Hi anyone knows how to solve this? My answer was 3. Is it correct?
Question:
If f:S_{3}-> Z_{2 }given by
f(a) = 1 if a is a 2-cycle
f(a) = 0 otherwise is a group homomorphism.
What is |ker(f)|?
Yes, it's 3. With such a small group, and with an explicit homomorphism given, you can explicitly see (and count) what the kernel is.
But there's a more insightful way to see that the kernel has order 3. It's what you'll need when things aren't so simple.
You should understand why this true:
$\displaystyle \text{Let } \phi : S_3 \rightarrow \mathbb{Z}_2 \text{ be *any* surjective homomorphism. Then } |ker(\phi)| = 3.$
You should also understand why this true:
$\displaystyle \text{Let } G_0, G_1 \text{ be *any* groups such that }|G_0| = 12,000,000, \ |G_1| = 1,000.$
$\displaystyle \text{Let }\phi : G_0 \rightarrow G_1 \text{ be any surjective homomorphism. Then } |ker(\phi)| = 12,000.$