I'm dealing with free abelian groups and I encountered the following question:
Let $A$ be an abelian free group with a basis $\{x_1, x_2, x_3\}.$
Let $B$ be subgroup of A generated by $x_1+x_2+4x_3$, $2x_1-x_2+2x_3$.
Find the order of the coset $(x_1+2x_3)+B$.
Approach:
By definition the order of the coset is the minimal k$\in$$N$ such that $k(x_1+2x_3)+B=B$ iff $k(x_1+2x_3)\in B$ iff there exist $m_1, m_2\in N$ such that $kx_1+2kx_3=m_1(x_1+x_2+4x_3)+m_2(2x_1-x_2+2x_3)$. Adding things together I get that $m_1=m_2$ and $(k-3m_1)x_1+(2k-6m_1)x_3=0$. Therefore choosing $m_1=1, k=3$ the equation holds and the order is 3.
Is that correct?
Best Answer
$3x_1+6x_3=(x_1+x_2+4x_3)+(2x_1-x_2+2x_3)\in B$ and $3(x_1+2x_3)=3x_1+6x_3$. From there it follows that it has order $3$. (Use that it's a torsion element in $\Bbb Z\oplus\Bbb Z_3$).