Kernels and homomorphism of groups

**funnyinga** · April 20th 2009, 08:36 PM

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

**Gamma** · April 20th 2009, 09:06 PM

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

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

**TheEmptySet** · April 20th 2009, 09:09 PM

Originally Posted by funnyinga

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

let $(a,b),(c,d) \in G \times H$

Then

$<br /> f[(a,b)\cdot(c,d)]=f[(a*c,b\circ d)]=a*c=f[(a,b)]*f[(c,d)]<br />$

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

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

**Gamma** · April 20th 2009, 09:29 PM

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 $\pi_i : G_1 \times G_2 \times ... \times G_n \rightarrow G_i$ by $\pi_i(g_1, g_2,..., g_i,..., g_n) = g_i$

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

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