# Finding a surjective homomorphism

I have to show that for all groups with $2007(=3^2\times223$) elements that there exists a surjective homomorphism to a group of 9 elements.

Obviously a group with 2007 elements has a subgroup of order 9 (sylow theorem) and I think this should help me with finding the homomorphism, but I'm stuck. I'm also thinking about using the fact that there also should be a subgroup of order 223 which is cyclic. But this didn't help me much either.

• Hint: Think about the kernel of this homomorphism, what would it have to look like? – Myself Dec 20 '14 at 16:02
• The kernel would have 223 elements I suppose – DeanTheMachine Dec 20 '14 at 16:04
• Right. So if this is to be a kernel, it had better be a normal subgroup. Is it? – rogerl Dec 20 '14 at 16:04
• But a subgroup of 223 elements of a group of order 2007 (for simplicity lets call the group of 2007 elements G), is always normal in G by sylow theorems – DeanTheMachine Dec 20 '14 at 16:06
• Then you are done, aren't you? Every normal subgroup is the kernel of the projection homomorphism. – rogerl Dec 20 '14 at 16:07

$223$ is a prime number. By Sylow theorems, if $N$ is the number of $223$-sylow subgroups, then $$N \equiv 1 \mod (223)$$ $$N \mid 3^2$$ This implies that $N =1$.
Thus there is an unique $223$-sylow , which is therefore normal. Suppose we call it H.
Consider $G/H$, it is a group of order $9$, and the projection $$\phi : G \to G/H$$ is a surjective homomorphism. The action of the homomorphism is $$\phi(g) = gH$$ i.e t sends $g$ in its lateral class in $G/H$.
• I think this answer would be to vague for my exam because you don't tell how you map different elements. In the comments of my question rogerl gave the same map (he made a typo). Which looked like this: $\phi: G \rightarrow G/H, g \mapsto g G/H$. But still I don't really understand what this map would do to any element in G – DeanTheMachine Dec 20 '14 at 19:44
• I have to understand quotient groups but I have a hard time understanding them. At the moment I think about quotient groups like this. Suppose G/N, with $N \triangleleft G$ then G/N is a group consisting of set. in which the unit element is the set consisting of the elements that are in N. And every other element is a set (of the size of N) with the elements $\{ gn_1,gn_2,.... \}$, so the elements of N multiplied (according to the multiplication of G) with g on the left hand. So I imagine it as $G/N=\{\{ n_1,n_2,....\},\{g_1n_1,g_2n_2,....\},....\}$ – DeanTheMachine Dec 20 '14 at 20:12