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I've spent a lot of time thinking about this question but I can't seem to come up with anything of substance. I tried to use some basic lemma for group homomorphisms, but nothing seemed to work. My intuition is to do a proof by contrapositive, considering $\psi$(G) and its cosets, but this also leads me to a dead end, unless of course I'm missing something.

It may be useful to note that the proof of the converse is on Stack Exchange, but it didn't seem to me that anything from there is useful in proving this statement.

I'd greatly appreciate any help with this, it's been bugging me a lot these past couple days. Thank you!

Sebastiano
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Joseph_Kopp
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  • Seems it's the same question as https://math.stackexchange.com/questions/1401538/a-group-homomorphism-from-a-simple-group-is-injective?rq=1. So... – TheWildCat Mar 05 '23 at 19:45
  • @TheWildCat It's certainly very similar, effectively identical probably. It seems the question was written very informally though, which is probably why I didn't find it when I looked. – Joseph_Kopp Mar 05 '23 at 20:00

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If $G$ is a simple group, then by definition $G$ has only two normal subgroups: either $G$ or the trivial subgroup $\{e\}$. Now given any group homomorphism $f\colon G\to G'$ where $G$ is a simple group, then the kernel of $f$, $\ker (f)\subseteq G$, is a normal subgroup of $G$, which means $\ker (f)$ should be either $G$ or trivial. If it's $G$, then $f$ is trivial; if $\ker (f)$ is trivial, then $f$ is injective.

TheWildCat
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  • Thank you so much! I really appreciate it, however I think the proof may be improved if the implication that f is injective were more detailed. Assume f(g1)=f(g2), g1=/=g2. Take f(g1g2^-1=f(g1)f(g2^-1)=f(g1)f(g2)^1=f(g2)f(g2)^-1=e' => kerf is not trivial. Therefore f must be injective if the kernel is trivial. – Joseph_Kopp Mar 05 '23 at 19:58
  • @Joseph_Kopp No need to "improve" it, because that can be found already here, see this post for example: Theorem: Let $\phi:G\to G'$ be a group homomorphism, then $\phi$ is $1-1$ $\iff$ $Ker;\phi ={e}$. It should also be in your book before the above exercise, so that you can use it. – Dietrich Burde Mar 05 '23 at 20:14