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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – user402494
    Nov 7, 2018 at 22:37
  • $\begingroup$ @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. $\endgroup$
    – John Douma
    Nov 7, 2018 at 22:39
  • $\begingroup$ 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 $G$. $\endgroup$
    – user402494
    Nov 7, 2018 at 23:08
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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.

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  • $\begingroup$ Forgive my ignorance but I am not sure what you mean when you say the group homomorphism is "k to 1". $\endgroup$
    – user402494
    Nov 7, 2018 at 22:22
  • $\begingroup$ A function is $k$ to $1$ if for every element in its codomain there are precisely $k$ elements in the domain mapped to it. $\endgroup$ Nov 8, 2018 at 14:37

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