On isomorphism of quotient groups of free abelian groups of finite rank

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Consider the free abelian group $mathbb Z^n$, with elements considered as row vectors. For every $Ain M_rtimes n(mathbb Z)$ , let $K_A$ be the subgroup of $mathbb Z^n$ generated by the row vectors of $A$ . Now let $Ain M_rtimes n(mathbb Z)$ and $B:=PAQ$ where $Pin GL_rtimes r(mathbb Z), Q in GL_ntimes n(mathbb Z)$. Then how to prove that $mathbb Z^n/K_A $ and $mathbb Z^n/K_B$ are isomorphic as groups ?



I think the map $f: mathbb Z^n/K_A to mathbb Z^n/K_B$ defined as $f(bar a +K_A)=bar a Q+K_B, forall bar a in mathbb Z^n$ is an isomorphism, but I am not sure (I can't even show whether this map is well-defined or not).



Please help.




modules abelian-groups group-isomorphism free-modules free-abelian-group




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  • That's basically it.
    – Lord Shark the Unknown
    Jul 28 at 17:13










  • @LordSharktheUnknown: Can you please at least provide a proof of the well-defined ness in an answer ?
    – user521337
    Jul 28 at 17:48














up vote
0
down vote

favorite












Consider the free abelian group $mathbb Z^n$, with elements considered as row vectors. For every $Ain M_rtimes n(mathbb Z)$ , let $K_A$ be the subgroup of $mathbb Z^n$ generated by the row vectors of $A$ . Now let $Ain M_rtimes n(mathbb Z)$ and $B:=PAQ$ where $Pin GL_rtimes r(mathbb Z), Q in GL_ntimes n(mathbb Z)$. Then how to prove that $mathbb Z^n/K_A $ and $mathbb Z^n/K_B$ are isomorphic as groups ?



I think the map $f: mathbb Z^n/K_A to mathbb Z^n/K_B$ defined as $f(bar a +K_A)=bar a Q+K_B, forall bar a in mathbb Z^n$ is an isomorphism, but I am not sure (I can't even show whether this map is well-defined or not).



Please help.




modules abelian-groups group-isomorphism free-modules free-abelian-group




share|cite|improve this question



















  • That's basically it.
    – Lord Shark the Unknown
    Jul 28 at 17:13










  • @LordSharktheUnknown: Can you please at least provide a proof of the well-defined ness in an answer ?
    – user521337
    Jul 28 at 17:48












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider the free abelian group $mathbb Z^n$, with elements considered as row vectors. For every $Ain M_rtimes n(mathbb Z)$ , let $K_A$ be the subgroup of $mathbb Z^n$ generated by the row vectors of $A$ . Now let $Ain M_rtimes n(mathbb Z)$ and $B:=PAQ$ where $Pin GL_rtimes r(mathbb Z), Q in GL_ntimes n(mathbb Z)$. Then how to prove that $mathbb Z^n/K_A $ and $mathbb Z^n/K_B$ are isomorphic as groups ?



I think the map $f: mathbb Z^n/K_A to mathbb Z^n/K_B$ defined as $f(bar a +K_A)=bar a Q+K_B, forall bar a in mathbb Z^n$ is an isomorphism, but I am not sure (I can't even show whether this map is well-defined or not).



Please help.




modules abelian-groups group-isomorphism free-modules free-abelian-group




share|cite|improve this question











Consider the free abelian group $mathbb Z^n$, with elements considered as row vectors. For every $Ain M_rtimes n(mathbb Z)$ , let $K_A$ be the subgroup of $mathbb Z^n$ generated by the row vectors of $A$ . Now let $Ain M_rtimes n(mathbb Z)$ and $B:=PAQ$ where $Pin GL_rtimes r(mathbb Z), Q in GL_ntimes n(mathbb Z)$. Then how to prove that $mathbb Z^n/K_A $ and $mathbb Z^n/K_B$ are isomorphic as groups ?



I think the map $f: mathbb Z^n/K_A to mathbb Z^n/K_B$ defined as $f(bar a +K_A)=bar a Q+K_B, forall bar a in mathbb Z^n$ is an isomorphism, but I am not sure (I can't even show whether this map is well-defined or not).



Please help.





modules abelian-groups group-isomorphism free-modules free-abelian-group





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 28 at 17:11









user521337

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  • That's basically it.
    – Lord Shark the Unknown
    Jul 28 at 17:13










  • @LordSharktheUnknown: Can you please at least provide a proof of the well-defined ness in an answer ?
    – user521337
    Jul 28 at 17:48
















  • That's basically it.
    – Lord Shark the Unknown
    Jul 28 at 17:13










  • @LordSharktheUnknown: Can you please at least provide a proof of the well-defined ness in an answer ?
    – user521337
    Jul 28 at 17:48















That's basically it.
– Lord Shark the Unknown
Jul 28 at 17:13




That's basically it.
– Lord Shark the Unknown
Jul 28 at 17:13












@LordSharktheUnknown: Can you please at least provide a proof of the well-defined ness in an answer ?
– user521337
Jul 28 at 17:48




@LordSharktheUnknown: Can you please at least provide a proof of the well-defined ness in an answer ?
– user521337
Jul 28 at 17:48















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