Let $K : G \to G'$ be a group homomorphism
Show that If order of $G$ is finite, then order of $K[G]$ is finite and is a divisor of order of $G$.
My professor used an one-to-one correspondence between $K[G]$ and kernel of $G$
ans therefore order of $K[G]$ = order of $G$ divided by order of kernel of $G$.
I don't understand why there is one-to-one relationship and it seems to me that order of $K[G]$ is equal to number of left cosets of kernel of $G$
Can anyone explain?
Best Answer
By the first isomorphism theorem, $K(G)\simeq G/\text{ker } K$. Then $|G/\text{ker }K|=|G|/|\text{ker }K|$ which is a divisor of $G$. See http://en.wikipedia.org/wiki/Isomorphism_theorem#First_isomorphism_theorem.