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Thread: Isomorphism (Help)

  1. #1
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    Isomorphism (Help)

    If $\displaystyle H$ and K are normal subgroups of a group $\displaystyle G$ with $\displaystyle HK=G$, prove that $\displaystyle G/(H \cap K) \simeq (G/H) \times (G/K)$.
    Hint: If $\displaystyle \phi:G \rightarrow (G/H) \times (G/K)$ is defined by $\displaystyle x \mapsto (xH,xK)$, then $\displaystyle ker \: \phi=H \cap K$; moreover, we have $\displaystyle G=HK$, so that $\displaystyle \displaystyle\cap_{a}aH=HK=\displaystyle\cap_{b}bK$.

    I have managed to prove that $\displaystyle ker \: \phi=H \cap K$. I'm having difficulty in proving that $\displaystyle \phi$ is surjective. This is crucial in proving the above isomorphism using 1st Isomorphism Theorem.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Markeur View Post
    If $\displaystyle H$ and K are normal subgroups of a group $\displaystyle G$ with $\displaystyle HK=G$, prove that $\displaystyle G/(H \cap K) \simeq (G/H) \times (G/K)$.
    Hint: If $\displaystyle \phi:G \rightarrow (G/H) \times (G/K)$ is defined by $\displaystyle x \mapsto (xH,xK)$, then $\displaystyle ker \: \phi=H \cap K$; moreover, we have $\displaystyle G=HK$, so that $\displaystyle \displaystyle\cap_{a}aH=HK=\displaystyle\cap_{b}bK$.

    I have managed to prove that $\displaystyle ker \: \phi=H \cap K$. I'm having difficulty in proving that $\displaystyle \phi$ is surjective. This is crucial in proving the above isomorphism using 1st Isomorphism Theorem.
    Let $\displaystyle \left(yH,zK\right)$ be in the codomain. Since $\displaystyle HK=G$ what can you say about $\displaystyle y,z$?
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  3. #3
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    Hey,

    The proof is as below:

    We want to show that $\displaystyle \forall (yH,zK) \in (G/H) \times (G/K)$, $\displaystyle \exists g \in G$ such that $\displaystyle \phi(g)=(gH,gK)$.

    Since $\displaystyle (yH,zK) \in (G/H) \times (G/K)$, $\displaystyle yH \in G/H$ and $\displaystyle zK \in G/K$.
    Since $\displaystyle G=HK$, $\displaystyle yH \in HK/H$ and $\displaystyle zK \in HK/K$.
    This implies that $\displaystyle y,z \in HK$.
    Since $\displaystyle G=HK$, $\displaystyle g \in HK$.
    This implies that $\displaystyle y=g$ and $\displaystyle z=g \: \forall y,z \in G$.
    Hence $\displaystyle \exists g \in G$ such that $\displaystyle \phi(g)=(gH,gK)$.
    Therefore $\displaystyle (G/H) \times (G/K) \subseteq Im \; \phi$.
    Clearly,$\displaystyle Im \; \phi \subseteq (G/H) \times (G/K)$.
    Finally, we conclude that $\displaystyle (G/H) \times (G/K)=Im \; \phi$.

    From there, we can establish the isomorphism using First Isomorphism Theorem.

    So, is the above proof correct?

    Thanks in advance.
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