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Given group G=MN where M and N are normal subgroup of G. I can show that $\frac{G}{M\cap N}$ isomorphic to $\frac{G}{M}\times\frac{G}{N}$. However I am a bit confused when I draw the lattice. I got the following diagram.

G

/ \

M N

\ /

$M\cap N$

If I mod out the $M\cap N$, I got following.

$\frac{G}{M\cap N}$

/ \

$\frac{M}{M\cap N}$ $\frac{N}{M\cap N}$

\ /

1 But $\frac{M}{M\cap N}$ is isomorphic to $\frac{MN}{N}$=$\frac{G}{N}$ and $\frac{N}{M\cap N}$ is isomorphic to $\frac{MN}{M}$=$\frac{G}{M}$ by diamond isomorphism theorem. How do I infer from above diagram that $\frac{G}{M\cap N}$ isomorphic to $\frac{G}{M}\times\frac{G}{N}$. Or did I do something wrong?

user45765
  • 8,500
  • You can't have shown $\frac{G}{M\cap N}\cong \frac{G}{M}\times\frac{G}{N}$; that needn't be true. Consider $M=N$ for instance. The first naturally injects into the second though. What are trying to do with the lattice anyway? All it says is that the isomorphic copies of $G/N$ and $G/M$ sitting inside $G/(M\cap N)$ generate the whole thing. Just because two subgroups generate a group doesn't mean the group is isomorphic to their direct product, though, so you can't infer $\frac{G}{M\cap N}\cong\frac{G}{M}\times\frac{G}{N}$. – anon Jul 06 '14 at 22:55
  • Why would $MN \cong G$? $M$ and $N$ can be two tiny little normal subgroups in a vast $G$. You're more likely to convince me that $\frac{MN}{N} \subseteq M$. – Eric Towers Jul 06 '14 at 22:58
  • Again, please tell us what you're aiming to do. You say in the beginning of the question that you can prove the isomorphism (although it's false in general; the additional hypothesis $M\cap N=1$ is necessary and sufficient). But at the end of your question you're asking how to infer the isomorphism from the work you've done. Which is it? @EricTowers OP says $G\color{Red}{=}MN$ is a given. The issue is not that they can be too small, it is that they can be too big (nontrivially intersecting, which makes OP's desired isomorphism false). – anon Jul 06 '14 at 23:00
  • In dummit foote Section 3.3 Excercise 7, It says G=MN as a given assumption – user45765 Jul 06 '14 at 23:30
  • I do not see from above lattice that $\frac{G}{M\cap N}$ is isomorphic to the product of quotient group G/N and G/M. – user45765 Jul 06 '14 at 23:32
  • I think what the book is saying is that M is different from N and they do not coincide or G=MN is a helpful statement if we assume M=N which reduces G=M. – user45765 Jul 06 '14 at 23:35
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    Ack, I am wrong, it is an isomorphism. Brainfreeze. You know that $G/(M\cap N)$ has two subgroups, call them $A$ and $B$, which are isomorphic copies of $G/N$ and $G/M$. You can tell that $A$ and $B$ are normal. Your lattice says that $A$ and $B$ generate the group $G/(M\cap N)$, which together with normality says $G/(M\cap N)=AB$. The lattice also says that $A,B$ are trivially intersecting. All of this implies $G/(M\cap N)$ is an internal direct product $A\times B$, and you're done. I think it'd be easier to exhibit an isomorphism $G/(M\cap N)\to G/M\times G/N$ though ($G=MN$ gives onto). – anon Jul 06 '14 at 23:53

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