# Homomorphisms from C_6 to C_4

1. Mar 9, 2009

### hsong9

1. The problem statement, all variables and given/known data

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

2. Relevant 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.

3. 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?

2. Mar 10, 2009

### foxjwill

Re: homomorphism

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 $$C_6\to C_4$$.

Last edited: Mar 10, 2009
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