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

Let $f:G \longrightarrow G'$ and $g:G' \longrightarrow G''$ 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 - but which property of a group is used in the second step? \begin{eqnarray} g(f(x \cdot y)) &=& g(f(x) \cdot f(y)) \\ \end{eqnarray}

Is it a cauchy function? Is there another property that lets a group operation $ \cdot $ be pushed outside of a function?

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    in f(xy)=f(x)f(y) is used the fact that f is homomorphism.... from that implies g(f(xy))=g(f(x)f(y)) – 2017-02-05

2 Answers 2

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A group homomorphism is a map from a group $G$ to another group $G'$ that preserves the "group structure" (homomorphism are in general "structur-preserving maps"). This structure is the multiplication for multiplicative groups (addition for addivite groups). A homomorphism $f$ has then the defining property that for any $g,h \in G$:

$$f(g\cdot h) = f(g) \cdot' f(h)$$

where $\cdot$ denotes multiplication in $G$ and $\cdot'$ multiplication in $G'$.

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Thanks. To me this notation made it clearer. So using this we have \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}

where $⋅''$ denotes multiplication in $G''$