4
$\begingroup$

For the group $G=\{a+b\sqrt 2 \mid a,b\in\mathbb Z\},$ and the subgroup $H=\{3m+2n\sqrt 2 \mid m,n \in\mathbb Z\},$ I want to prove that the index of $H$ in $G$ is six.

On this problem I have guessed that the cosets are of the form $(3m)(2n), (3m)(2n), (3m+1)(2n), (3m+1)(2n+1), (3m+2)(2n), (3m+2)(2n+1)$

And that they act like $\mathbb Z_3 \times\mathbb Z_2$ and its cosets?

How can I proceed in proving that the index is indeed six?

$\endgroup$
4
  • $\begingroup$ Can you find a homomorphism from $G$ to $\mathbb Z_3 \times \mathbb Z_2$? $\endgroup$
    – user169852
    Aug 28, 2018 at 21:04
  • 1
    $\begingroup$ Note proper MathJax usage, as in my edits to this question. In particular, notice the difference between$\quad$ $Z_2$X$Z_2$ and $Z_2\times Z_2.$ I took that one step further and made it $\mathbb Z_2\times\mathbb Z_2.$ And in the expression $$ G=\{ a+b\sqrt 2\mid a,b \in \mathbb Z\} $$ the whole thing is between one pair of dollar signs or double dollar signs. Rather than $\sqrt(pqrst)$ one writes $\sqrt{pqrst}.$ And note the use of \mid. $\qquad$ $\endgroup$ Aug 28, 2018 at 21:04
  • $\begingroup$ f(3m+x,2n+y)=(xmod3,ymod2)? $\endgroup$ Aug 28, 2018 at 21:06
  • $\begingroup$ You can forget all the $\sqrt{2}$ stuff, and just write $G=\mathbb{Z}\times\mathbb{Z}$, generated by $a$ and $b$, and $H$ is the generated by $3a$ and $2b$. Does this help? $\endgroup$
    – Steve D
    Aug 28, 2018 at 21:17

1 Answer 1

1
$\begingroup$

Define a surjective homomorphism $h:G\to \mathbb Z_3×\mathbb Z_2$ by $h(a+b\sqrt2)=(a\pmod3,b\pmod2)$.

Then $H=\operatorname{ker} h$.

By the first isomorphism theorem, $G/H\cong\mathbb Z_3×\mathbb Z_2$.

Thus $[G:H]=\mid G/H\mid=\mid\mathbb Z_3×\mathbb Z_2\mid=6$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .