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Let $G$ be a finite group with normal subgroups $F$ and $H$. Define $FH = \{fh; f \in F, h \in H\}$.

If $F \cap H = \{id\}$, then $FH/H \cong F$.

Proof:

Consider the quotient map $q: FH \rightarrow FH/H$. This map is clearly surjective. Now if $q(fh) = 1$ then $f \in F \cap H = \{id\}$ so the kernel of this map is trivial.

I understand the proof up until this point but it does not go on to say why we get $FH/H \cong F$. How do I finish up the proof? Does it have anything to do with the fact that $FH$ is a normal subgroup as well?

lucid_mathing
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  • Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Jan 25 '25 at 23:58
  • The claim that the kernel of $q$ is trivial is wrong. The kernel is of course $H$. We need to restrict $q$ to $F$ (so that the kernel becomes $F \cap H$), and then apply the 1st isomorphism theorem. The question itself is a special case of the 2nd isomorphism theorem - hence I think it should be closed as a duplicate. – Martin Brandenburg Jan 25 '25 at 23:59
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    The proof is explained in https://math.stackexchange.com/questions/1738334/ and https://math.stackexchange.com/questions/3160013 and https://math.stackexchange.com/questions/3242991 and many more posts. Also here: https://proofwiki.org/wiki/Second_Isomorphism_Theorem/Groups – Martin Brandenburg Jan 26 '25 at 00:03

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