$H,G$ are finite groups, and $\gcd(|H|,|G|)=1$.
I need to prove that if $\varphi:G\to H$ is homomorphism it must be the Trivial Homomorphism.

My try:
I assume that $\varphi:G\to H$ is homomorphism. By Lagrange Theorem we know that:
$ord(a)\mid |G|,\;ord(\varphi(a))\mid |H|$ (for any $a\in G$).
We know that $ord(\varphi(a))\mid ord(a) $,
But, because $\gcd(|H|,|G|)=1\;\Rightarrow\;ord(a)=1$ (because $\gcd(ord(a),ord(\varphi(a))=1$).

This is why $\forall a\in G,\varphi(a)=e_H$, and this prove that $\varphi:G\to H$ is the Trivial Homomorphism.

My proof is OK? Or I miss something?
Thank you!

  • 1
    $\begingroup$ I think you mean that $\text{ord}(\varphi(a)) | \text{ord}(a)$ and not the other way around, and thus $\text{ord}(\varphi(a)) = 1$. $\endgroup$ Dec 9, 2013 at 20:34
  • $\begingroup$ @universalset - Yes, thank you!!, and this is why $\varphi(a)=1$? $\endgroup$
    – CS1
    Dec 9, 2013 at 20:41

1 Answer 1


With the correction I gave in my comment, your proof is correct. That is, you want to change the line $\text{ord}(a)|\text{ord}(\varphi(a))$ to $\text{ord}(\varphi(a))\ |\ \text{ord}(a)$. So we have $\text{ord}(\varphi(a))\ |\ |G|$, and thus $\text{ord}(\varphi(a))\ |\ \gcd(|G|, |H|) = 1$.

The only element of $H$ with order $1$ is the identity, so $\varphi(a) = e_H$ for every $a \in G$.

  • $\begingroup$ I fix it!! Thank you! $\endgroup$
    – CS1
    Dec 9, 2013 at 20:47

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