Hi anyone knows how to solve this? My answer was 3. Is it correct?
If f:S3-> Z2 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:
You should also understand why this true: