyclic group G of order n, writtten multiplicatively, the equation xm = e has exactly m solutions x in G for each positive integers m that divides n.

Attempt...

Proof:

Let G be a finite cyclic group of order n, and suppose m is a positive integer that divides n. Let x be an element in G. Then, x generates a cyclic subgroup of G that contains (n/m) elements since the gcd(n,m)=m [we know this by a theorem presented in the section0.
Hence,

(1) (xm)n/m= xn= e.

The positive integer m was picked arbitrarily. Thus, equation (1) will hold for each m that divides n. Hence, xm=e has exactly m solutions.

QED??

Does this hold or am I missing something?

2. Jun 14, 2012

### algebrat

Absorbed into next post

Last edited: Jun 14, 2012
3. Jun 14, 2012

### algebrat

The proof starts out bad, x is arbitrary, it could be the identity. I think you want to start by selecting the generator x of G (remind me why such exists). Then let z=xn/,
The rest looks like you're on the right track. Might write zm=(xn/m)m=xn=e. Show that zk≠e for 0<k<m. (I think that's the hard part:)

Last edited: Jun 14, 2012
4. Jun 14, 2012

### jmjlt88

Thank you so much!

Suppose G is generated by a.

Take x element of G. Then x can be written as as for some integer s.

Then,

1) (xm)n/m= ((as)n)n/m = (as)n= e.

This shows at least m solutions. Thus, we still need to show that there are exactly m solutions. Let 0<k<m. We must show xk≠e.

I'll work on this tonight!! =)

5. Jun 14, 2012

### jmjlt88

Off the cuff, gcd(n,k) does not equal k (dissimilar to before). So, our nice equation in (1) wont work... Is this the right idea?

6. Jun 14, 2012

### jmjlt88

Is the following, perhaps, a better idea?

Proof:

Since G is a finite cyclic group of order n, it is isomorphic to Zn. [Theorem in book]. Let m be positive integer that divides n. Thus, n=km for some positive integer k. We want to show that there are exactly x solution to the equation xm=e, but since we are in Z, the equation becomes mx=0. The identity element 0 serves as a solution since m(0)=0. Also, since n=km, m(k)=n=0. Then every multiple of k will also make our equation 0 since mk=0. Thus, 0, k, 2k, 3k, .... , (m-1)k will also be soltions to our equation. Hence, there are exactly m solutions.

QED

7. Jun 14, 2012

### algebrat

It is hard to say where you should go next, so I'll just point out where I see plain old mistakes.

That sounds weird to me. I think you're given that G is finite cyclic, so somehow (you fill in the details), that implies that it is generated by some element.

So Here it looks like you're assuming G is generated by a (you need to say these things explicitly). But x still looks arbitrary, s could be 0, and x=e. You need to be more specific in your choices, but if you're just searching for the proof, that's fine, keep looking for the logic to match up. Right now, it's not coming together yet. You want small holes in your logic, not big ones. In a sense, you ned your proof converge to a correct one, so keep writing drafts, and keep checking for the flaws. That is university math (in my opinion.

You need to be much more thoughtful here, you're first attempt at this particular expression did not make sense to me either. I would expect something more like x^m=e, which shows there are at most, not at least, m solutions. But that is still not the correct thing to be writing, because I do not see the connection between s, n and m.

If you can show these last two sentences, then you have at least m solutions (including k=0 in the end).

That is the best idea. This problem is a very good opportunity for you to develop a long sequence of arguments. Enjoy!