# Thread: Kernels and homomorphism of groups

1. ## Kernels and homomorphism of groups

Let $\displaystyle (G,*)$ and $\displaystyle (H, \circ)$ be any two groups, and let $\displaystyle f:G$ x $\displaystyle H \rightarrow \ G$ be given by $\displaystyle f(g,h) = g$. Prove that f is a homomorphism of groups and give ker f.

2. we show it is a homomorphism.
Let $\displaystyle (g_1,h_1),(g_2,h_2) \in G \times H$
$\displaystyle f((g_1,h_1)(g_2,h_2))=f((g_1 * g_2, h_1 \circ h_2))=g_1 * g_2 = f((g_1,h_1))*f((g_2,h_2))$

$\displaystyle Ker(f) = \{(g,h)\in G \times H | f(g,h) = 0 \} = \{(g,h)\in G \times H | g=0 \}= \{0\} \times H$

3. Originally Posted by funnyinga
Let $\displaystyle (G,*)$ and $\displaystyle (H, \circ)$ be any two groups, and let $\displaystyle f:G$ x $\displaystyle H \rightarrow \ G$ be given by $\displaystyle f(g,h) = g$. Prove that f is a homomorphism of groups and give ker f.
let $\displaystyle (a,b),(c,d) \in G \times H$

Then

$\displaystyle f[(a,b)\cdot(c,d)]=f[(a*c,b\circ d)]=a*c=f[(a,b)]*f[(c,d)]$

So f is a homomorphism

ker(f) is the set of all elements that get mapped to 0

since the 2nd coordinate does not affect the immage of f

$\displaystyle ker(f)=\{ (0,b):b \in H\}$

4. ## Generalization

I was just thinking it might be of interest to you to note that this same concept can be generalized to multiple products and considering the mapping $\displaystyle \pi_i : G_1 \times G_2 \times ... \times G_n \rightarrow G_i$ by $\displaystyle \pi_i(g_1, g_2,..., g_i,..., g_n) = g_i$

This is a homomorphism and the kernel of this is $\displaystyle G_1 \times G_2 \times ... \times \{0\} \times G_{i+1} \times ... \times G_n$. And if you apply the first isomorphism theorem you know that $\displaystyle \frac{G_1 \times G_2 \times ... \times G_n}{ker(\pi_i)} \cong Im(\pi_i) = G_i$