0
$\begingroup$

In Robert Ash Abstract Algebra chapter 4 section 4.6.6 'Abelian Group Specified by Generators and Relations', I don't understand the sentence ' This process yields the abelian group $G=F/K$'. This process refers to what?

Also, after we transform the matrix into its smith normal form, why we can deduce that $F/K \cong \mathbb{Z}_2 \bigoplus \mathbb{Z}_4 \bigoplus \mathbb{Z}$ ? I kind of feel that we get this from fundamental decomposition theorem, but where we get the ideals? Can someone explan this to me?

Remark: Suppose that we have a free abelian group $F$ with basis $x_1.x_2.x_3$ and the following constraints on the $x_i$ $$2x_1+2x_2+8x_3=0$$ and $$-2x_1+2x_2+4x_3=0$$ What we are doing is forming a ' submodule of relations' $K$ with generators $u_1=2x_1+2x_2+8x_3$ and $u_2=-2x_1+2x_2+4x_3$ and we are identifying every element in $K$ with zero. This process yields the abelian group $G=F / K$, which is generated by $x_1+K$, $x_2+K$, $x_3+K$. 'this part discuss how we obtain the matrix in smith normal form, so I skip it'. This we have a new basis $y_1,y_2,y_3$ and new generators $2y_1,4y_2$ for $K$. The quotient group $F/K$ is generated by $y_1+k$,$y_2+K$ and $y_3+K$, with $2(y_1+K)=4(y_2+K)=0+K$. Hence, we have $F/K \cong \mathbb{Z}_2 \bigoplus \mathbb{Z}_4 \bigoplus \mathbb{Z}$

$\endgroup$
  • 2
    $\begingroup$ Please provide the actual passage. $\endgroup$ – Alex Youcis Sep 8 '13 at 8:58
  • $\begingroup$ The process he refers to is nothing more than taking the quotient group--identifying $K$ with zero, as well as any other pursuant identities. Do you know about Smith normal forms? Part of the theory is that once you obtain the SNF, you can read the decomposition of the quotient into a product of cyclic groups--the orders just being the diagonal entries. $\endgroup$ – Alex Youcis Sep 8 '13 at 9:40
  • $\begingroup$ @Idonknow : I think that when you specify the relations, then it is not a free abelian group anymore. So you should say: "Suppose we have an abelian group with basis ..., specified by the following constraints:" $\endgroup$ – Manolito Pérez Apr 8 '14 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.