1
$\begingroup$

Let M be the matrix: $$ M = \begin{bmatrix} 2& 2& 0& 4\\ 0& 1& 3& 5\\ 2& 5& 3& 1\\ 3& 1& 7& 3\end{bmatrix} $$ And H be the sub group of $\mathbb {Z}^4$ created by the matrix' columns.

I need to display $\mathbb{Z}^4/H$ as a direct sum of cylic groups.

How can I do that ?

(I'm sorry but I don't know how to use the math formulas)

  • 0
    I'm not very good at algebra, but hope I used the right symbols. – 2017-02-25
  • 0
    That's not the right symbols :( It's not the Z4 you wrote, it's Z^4 – 2017-02-25
  • 0
    The discussion in http://web.science.mq.edu.au/~chris/groups/CHAP10%20Finitely-Generated%20Abelian%20Groups.pdf should show you what to do. – 2017-02-25
  • 0
    Ok, Z^4 is on the way in the review queue. – 2017-02-25
  • 0
    @JazzyMatrix this is not true. A similar example is $2\mathbb{Z}\cong\mathbb{Z}$ but $\mathbb{Z}/2\mathbb{Z}$ is non-trivial. In this case you can column reduce the matrix (swapping columns and subtracting multiples of one column from another only) to obtain a diagonal matrix. This gives a more natural basis for $H$ so you can easily calculate the quotient – 2017-02-25
  • 0
    @RobertChamberlain Thanks for the correction -- deleted my comment – 2017-02-25
  • 0
    So, have you had a look at those notes, MathEnthusiast? – 2017-02-26
  • 0
    I have looked and I was able to get the matrix to the form: 1 0 0 0 0 1 0 0 0 0 2 0 0 0 0 192 Does that mean that Z^4/H is isomorphic to Z2 + Z192 ? Or is H isomorphic to Z2 + Z192 ? – 2017-03-01
  • 0
    The matrix should be $$ M = \begin{bmatrix} 1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 192\end{bmatrix} $$ – 2017-03-01

0 Answers 0