If $N_1,N_2$ are normal subgroups of a group $G$, is $G/N_1\cap N_2\cong G/N_1\times_{G/N_1N_2} G/N_2$?
Is the quotient of a group $G$ by the intersection of normal subgroups $N_1\cap N_2$ isomorphic to the fiber product $G/N_1\times_{G/N_1N_2} G/N_2$?
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abstract-algebra
group-theory
1 Answers
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Consider the natural quotient maps $G\to G/N_1$ and $G\to G/N_2$ and the map $G\to G/N_1\times G/N_2$ induced by these.
The kernel of this map is the intersection of the kernels of $G\to G/N_1$ and $G\to G/N_2$ - that is $N_1\cap N_2$
Clearly if we extend each quotient map to $G/N_1N_2$ (e.g. $G\to G/N_1\to G/N_1N_2$) we obtain the natural quotient map $G\to G/N_1N_2$. In particular, the image of $G\to G/N_1\times G/N_2$ is contained in $G/N_1\times_{G/N_1N_2} G/N_2$.
To see the image is $G/N_1\times_{G/N_1N_2} G/N_2$, consider $(gN_1,hN_2)\in G/N_1\times_{G/N_1N_2} G/N_2$ for $g,h\in G$. By definition $gN_1N_2=hN_1N_2$ so $h\in gN_1N_2$. Therefore there is some $x\in N_1$, $y\in N_2$ with $gxy=h$. The image under $G\to G/N_1\times G/N_2$ of $gx$ is $(gxN_1,gxN_2)=(gN_1,hy^{-1}N_2)=(gN_1,hN_2)$. Hence the image is indeed $G/N_1\times_{G/N_1N_2} G/N_2$.
By the first isomorphism theorem then $G/(N_1\cap N_2)\cong G/N_1\times_{G/N_1N_2} G/N_2$.