Suppose $f:G\longrightarrow H$ is a group homomorphism with $H$, $Ker(f)$ finite. Is $G$ finite?

Question

Let $f:G\longrightarrow H$ be a group homomorphism with $G$ not necessarily a finite group, but $H$ is a finite group. By the first isomorphism theorem we have:

$\frac{G}{Ker(f)}\cong Im(f)$.

Suppose further that we know that $Ker(f)$ is finite. Is it now possible to conclude that $G$ is a finite group?

I am currently under the impression that lagrange's theorem can't be used, since it assumes the very thing we are trying to prove. Perhaps I am missing something obvious. Any help would be vastly appreciated.

Answer 1

The kernel is one of the cosets in the quotient group and all cosets are the same size. Since the image is finite, there are a finite number of cosets. A finite number of cosets, each of a finite size implies that there are a finite number of elements in total.

Answer 2

Generally, a group homomorphism $G\to H$ is always $k$ to $1$, where $k$ is the order of the kernel. It's quite easy to prove this fact. Now, using that fact, the given homomorphism $G\to H$ is $k$ to $1$ where $k$ is finite. If $G$ is infinite, then the image must be infinite too, so if $H$ is known to be finite, then $G$ must be finite too.