# Defining a bijection from kernel of homomorphism to Image of G. (NO isomorphism theorems)

I am trying to prove that given a homomorphism from group $G$ to group $G'$ $f: G \to G'$,

1. If $G$ is finite then $|im(f)|$ is finite and divides $|G|$
2. If $G'$ is finite then $|im(f)|$ is finite and divides $|G'|$

I know $|im(f)|$ is finite as it can be at most $|G|$, also finite. I know $ker(f) \leq G$ so $|ker(f)|$ divides $|G|$ by Lagrange's theorem.

I also know that the cosets of the kernel are the subsets of $G$ whose elements are mapped to the same element in $im(f)$. I need help for the first one as I cant seem to make a well-defined bijective function with the collection of cosets of $ker(f)$ as a domain and the image as a codomain. Do I need a function about the image and the kernel and make the function utilize the cosets and use counting somehow instead? I'm new to cosets so I have trouble utilizing them. I know that the order is the same for the kernel and its cosets but is using a function from an arbitrary coset of the kernel to the image of f somehow easier? I am mind-boggled at this point.

For the second one I know that $im(f)$ is a subgroup of $G'$, so $im(f)$ is a subset of $G'$ and is also finite so by lagranges thm $|im(f)|$ divides $|G'|$ But if I am wrong please point it out.

• What is $G'$? And how are your two statements different from each other? – zipirovich May 10 '18 at 23:43

For $h\in im(f)$ let $g_h\in G$ be some element in $f^{-1}({h})$. Then I claim that any element in $f^{-1}({h})$ can be written as $g_h k$ where $k$ belongs to $\ker(f)$. Conversely $f(g_hk)=f(g_h)f(k)=f(g_h)=h$ so that any element written as $g_h k$ where $k$ belongs to $\ker(f)$ is in $f^{-1}({h})$. So that we have : $$f^{-1}({h})=g_h\ker(f)$$
Remark that this where the cosets come into play. What we have proved here is that $f$ induces a natural bijection between $G/\ker(f)$ (the set of cosets) and $im(f)$.
Finally remark that right multiplication by $g_h$ induces a bijection between $\ker(f)=g_h \ker(f)$. Now you can conclude using the following disjoint union of $G$: $$G=\bigcup_{h\in G'}f^{-1}({h})=\bigcup_{h\in im(f)}f^{-1}({h})$$