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

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.

## 2 Answers

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.

• My intuition agrees with this answer, but my concern is when we say "since the image is finite, there are a finite number of cosets". Currently, the way I am rationalizing this is by dividing $|Im(f)|$ by $|Ker(f)|$ and concluding that the resulting number is finite. For some reason I think I am not allowed to do this, but I am not sure. Nov 7, 2018 at 22:37
• @uaa209 The image is a subset of a finite set so it is finite. The quotient group consists of cosets and this set of cosets is isomorphic to the image. That's how I justify that. Nov 7, 2018 at 22:39
• Thanks for your help, I think I understand it now. I've not been completely recognising that the number of cosets in the quotient group $\frac{G}{Ker(f)}$ is equal to $|Im(f)|$ (due to the isomorphism). And from there we take the product of the "number of cosets" with the "size of each coset" to yield some finite size of $G$. Nov 7, 2018 at 23:08

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.

• Forgive my ignorance but I am not sure what you mean when you say the group homomorphism is "k to 1". Nov 7, 2018 at 22:22
• A function is $k$ to $1$ if for every element in its codomain there are precisely $k$ elements in the domain mapped to it. Nov 8, 2018 at 14:37