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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.

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2 Answers 2

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.

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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 $\mathbb Z_{15}$.

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