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What is a "natural isomorphism" taking $G/(H\cap K)\to G/H\times G/K$ where $G$ is a group and $H,K$ are its subgroups? I don't quite know how to split $g(H\cap K)$. Would somebody out there help me out?

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The natural choice is $gH\cap K \mapsto (gH,gK)$ –  Bill Cook Feb 15 '12 at 18:47
    
... and the hard part is proving that it is surjective. –  lhf Feb 15 '12 at 18:53
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This cannot be true without further hypotheses. Take for instance $G$ finite and $H=K=1$. –  lhf Feb 15 '12 at 18:55
    
There's definitely a natural homomorphism, but not a natural isomorphism. –  Thomas Andrews Feb 15 '12 at 18:57
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Since you are a @Newbie, you should know that some authors use the term "isomorphism" for a one-to-one homomorphism. Although it might be more accurate to say they "used" to use the term that way. Now "isomorphism" almost univerisally means one-to-one and onto homomorphisms. The older use of the word has been replaced by "monomorphism" (a one-to-one homomorphism). –  Bill Cook Feb 15 '12 at 19:03

2 Answers 2

up vote 5 down vote accepted

Take the map $\varphi: G \rightarrow G/H \times G/K, g \mapsto (gH, gK)$ and let $g,h \in G$. We have $\varphi(g) = \varphi(h)$ if and only if $gH = hH$ and $gK = hK$, i.e $gh^{-1} \in H \cap K$. This shows that the desired map $$G/(H \cap K) \rightarrow G/H \times G/K, g(H \cap K) \mapsto (gH,gK)$$ is a well defined isomorphism onto its image (if you take $H = \{1\} = K$, you don't get a surjection), where "isomorphism" could mean isomorphism of sets, of G-sets or of groups, if $H$ and $K$ are assumed to be normal. Of course the proof is easier in the later case as was shown in another answer.

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Assuming that $H$ and $K$ are normal subgroups, the natural projection $G \to G/H\times G/K$ has kernel $H\cap K$ and so induces a natural injective homomorphism $G/(H\cap K)\to G/H\times G/K$.

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