If and has order k, prove that has order m, where m|k.

Also, if gcd(|G|,|H|)=1, then for all

Printable View

- March 10th 2009, 11:02 PMdidact273Homomorphism proof
If and has order k, prove that has order m, where m|k.

Also, if gcd(|G|,|H|)=1, then for all - March 11th 2009, 12:27 AMclic-clac
Hi

You know that is a group homomorphism, and that has order Let's use this:

*(where**is the identity element of G)*

Moreover, since is a homomorphism,

So we get What can you deduce?

Now, the second question. In a group, elements orders always divide the group order. If then the only common positive divisor between and will be . For any in we want to prove that i.e. that has order . Some idea? (Using the first result of course) - March 11th 2009, 12:33 AMaliceinwonderland