2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is $G'$? And how are your two statements different from each other? $\endgroup$ – zipirovich May 10 '18 at 23:43
1
$\begingroup$

For the second point you are all right.

For the first point. As you hinted in your question, you should use cosets.

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 $G$: $$G=\bigcup_{h\in G'}f^{-1}({h})=\bigcup_{h\in im(f)}f^{-1}({h})$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.