# Prove that $g \circ f:G_1 \rightarrow G_3$ is a group homomorphisms.

Let $(G_1,*_1,(G_2,*_2)\ and\ (G_3,*_3)$ be groups and let $f: G_1 \rightarrow G_2\ and\ g:G_2 \rightarrow G_3$ be group homomorphisms.

Prove that $g \circ f:G_1 \rightarrow G_3$ is a group homomorphisms.

The solution I had was nowhere near the book solution and so I would like to ask if anyone could explain the solution to me, which is proposed in the book:

$$(g\circ f)(x*_1y)= g(f(x*_1y)) = g(f(x)*_2f(y)) = g(f(x))*_3g(f(y)) = (g\circ f)(x)*_3(g \circ f)(y)$$

It is therefore proven that $g \circ f:G_1 \rightarrow G_3$ is a group homomorphism.

The part that I don't get is this one: $g(f(x)*_2f(y)) = g(f(x))*_3g(f(y))$.

I am aware that if $f:G \rightarrow H$ is a group homomorphism, then the following rules apply:

1. $f(e_G) = e_H$ (e = neutral element)
2. If $g \in G$ has an invers $g^-1$ to g in G, then $f(g^-1) = f(g)^-1$ Where $f(g)^-1$ is the inverse Element to $f(g) \ in \ H$.
3. If $f$ is invertible, then the inverse transformation $\ f^{-1}$ is also a group homomorphism.

However, I am still unable to connect the dots as to why this leads to the solution proposed in the book.

Thank you for any help provided in explaining this to me like I am five!

• You don't need properties (1,2,3) but the definition of homomorphism. i.e. $g(a *_2 b)=g(a) *_3 g(b)$. Dec 10, 2017 at 8:59
• observe that $f(x),f(y)\in G_2$ and $g:G_2\to G_3$ is an homomorphism Dec 10, 2017 at 8:59
• The part you are asking about is the definition of being a homomorphism. Dec 10, 2017 at 9:00

For $a,b\in G_1$, it is $g(f(a\ast_1 b))\stackrel{\text{f group hom.}}{=}g(f(a)\ast_2 f(b))\stackrel{\text{g group hom.}}{=}g(f(a))\ast_3 g(f(b))=g\circ f(a)\ast_3 g\circ f(b)$