If $\displaystyle f:G \rightarrow H $ and $\displaystyle x \in G $ has order k, prove that $\displaystyle f(x) \in H $ has order m, where m|k.

Also, if gcd(|G|,|H|)=1, then $\displaystyle f(x)=1 $ for all $\displaystyle x \in G$

Printable View

- Mar 10th 2009, 10:02 PMdidact273Homomorphism proof
If $\displaystyle f:G \rightarrow H $ and $\displaystyle x \in G $ has order k, prove that $\displaystyle f(x) \in H $ has order m, where m|k.

Also, if gcd(|G|,|H|)=1, then $\displaystyle f(x)=1 $ for all $\displaystyle x \in G$ - Mar 10th 2009, 11:27 PMclic-clac
Hi

You know that $\displaystyle f$ is a group homomorphism, and that $\displaystyle x$ has order $\displaystyle k.$ Let's use this:

$\displaystyle f(e_G)=f(x^m)=f(\underbrace{x...x}_{\text{m times}})=\underbrace{f(x)...f(x)}_{\text{m times}}=f(x)^m$*(where*$\displaystyle e_G$*is the identity element of G)*

Moreover, since $\displaystyle f$ is a homomorphism, $\displaystyle f(e_G)=e_F.$

So we get $\displaystyle f(x)^m=e_F.$ What can you deduce?

Now, the second question. In a group, elements orders always divide the group order. If $\displaystyle \text{gcd}(|G|,|H|)=1,$ then the only common positive divisor between $\displaystyle |G|$ and $\displaystyle |H|$ will be $\displaystyle 1$. For any $\displaystyle x$ in $\displaystyle G,$ we want to prove that $\displaystyle f(x)=e_F,$ i.e. that $\displaystyle f(x)$ has order $\displaystyle 1$. Some idea? (Using the first result of course) - Mar 10th 2009, 11:33 PMaliceinwonderland