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.)

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

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

$g : G' \to G''$


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