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)|?

Printable View

- October 27th 2012, 12:10 AMweijing85Group Homomorphism
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)|? - October 27th 2012, 12:56 AMjohnsomeoneRe: Group Homomorphism
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: