1
$\begingroup$

Prove that the composition of two group homomorphisms is a group homomorphism.

Let $f:G \longrightarrow G'$ and $:G' \longrightarrow G''$ be two group homomorphisms.

Let $x$ and $y$ be two arbitrary elements of $G$. Then,

\begin{eqnarray} (g \circ f)(x \cdot y) &=& g(f(x \cdot y)) \\ &=& g(f(x) \cdot f(y)) \\ &=& g(f(x)) \cdot g(f(y)) \\ &=& (g \circ f)(x) \cdot (g \circ f)(y) \end{eqnarray}

This completes the proof.

(It may have been a poor choice to use $\cdot$ to denote the group operations in different groups, but other than that, I think it's fine.)

$\endgroup$
1
  • $\begingroup$ Your proof is okay. $\endgroup$
    – drhab
    Jun 3 '14 at 11:18
3
$\begingroup$

I agree with drhab and yourself. Aside from the use of $\cdot$ to denote different group operations, the proof looks fine. If you're struggling to think of a suitable symbol to denote a different group operation, $\ast$ is commonly used.

Also, in your post

$: G' \to G''$

should be

$g : G' \to G''$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.