Prove $G/(Mcap N) cong M/(Mcap N) times N/(Mcap N)$ where $G=MN$ and $M,Ntriangleleft G$
$begingroup$
If we consider the map $phi: Mtimes N rightarrow M/(Mcap N) times N/(Mcap N)$, I was able to show that this is onto and the kernel of the map is $(Mcap N) times (Mcap N)$ and hence by first isom. theorem,
$$(Mtimes N)/bigl((Mcap N)times (Mcap N)bigr) cong bigl(M/(Mcap N)bigr) times bigl(N/(Mcap N)bigr).$$
But how would I go about showing
$$(Mtimes N)/bigl((Mcap N)times (Mcap N)bigr) cong MN/(Mcap N)$$ to prove the final result? Appreciate your help.
abstract-algebra group-theory normal-subgroups group-isomorphism
edited Dec 4 '18 at 0:00
Andrews
3831317
asked Dec 3 '18 at 19:00
manifoldedmanifolded
615
$endgroup$
add a comment |
$begingroup$
If we consider the map $phi: Mtimes N rightarrow M/(Mcap N) times N/(Mcap N)$, I was able to show that this is onto and the kernel of the map is $(Mcap N) times (Mcap N)$ and hence by first isom. theorem,
$$(Mtimes N)/bigl((Mcap N)times (Mcap N)bigr) cong bigl(M/(Mcap N)bigr) times bigl(N/(Mcap N)bigr).$$
But how would I go about showing
$$(Mtimes N)/bigl((Mcap N)times (Mcap N)bigr) cong MN/(Mcap N)$$ to prove the final result? Appreciate your help.
abstract-algebra group-theory normal-subgroups group-isomorphism
edited Dec 4 '18 at 0:00
Andrews
3831317
asked Dec 3 '18 at 19:00
manifoldedmanifolded
615
$endgroup$
$begingroup$
You can do it, but it'll take some work because you need to go from $Mtimes N$ to $MN= G$ first. Better is to consider the maps from $G=MN$ to $G/N$ and to $G/M$, use them to obtain a map to $(G/N)times (G/M)$, and then use the isomorphism theorem that tells you that $HK/K cong H/Hcap K$.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:41
$begingroup$
Didn't you just ask this? math.stackexchange.com/questions/3023760
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:44
$begingroup$
@ArturoMagidin Reading the answers there, it appears that this is a follow up question to attempting to follow one of the answers there. Nonetheless, the previous question should have been linked, and the OP should have made clear that this is a distinct follow up question.
$endgroup$
– jgon
Dec 3 '18 at 21:57
add a comment |
$begingroup$
If we consider the map $phi: Mtimes N rightarrow M/(Mcap N) times N/(Mcap N)$, I was able to show that this is onto and the kernel of the map is $(Mcap N) times (Mcap N)$ and hence by first isom. theorem,
$$(Mtimes N)/bigl((Mcap N)times (Mcap N)bigr) cong bigl(M/(Mcap N)bigr) times bigl(N/(Mcap N)bigr).$$
But how would I go about showing
$$(Mtimes N)/bigl((Mcap N)times (Mcap N)bigr) cong MN/(Mcap N)$$ to prove the final result? Appreciate your help.
abstract-algebra group-theory normal-subgroups group-isomorphism
edited Dec 4 '18 at 0:00
Andrews
3831317
asked Dec 3 '18 at 19:00
manifoldedmanifolded
615
$endgroup$
If we consider the map $phi: Mtimes N rightarrow M/(Mcap N) times N/(Mcap N)$, I was able to show that this is onto and the kernel of the map is $(Mcap N) times (Mcap N)$ and hence by first isom. theorem,
$$(Mtimes N)/bigl((Mcap N)times (Mcap N)bigr) cong bigl(M/(Mcap N)bigr) times bigl(N/(Mcap N)bigr).$$
But how would I go about showing
$$(Mtimes N)/bigl((Mcap N)times (Mcap N)bigr) cong MN/(Mcap N)$$ to prove the final result? Appreciate your help.
abstract-algebra group-theory normal-subgroups group-isomorphism
abstract-algebra group-theory normal-subgroups group-isomorphism
edited Dec 4 '18 at 0:00
Andrews
3831317
asked Dec 3 '18 at 19:00
manifoldedmanifolded
615
edited Dec 4 '18 at 0:00
Andrews
3831317
asked Dec 3 '18 at 19:00
manifoldedmanifolded
615
edited Dec 4 '18 at 0:00
Andrews
3831317
edited Dec 4 '18 at 0:00
Andrews
3831317
edited Dec 4 '18 at 0:00
Andrews
3831317
3831317
asked Dec 3 '18 at 19:00
manifoldedmanifolded
615
asked Dec 3 '18 at 19:00
manifoldedmanifolded
615
asked Dec 3 '18 at 19:00
manifoldedmanifolded
615
615
$begingroup$
You can do it, but it'll take some work because you need to go from $Mtimes N$ to $MN= G$ first. Better is to consider the maps from $G=MN$ to $G/N$ and to $G/M$, use them to obtain a map to $(G/N)times (G/M)$, and then use the isomorphism theorem that tells you that $HK/K cong H/Hcap K$.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:41
$begingroup$
Didn't you just ask this? math.stackexchange.com/questions/3023760
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:44
$begingroup$
@ArturoMagidin Reading the answers there, it appears that this is a follow up question to attempting to follow one of the answers there. Nonetheless, the previous question should have been linked, and the OP should have made clear that this is a distinct follow up question.
$endgroup$
– jgon
Dec 3 '18 at 21:57
add a comment |
$begingroup$
You can do it, but it'll take some work because you need to go from $Mtimes N$ to $MN= G$ first. Better is to consider the maps from $G=MN$ to $G/N$ and to $G/M$, use them to obtain a map to $(G/N)times (G/M)$, and then use the isomorphism theorem that tells you that $HK/K cong H/Hcap K$.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:41
$begingroup$
Didn't you just ask this? math.stackexchange.com/questions/3023760
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:44
$begingroup$
@ArturoMagidin Reading the answers there, it appears that this is a follow up question to attempting to follow one of the answers there. Nonetheless, the previous question should have been linked, and the OP should have made clear that this is a distinct follow up question.
$endgroup$
– jgon
Dec 3 '18 at 21:57
$begingroup$
You can do it, but it'll take some work because you need to go from $Mtimes N$ to $MN= G$ first. Better is to consider the maps from $G=MN$ to $G/N$ and to $G/M$, use them to obtain a map to $(G/N)times (G/M)$, and then use the isomorphism theorem that tells you that $HK/K cong H/Hcap K$.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:41
$begingroup$
You can do it, but it'll take some work because you need to go from $Mtimes N$ to $MN= G$ first. Better is to consider the maps from $G=MN$ to $G/N$ and to $G/M$, use them to obtain a map to $(G/N)times (G/M)$, and then use the isomorphism theorem that tells you that $HK/K cong H/Hcap K$.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:41
$begingroup$
Didn't you just ask this? math.stackexchange.com/questions/3023760
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:44
$begingroup$
Didn't you just ask this? math.stackexchange.com/questions/3023760
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:44
$begingroup$
@ArturoMagidin Reading the answers there, it appears that this is a follow up question to attempting to follow one of the answers there. Nonetheless, the previous question should have been linked, and the OP should have made clear that this is a distinct follow up question.
$endgroup$
– jgon
Dec 3 '18 at 21:57
$begingroup$
@ArturoMagidin Reading the answers there, it appears that this is a follow up question to attempting to follow one of the answers there. Nonetheless, the previous question should have been linked, and the OP should have made clear that this is a distinct follow up question.
$endgroup$
– jgon
Dec 3 '18 at 21:57
add a comment |
1 Answer
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$begingroup$
Use first isomorphism theorem again. Can you show that $Mtimes Nto G/(Mcap N)$ given by $(m,n)mapsto mn(Mcap N)$ is a group homomorphism and has kernel $(Mcap N)times (Mcap N)$?
Second edit after reading comments on Servaes' answer on the first question you asked related to this Servaes appears to have suggested this homomorphism in a prior version of his answer according to the comments below it and a skim of revision history. Since you couldn't see why it is a homomorphism, I'll add the following hint: Using that $M$ and $N$ are normal, can you show that the element $mnm^{-1}n^{-1}$ is in $Mcap N$ when $min M$, $nin N$. Can you see that this implies that $M/Mcap N$ and $N/Mcap N$ commute with each other in $G/Mcap N$?
Edit Or alternatively you can follow Arturo's hint.
answered Dec 3 '18 at 19:43
jgonjgon
13.5k22041
$endgroup$
add a comment |
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$begingroup$
Use first isomorphism theorem again. Can you show that $Mtimes Nto G/(Mcap N)$ given by $(m,n)mapsto mn(Mcap N)$ is a group homomorphism and has kernel $(Mcap N)times (Mcap N)$?
Second edit after reading comments on Servaes' answer on the first question you asked related to this Servaes appears to have suggested this homomorphism in a prior version of his answer according to the comments below it and a skim of revision history. Since you couldn't see why it is a homomorphism, I'll add the following hint: Using that $M$ and $N$ are normal, can you show that the element $mnm^{-1}n^{-1}$ is in $Mcap N$ when $min M$, $nin N$. Can you see that this implies that $M/Mcap N$ and $N/Mcap N$ commute with each other in $G/Mcap N$?
Edit Or alternatively you can follow Arturo's hint.
answered Dec 3 '18 at 19:43
jgonjgon
13.5k22041
$endgroup$
add a comment |
$begingroup$
Use first isomorphism theorem again. Can you show that $Mtimes Nto G/(Mcap N)$ given by $(m,n)mapsto mn(Mcap N)$ is a group homomorphism and has kernel $(Mcap N)times (Mcap N)$?
Second edit after reading comments on Servaes' answer on the first question you asked related to this Servaes appears to have suggested this homomorphism in a prior version of his answer according to the comments below it and a skim of revision history. Since you couldn't see why it is a homomorphism, I'll add the following hint: Using that $M$ and $N$ are normal, can you show that the element $mnm^{-1}n^{-1}$ is in $Mcap N$ when $min M$, $nin N$. Can you see that this implies that $M/Mcap N$ and $N/Mcap N$ commute with each other in $G/Mcap N$?
Edit Or alternatively you can follow Arturo's hint.
answered Dec 3 '18 at 19:43
jgonjgon
13.5k22041
$endgroup$
add a comment |
$begingroup$
Use first isomorphism theorem again. Can you show that $Mtimes Nto G/(Mcap N)$ given by $(m,n)mapsto mn(Mcap N)$ is a group homomorphism and has kernel $(Mcap N)times (Mcap N)$?
Second edit after reading comments on Servaes' answer on the first question you asked related to this Servaes appears to have suggested this homomorphism in a prior version of his answer according to the comments below it and a skim of revision history. Since you couldn't see why it is a homomorphism, I'll add the following hint: Using that $M$ and $N$ are normal, can you show that the element $mnm^{-1}n^{-1}$ is in $Mcap N$ when $min M$, $nin N$. Can you see that this implies that $M/Mcap N$ and $N/Mcap N$ commute with each other in $G/Mcap N$?
Edit Or alternatively you can follow Arturo's hint.
answered Dec 3 '18 at 19:43
jgonjgon
13.5k22041
$endgroup$
Use first isomorphism theorem again. Can you show that $Mtimes Nto G/(Mcap N)$ given by $(m,n)mapsto mn(Mcap N)$ is a group homomorphism and has kernel $(Mcap N)times (Mcap N)$?
Second edit after reading comments on Servaes' answer on the first question you asked related to this Servaes appears to have suggested this homomorphism in a prior version of his answer according to the comments below it and a skim of revision history. Since you couldn't see why it is a homomorphism, I'll add the following hint: Using that $M$ and $N$ are normal, can you show that the element $mnm^{-1}n^{-1}$ is in $Mcap N$ when $min M$, $nin N$. Can you see that this implies that $M/Mcap N$ and $N/Mcap N$ commute with each other in $G/Mcap N$?
Edit Or alternatively you can follow Arturo's hint.
answered Dec 3 '18 at 19:43
jgonjgon
13.5k22041
edited Dec 3 '18 at 22:07
answered Dec 3 '18 at 19:43
jgonjgon
13.5k22041
answered Dec 3 '18 at 19:43
jgonjgon
13.5k22041
answered Dec 3 '18 at 19:43
jgonjgon
13.5k22041
13.5k22041
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$begingroup$
You can do it, but it'll take some work because you need to go from $Mtimes N$ to $MN= G$ first. Better is to consider the maps from $G=MN$ to $G/N$ and to $G/M$, use them to obtain a map to $(G/N)times (G/M)$, and then use the isomorphism theorem that tells you that $HK/K cong H/Hcap K$.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:41
$begingroup$
Didn't you just ask this? math.stackexchange.com/questions/3023760
$endgroup$
– Arturo Magidin
Dec 3 '18 at 19:44
$begingroup$
@ArturoMagidin Reading the answers there, it appears that this is a follow up question to attempting to follow one of the answers there. Nonetheless, the previous question should have been linked, and the OP should have made clear that this is a distinct follow up question.
$endgroup$
– jgon
Dec 3 '18 at 21:57