Are there any homomorphisms from C_6 to C_4?

[hsong9]

Homework Statement

Show that there are exactly two homomorphisms f:C_(6) --> C_(4)

Homework Equations

Theorem.
let f: G -> G1 and h: G -> G1 be homomorphisms and assume that G=<X> is generaed by a subset X. Then f = h if and only if f(x) = h(x) for all x in X.

The Attempt at a Solution

C6 = <g>, |g| = 6. The divisors of 6 are 1,2,3,6
C4 = <g'>, |g'| = 4, the divisors of 4 are 1,2,4
only 1 and 2 of C6 are the divisors of C4.
so there are exactly two homomorphism.

right?

 

[foxjwill]

I'm pretty sure that there aren't any homomorphisms from C_6\to C_4:

To see this, consider, without loss of generality, the groups \mathbb{Z}_6:=\{0,1,2,3,4,5\} and \mathbb{Z}_4:=\{0,1,2,3\} under addition modulo 6 and 4, respectively. By the fundamental homomorphism theorem for groups, for any homomorphism h from \mathbb{Z}_6 to some group H,

\mathbb{Z}_6/N\cong H​

for some N\lhd \mathbb{Z}_6. But the only subgroups of \mathbb{Z}_6 (all of which are normal, since \mathbb{Z}_6 is abelian) are

\{0,1,2,3,4,5\}, \{0,2,4\}, \{0,3\}, \{0\}.​

So, taking N to be one of these subgroups,

|G/N|=1,2,3,\text{ or }6.​

Thus, the order of any homomorphic image of \mathbb{Z}_6 must be one of these. Specifically, the order of any homomorphic image cannot be 4. Thus, there is no homomorphism from \mathbb{Z}_6\to \mathbb{Z}_4 and by extension from C_6\to C_4.