Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $K$ be the additive group of $\mathbb Z\oplus \mathbb Z$. If $A = \left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$ is an $2\times 2$ matrix where $a, b, c, d$, are in $\mathbb Z$, then $HA*=\langle(a,b), (c,d)\rangle$ is the subgroup generated by rows of matrix $A$.

1) Let $A= \left(\begin{array}{cc} 3 & 1\\ 0 & 5 \end{array}\right)$. Show that $K/HA*$ has order 15.

2) Let $B= \left(\begin{array}{cc} 3 & 1\\ 6 & 7 \end{array}\right)$. Show that $K/HB*$ has order 15.

3) Let $C= \left(\begin{array}{cc} 9 & 8\\ 6 & 7 \end{array}\right)$. Show that $K/HC*$ has order 15.

My thinking: I can use something that restricts the quotient group but I am not sure how to do it?

Thank you for helping out.

share|cite|improve this question

You should be able to see how to get the second and third fom the first by using row operations. To see why the first gives an isomorphism to $\mathbb Z_3 \oplus \mathbb Z_5$ is trickier. You need to use the fact that $3$ and $5$ are relatively prime to get rid of the $1$ somehow.

share|cite|improve this answer
I have tried but I cannot get anywhere? Can you please explain the hint – Mark Rutherford May 5 '13 at 22:39
Do you see how to use row operations (which can be interpreted like a change of basis in linear algebra) to get the second two isomorphic to the first one? – Ted Shifrin May 6 '13 at 2:42

The Smith Normal Form of your matrices is the same: $\left(\begin{array}{cc} 1 & 0\\ 0 & 15 \end{array}\right).$ This shows that all three factor groups are isomorphic to $\mathbb Z_{15}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.