There are 2 homomorphisms: $f(x)=(4x,6x,2x)$ and $g(x,y,z)=(5x-5y+5z,10x-10y+10z)$. Find a group $\ker(g) /{\rm im}(f)$.

Let $\mathbb{Z} \to \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ be homomorphisms given by:

• $f(x) = (4x, 6x, 2x)$
• $g(x, y, z) = (5x - 5y +5z, 10x - 10y + 10z)$

Check that $g \circ f$ is a trivial homomorphisms and find the group $\ker(g) /{\rm im}(f)$.

I know how to do the first part:

$f(x) = (4x, 6x, 2x)$

$g(x, y, z) = (5x - 5y +5z, 10x - 10y + 10z)$

$g \circ f = g(4x, 6x, 2x) = (20x - 30x +10x, 40x - 60x + 20x) = (0,0)$

So $g \circ f$ is a trivial homomorphisms.

However, I don't know how to find a group $\ker(g) /{\rm im }(f)$. Any help would be much appreciated.

Solution 1:

First construct a basis for $K=\{(x,y,z)|x+z=y\}$: $$e_1=(1,1,0),\qquad e_2=(1,0,-1).$$ Given any element of $K$ we can subtract an appropriate multiple of $e_1$, to be left with a vector whose middle co-ordinate is $0$, so we have a multiple of $e_2$. Thus $e_1,e_2$ span $K$ and are clearly linearly independent.

We have $$x=(4,6,2)=6e_1-2e_2.$$

It is now helpful to switch to a new basis of $K$: $$f_1=3e_1-e_2,\qquad f_2=e_1.$$

Then $x=2f_1$ and $$\mathbb{Z}\{ f_1,f_2\}/2\mathbb{Z}f_1\cong \mathbb{Z} f_1/2\mathbb{Z}f_1\oplus \mathbb{Z} f_2\cong \mathbb{Z}\oplus\mathbb{Z} /2\mathbb{Z}.$$