Homomorphism proof

**didact273** · March 10th 2009, 10:02 PM

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

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

**clic-clac** · March 10th 2009, 11:27 PM

Hi

You know that $f$ is a group homomorphism, and that $x$ has order $k.$ Let's use this:
$f(e_G)=f(x^m)=f(\underbrace{x...x}_{\text{m times}})=\underbrace{f(x)...f(x)}_{\text{m times}}=f(x)^m$ (where $e_G$ is the identity element of G)

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

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

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

**aliceinwonderland** · March 10th 2009, 11:33 PM

Originally Posted by didact273

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

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

Since m|k, k=mx where x is a positive integer.
By Lagrange's theorem, mx | |G| and m | |H|. Since |G| and |H| are coprime, there is no common factor other than 1. Thus, m is 1. We conclude that $f(x)=1$ for all $x \in G$ .

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