# Thread: Problem on homomorphism

1. ## Problem on homomorphism

Q: Let G, H be cyclic groups, generated by elements x, y. Determine the condition on the orders m,n of x and y so that the map sending x^i --> y^i is a group homomorphism.

Ans: For m<n a function which sends x^i --> y^i can not be well defined; but for m>or=n there is no such problem.

So, the required condition is m > or = n.

AM I RIGHT?

2. ## Re: Problem on homomorphism

Originally Posted by arindamnaskr
Q: Let G, H be cyclic groups, generated by elements x, y. Determine the condition on the orders m,n of x and y so that the map sending x^i --> y^i is a group homomorphism.

Ans: For m<n a function which sends x^i --> y^i can not be well defined; but for m>or=n there is no such problem.

So, the required condition is m > or = n.

AM I RIGHT?

Your condition is true but incomplete.

Suppose we call the map f.
Then for any i, we have

$x^i = x^{m+i}$

Since a map is required to have unique images, that means:

$f(x^i) = f(x^{m+i})$

What can you deduce from that (using the homomorphism property)?

3. ## Re: Problem on homomorphism

f(x^i)= f(x^m+i) = f(x^m).f(x^i)=y^m.y^i=y^m+i.
But where are we going from here?

4. ## Re: Problem on homomorphism

Originally Posted by arindamnaskr
f(x^i)= f(x^m+i) = f(x^m).f(x^i)=y^m.y^i=y^m+i.
But where are we going from here?
You should have:

$\displaystyle y^i = f(x^i) \overset{\boxed{|x|=m}}{=} f(x^m \cdot x^i) \overset{\boxed{homomorphism}}{=} f(x^m) \cdot f(x^i) = y^m \cdot y^i$

This will only hold true if $y^m = e$ while at the same time we know that $|y|=n$.
What does that mean for m?

5. ## Re: Problem on homomorphism

That means m= pn, where p=1,2,3,.....
Anything else?

thanks..

6. ## Re: Problem on homomorphism

Originally Posted by arindamnaskr
That means m= pn, where p=1,2,3,.....
Anything else?

thanks..
Yes.
So n must be a divider of m.
That's it!

7. ## Re: Problem on homomorphism

Originally Posted by ILikeSerena
Yes.
So n must be a divider of n.
That's it!
Thank you again