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Let $A$ be a free abelian group with basis $x_1,x_2,x_3$ and let $B$ be a subgroup of A generated by $x_1+x_2+4x_3, 2x_1-x_1+2x_3$. In the group $A/B$ find the order of the coset $(x_1+2x_3)+B$.

How can I find this order? Just so you can reply more directly, I shall find a basis $f_1,f_2,f_3$ such that $d_1\cdot f_1,d_2\cdot f_2,d_3\cdot f_3$ is a basis of $B$ where $d_i|d_{i+1}$.

$Attempt$: so by $Smith$ algorithm, $\begin{pmatrix}1 & 2 \\1 & -1 \\4 & 2 \\\end{pmatrix}$ $\underrightarrow{R_2\to R_2-R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\4 & 2 \\\end{pmatrix}$ $\underrightarrow{R_3\to R_3-4R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\0 & -6 \\\end{pmatrix}$ $\underrightarrow{R_3\to R_3-2R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\0 & 0 \\\end{pmatrix}$ $\underrightarrow{C_2\to C_2-2C_1} \begin{pmatrix}1 & 0 \\0 & -3 \\0 & 0 \\\end{pmatrix}$$\underrightarrow{C_2\to -C_2} \begin{pmatrix}1 & 0 \\0 & 3 \\0 & 0 \\\end{pmatrix}$ and therefore, $d_1=1, \space d_2=3, \space d_3=0.$

Therefore, while $M= \begin{pmatrix}1 & 2 \\1 & -1 \\4 & 2 \\\end{pmatrix}$, there exist $Q,P\in GL_n(\Bbb{Z})$ such that $QMP^{-1}= \begin{pmatrix}1 & 0& 0 \\0 & 3 & 0\\0 & 0 & 0\\\end{pmatrix}$ and $P^{-1}$ columns are a basis of $A$ that fulfills the aforementioned requirements. One way to find $P^{-1}$ is to classify the identity element where the classification is the inverse of the row operations I used above in the opposite direction: from the end to the start: $\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\0 & 0 & 1\\\end{pmatrix}$ $\underrightarrow{R_3\to R_3 + 2R_1}\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\0 & 2 & 1\\\end{pmatrix}$ $\underrightarrow{R_3\to R_3+4R_1}\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\4 & 2 & 1\\\end{pmatrix}$ $\underrightarrow{R_2\to R_2+R_1}\begin{pmatrix}1 & 0& 0 \\1 & 1 & 0\\4 & 2 & 1\\\end{pmatrix}$ $\Rightarrow \{x_1+x_2+4x_3, x_2+2x_3 , x_3\}$ is the desired basis, where $\Rightarrow \{x_1+x_2+4x_3, 3\cdot (x_2+2x_3 )\}$ is a basis of $B$. Furthermore: $A/B\cong \Bbb{Z}_{f_1}\times \Bbb{Z}_{f_2}\times \Bbb{Z}_{f_3}/1\cdot \Bbb{Z}_{f_1}\times 3\cdot \Bbb{Z}_{f_2}\times 0\cdot \Bbb{Z}_{f_3} \cong \Bbb{Z}/\Bbb{Z}\times \Bbb{Z}/3\Bbb{Z}\cong \Bbb{Z}/3\Bbb{Z}$. How shall I continue?

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  • $\begingroup$ Since when do cosets have orders? $\endgroup$ – Donna Jan 25 '15 at 23:21
  • $\begingroup$ Oh, since they are elements in the quotient group.. $\endgroup$ – Donna Jan 25 '15 at 23:21
  • $\begingroup$ Okay... Any notes on my attempt below? $\endgroup$ – Donna Jan 25 '15 at 23:56
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If you stick to the "$QAP^{-1}$ convention" and do things horizontally, then $(Q^{-1})^{\top}$ is what gives you the basis and $P^{-1})^{\top}$ is what gives you the generators. To avoid transposing, take $\begin{pmatrix}1&1&4\\2&-1&-2\end{pmatrix}$.

Using elementary row and column operations and recording them in matrix form, I get that $$QAP^{-1}=\begin{pmatrix}1&0&0\\0&3&0\end{pmatrix}$$

where $Q=\begin{pmatrix}1&0\\-2&1\end{pmatrix}$ and $P=\begin{pmatrix}1&1&4\\0&-1&-2\\0&0&1\end{pmatrix}$, since $P^{-1}=\begin{pmatrix}1&1&-2\\0&-1&2\\0&0&1\end{pmatrix}$.

As it is known, our good generators are now given by $(f_1',f_2')=Q(f_1,f_2)^{\top}$ where $f_1=e_1+e_2+4e_3$ and $f_2=2e_1-e_2+2e_3$, so that $f_1'=f_1$ and $f_2'= -3e_2-6e_3=-3(e_2+2e_3)$.

And our good basis is given by $(e_1',e_2',e_3')=P(e_1,e_2,e_3)^{\top}$, so that $e_1'=f_1'=f_1$ and $e_2'=-e_2-2e_3$ and $f_2'=3e_2'$.

Now $x_1+2x_3=(1,0,2)= (1,1,4)+(0,-1,-2)=e_1'+e_2'$. I say this has order $3$ in the quotient.

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  • $\begingroup$ But we were taught in class that we can do it vertically,,, $\endgroup$ – Donna Jan 26 '15 at 1:32
  • $\begingroup$ The algorithm I used is not questionable as it was given by my teacher but I shall rethink it... Thanks for replying $\endgroup$ – Donna Jan 26 '15 at 1:33
  • $\begingroup$ @donnaIceberg Yes, but you should keep correct track of what the matrices you obtain are telling you. In your case, you need to invert the role and transpose every single thing. $\endgroup$ – Pedro Tamaroff Jan 26 '15 at 1:34
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    $\begingroup$ @donnaIceberg I am not saying your algorithm is incorrect. I have edited my post. $\endgroup$ – Pedro Tamaroff Jan 26 '15 at 1:37
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    $\begingroup$ @donnaIceberg Your proof is correct. I have arrived at the same results as you did. Kudos! $\endgroup$ – Pedro Tamaroff Jan 26 '15 at 1:40
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$Proof$: since no one has yet answered, I shall make an attempt and would appreciate your correction: Let us denote the basis $\{x_1+x_2+4x_3=a,\ 3 \cdot (x_2+2x_3)=b\}$ then $3a-b=3x_2+6x_6$. Then obviously, since $3x_2+6x_6\in B$ then the order of $(x_1+2x_3)+B$ is $3$, as $(x_1+2x_3)+B\ne B$, $2(x_1+2x_3)+B\ne B$ but $3(x_1+2x_3)+B=B= identity_{A/B}$.

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  • $\begingroup$ Or should I say that the order should divide 3 but since it isn't 1, it has to be 3? $\endgroup$ – Donna Jan 25 '15 at 23:38

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