Number of elements mapped to $f(a)$ where $f$ is a group homomorphism

Question

Let $G$ and $G'$ be any two groups.

Let $f$ : $G$ $\rightarrow$ $G'$ be a group homomorphism.

Let $K$ denotes the kernel of $f$.

Let $a$ be any arbitrary element of $G$.

I need to show that there are total $m$ elements mapped to $f$($a$) where $m$ is the order of $K$.

We know that $aK$ = {$ak$ : $k \in K$}

Consider $f$($ak$) = $f$($a$)$f$($k$) = $f$($a$) (as $f$ is a homomorphism and $k \in K$)

Using the fact that $O(aK)$ = $O(K)$, we get that there are atleast $m$ elements mapped to $f(a)$.

Now how can I show that there are no other elements mapped to $f(a)$?

Answer 1

Let $x$ be any element of $G$ for which $f(x) = f(a)$. Then we obtain $$ \big( f(a) \big)^{ - 1 } f(x) = e^\prime, $$ where $e^\prime$ is the identity element of $G^\prime$, that is, $$ f \left( a^{-1} \right) f(x) = e^\prime, $$ or $$ f \left( a^{-1} x \right) = e^\prime, $$ and so $a^{-1} x \in K$, which shows that $$ x = a \left( a^{-1} x \right) \in aK. $$

Hope this helps.