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I know that the second isomorphism theorem for groups doesn't hold for topological groups, the version that I have for the second isomorphism theorem is:

  • If $G$ is a group, $H$ a subgroup of $G$ and $N$ a normal subgroup of $G$, Then $ (HN)/N \cong H/(H\cap N)$.

Under wich conditions we have that the theorem holds for topological groups?,

On the other side the third isomorphism theorem holds for topological groups.

  • If $H$ and $K$ are normal subgroups of a topological group $G$ and $K$ is a subgroup of $H$, then $G/H \cong (G/K) /(H/K)$

In the prove for ordinary groups, the homomorphism $\phi: G \longrightarrow (G/K)/(H/K)$ defined by $ \phi (a)= aK(H/K)$ is surjective with kernel $H$, then by the first isomorphism theorem we have the isomorphism.

In the case for topological groups, we need to have that the homomorphism is an open map to use the first isomorphism theorem, how can I prove that the homomorphism is open?.

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  • $\begingroup$ Don't you need closed normal subgroups? $\endgroup$
    – lhf
    May 2 '16 at 21:46
  • $\begingroup$ For the second isomorphism theorem? $\endgroup$
    – Giotaker
    May 2 '16 at 21:50
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Quotient maps $f:G\to G/N$ are always open. Let $V\subseteq G$ be open; we need to show $f(V)$ is open. By definition this is equivalent to $f^{-1}(f(V))$ being open. But $f^{-1}(f(V))=VN=\bigcup_{n\in N}Vn$, which is a union of open sets (since each $Vn$ is open, as a shift of an open sets), hence open.

Therefore the third isomorphism theorem holds for topological groups.

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