Show that $(Mtimes N)/Rcong M$.












1












$begingroup$


If there are two groups, $M$ (with multiplication $cdot_M$) and $N$ (with multiplication $cdot_N$) and we define a new group $M times N$ with multiplication such that
$$
(m,n)(m',n') = (m cdot_M m',n cdot_N n')
$$
There can be a normal subgroup $R$ of $M times N$ such that




  • $eR = text{identity element}$


  • $R = {(eR, r) | r in R}$



Show that



$$
frac{M times N}{R} simeq M
$$






abstract-algebra group-theory normal-subgroups group-isomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:24












  • $begingroup$
    What about R - the normal subgroup?
    $endgroup$
    – user628956
    Dec 24 '18 at 4:25










  • $begingroup$
    Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:28












  • $begingroup$
    Also I hastily ended up answering this, but please provide what you tried next time.
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:30
















1












$begingroup$


If there are two groups, $M$ (with multiplication $cdot_M$) and $N$ (with multiplication $cdot_N$) and we define a new group $M times N$ with multiplication such that
$$
(m,n)(m',n') = (m cdot_M m',n cdot_N n')
$$
There can be a normal subgroup $R$ of $M times N$ such that




  • $eR = text{identity element}$


  • $R = {(eR, r) | r in R}$



Show that



$$
frac{M times N}{R} simeq M
$$






abstract-algebra group-theory normal-subgroups group-isomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:24












  • $begingroup$
    What about R - the normal subgroup?
    $endgroup$
    – user628956
    Dec 24 '18 at 4:25










  • $begingroup$
    Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:28












  • $begingroup$
    Also I hastily ended up answering this, but please provide what you tried next time.
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:30














1












1








1





$begingroup$


If there are two groups, $M$ (with multiplication $cdot_M$) and $N$ (with multiplication $cdot_N$) and we define a new group $M times N$ with multiplication such that
$$
(m,n)(m',n') = (m cdot_M m',n cdot_N n')
$$
There can be a normal subgroup $R$ of $M times N$ such that




  • $eR = text{identity element}$


  • $R = {(eR, r) | r in R}$



Show that



$$
frac{M times N}{R} simeq M
$$






abstract-algebra group-theory normal-subgroups group-isomorphism







share|cite|improve this question











$endgroup$




If there are two groups, $M$ (with multiplication $cdot_M$) and $N$ (with multiplication $cdot_N$) and we define a new group $M times N$ with multiplication such that
$$
(m,n)(m',n') = (m cdot_M m',n cdot_N n')
$$
There can be a normal subgroup $R$ of $M times N$ such that




  • $eR = text{identity element}$


  • $R = {(eR, r) | r in R}$



Show that



$$
frac{M times N}{R} simeq M
$$






abstract-algebra group-theory normal-subgroups group-isomorphism




abstract-algebra group-theory normal-subgroups group-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 11:21









Shaun

11.1k113688




11.1k113688










asked Dec 24 '18 at 4:20









user628956user628956

203




203












  • $begingroup$
    Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:24












  • $begingroup$
    What about R - the normal subgroup?
    $endgroup$
    – user628956
    Dec 24 '18 at 4:25










  • $begingroup$
    Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:28












  • $begingroup$
    Also I hastily ended up answering this, but please provide what you tried next time.
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:30


















  • $begingroup$
    Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:24












  • $begingroup$
    What about R - the normal subgroup?
    $endgroup$
    – user628956
    Dec 24 '18 at 4:25










  • $begingroup$
    Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:28












  • $begingroup$
    Also I hastily ended up answering this, but please provide what you tried next time.
    $endgroup$
    – TrostAft
    Dec 24 '18 at 4:30
















$begingroup$
Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
$endgroup$
– TrostAft
Dec 24 '18 at 4:24






$begingroup$
Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
$endgroup$
– TrostAft
Dec 24 '18 at 4:24














$begingroup$
What about R - the normal subgroup?
$endgroup$
– user628956
Dec 24 '18 at 4:25




$begingroup$
What about R - the normal subgroup?
$endgroup$
– user628956
Dec 24 '18 at 4:25












$begingroup$
Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
$endgroup$
– TrostAft
Dec 24 '18 at 4:28






$begingroup$
Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
$endgroup$
– TrostAft
Dec 24 '18 at 4:28














$begingroup$
Also I hastily ended up answering this, but please provide what you tried next time.
$endgroup$
– TrostAft
Dec 24 '18 at 4:30




$begingroup$
Also I hastily ended up answering this, but please provide what you tried next time.
$endgroup$
– TrostAft
Dec 24 '18 at 4:30










1 Answer
1






active

oldest

votes


















2












$begingroup$

Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:



$$
(M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
$$



You can check it to be a group homomorphism.






share|cite|improve this answer









$endgroup$














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    $begingroup$

    Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:



    $$
    (M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
    $$



    You can check it to be a group homomorphism.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:



      $$
      (M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
      $$



      You can check it to be a group homomorphism.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:



        $$
        (M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
        $$



        You can check it to be a group homomorphism.






        share|cite|improve this answer









        $endgroup$



        Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:



        $$
        (M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
        $$



        You can check it to be a group homomorphism.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 4:26









        TrostAftTrostAft

        433412




        433412






























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