$A$ is an abelian free group with base $x_1,x_2,x_3$. Let $B$ be the subgroup generated by $x_1+x_2+4x_3,2x_1-x_2+2x_3$. In the group $A/B$ find the order of the coset $(x_1+2x_3)+B$.
This type of questions is common in the exems that i study for them know.
somebody can show me an algorithm for this type of questions and deep explanation?
I sincerely thank in advance.
I don't think this is a "deep explanation," but if you set up the equation $$ m(x_1 + 2x_3) = a(x_1 + x_2 + 4x_3) + b(2x_1 - x_2 + 2x_3) $$ you obtain the relations $m = a+2b$, $0 = a - b$, and $2m = 4a+2b$. This system reduces to $m = 3a$. There's probably a slick way to do it using matrices.
Edit: A slick method using matrices, a little under 3 years too late! The key ingredient is the Smith normal form of a matrix: see here for more on the general theory.
Computing the Smith normal form of the relations matrix $M$ (whose columns are the coefficients of the generators of $B$), we find $$ \underbrace{ \begin{pmatrix} 0 & 1 & 0 \\ 1 & -1 & 0 \\ 2 & 2 & -1 \end{pmatrix} }_P \underbrace{ \begin{pmatrix} 1 & 2\\ 1 & -1\\ 4 & 2 \end{pmatrix} }_M \underbrace{ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} }_Q = \underbrace{ \begin{pmatrix} 1 & 0 \\ 0 & 3 \\ 0 & 0 \end{pmatrix} }_D \, . $$ This gives us the isomorphism $$ \frac{A}{B} \cong \frac{\mathbb{Z} w_1 \oplus \mathbb{Z} w_2 \oplus \mathbb{Z} w_3}{\mathbb{Z} w_1 \oplus \mathbb{Z} 3 w_2} \cong \frac{\mathbb{Z}}{3\mathbb{Z}}w_2 \oplus \mathbb{Z} w_3 \cong \frac{\mathbb{Z}}{3\mathbb{Z}} \oplus \mathbb{Z} \, . $$ Moreover, the matrix $P$ tells us the basis $\{w_1, w_2, w_3\}$ with respect to which we have this isomorphism. Since $$ P^{-1}= \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 4 & 2 & -1 \end{pmatrix} $$ then $$ w_1 = x_1 + x_2 + 4 x_3,\quad w_2 = x_1 + 2 x_3,\quad w_3 = -x_3 $$ where $\{x_1,x_2,x_3\}$ is the standard basis for $\mathbb{Z}^3$. The element $w_2 = x_1 + 2 x_3$ should look familiar: it is the element whose order in the quotient we are trying to compute! Since the image of $w_2$ generates the factor $\mathbb{Z}/3\mathbb{Z}$, then it has order $3$, as we previously found.
