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 (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.
we show it is a homomorphism.
Let $\displaystyle (g_1,h_1),(g_2,h_2) \in G \times H$
$\displaystyle f((g_1,h_1)(g_2,h_2))=f((g_1 * g_2, h_1 \circ h_2))=g_1 * g_2 = f((g_1,h_1))*f((g_2,h_2))$
$\displaystyle Ker(f) = \{(g,h)\in G \times H | f(g,h) = 0 \} = \{(g,h)\in G \times H | g=0 \}= \{0\} \times H$
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
$\displaystyle ker(f)=\{ (0,b):b \in H\}$
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 $\displaystyle \frac{G_1 \times G_2 \times ... \times G_n}{ker(\pi_i)} \cong Im(\pi_i) = G_i$