Abstract Algebra: Homomorphisms?

Question

Suppose that phi is a homomorphism from Z_42 onto a group of order 6. What is the kernal of phi?

Answer 1

phi : Z_42 -> G, where |G| = 6

ker(phi) is a normal subgroup of Z_42. Also, every subgroup of Z_42 is normal because Z_42 is abelian. The subgroups of Z_42 are:

Z_42
Z_21
Z_14
Z_7
Z_6
Z_3
Z_2
0 (the trivial group)

Since phi is a surjective homomorphism, we know that
Z_42 / ker(phi) = G

In words: Z_42 modulo the kernel of phi is isomorphic to the group G.

Since |G| = 6, we must have
|Z_42 / ker(phi)| = 6 also

So then
|Z_42 / ker(phi)|
= |Z_42| / |ker(phi)|
= 42 / |ker(phi)|
= 6

means |ker(phi)| = 7.

Since ker(phi) is a normal subgroup of Z_42 of order 7, it must be Z_7.

Hope this helps.