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We looked briefly at this example in class but I'm not quite sure how to proceed, and I can't find examples of this in any textbooks I have (Dummit & Foote and Nicholson).

Suppose we have $H = \langle(1,1) , (1,-1)\rangle \le G = \mathbb{Z}^2$ for groups $H$ and $G$. Find $|G:H|$.

I think I'd have to take the standard basis for $G$ and then express the elements in $H$ as some combination of this basis. The goal (from what I understood, at least) seems to be to express $G$ and $H$ as direct products and then look at the order of the quotient (since that's equal to $|G:H|$).

Am I on the right track here? How would I go about actually showing all the work for this question? Thanks for reading.

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    $\begingroup$ The fastest way is probably to compute the (absolute value of) the determinant of $\begin{pmatrix} 1&1\\1&-1\end{pmatrix}$; the index is thus $2$. $\endgroup$ – user8268 Feb 9 '12 at 22:26
  • $\begingroup$ Wow, that's slick. I haven't seen that idea before but I'll check it out. I assume it generalizes to groups G of rank > 2? (It wouldn't be much good otherwise, now would it? Haha :) $\endgroup$ – user20682 Feb 9 '12 at 22:33
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    $\begingroup$ Note: You are computing the index of a subgroup of a free abelian group. Computing indices of subgroups of free abelian groups is a very different task than computing indices of subgroups of free groups. $\endgroup$ – Arturo Magidin Feb 9 '12 at 22:49
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    $\begingroup$ Given any subgroup of $\mathbb{Z}^n$ given by a set of generators, you can compute the Smith Normal Form of the matrix whose columns are the generators. This will give you a fairly straightforward way of computing the index. One way to see user8268's method is that the Smith Normal form of that matrix has a $1$ in the upper left corner, and $\det(A)$ in the bottom right, and the subgroup generated by $(1,0)$ and $(x,y)$ has index $|y|$. $\endgroup$ – Arturo Magidin Feb 9 '12 at 22:51
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Let's restrict to the case where we have a subgroup of $\mathbb{Z}^n$ generated by at least $n$ given vectors (if it is generated by fewer than $n$ vectors, then the index is infinite).

Write the vectors as columns of an $n\times m$ matrix, $n\leq m$. Then you can compute the Smith normal form of the matrix, which amounts to finding an automorphism of $\mathbb{Z}^n$ and a generating set for $B$ in which the generators of $B$ are scalar multiples of the standard basis vectors of $\mathbb{Z}^n$. In that situation, one can read off the index from the Smith normal form: it will be the product of the (absolute values of the) first $n$ diagonal entries if they are all nonzero, and the index will be infinite if the matrix has more than $m-n$ zeros in the entries.

If $n=m$ (so $B$ is generated by $n$ vectors from $\mathbb{Z}^n$ and the original matrix is a square), then the Smith normal form of the matrix has the same determinant as the original matrix; this determinant is just the product of the diagonal entries in the Smith normal form... which happens to be the index of $B$ in $\mathbb{Z}^n$. So computing the determinant of $B$ will do the trick: if the determinant is $0$, then the index is infinite; if the determinant is nonzero, then the (absolute value of the) determinant is the index of $B$.

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    $\begingroup$ And the Smith Normal Form also immediately gives you the structure of $G/H$ as a direct sum of cyclic groups. $\endgroup$ – Derek Holt Feb 10 '12 at 1:42
  • $\begingroup$ Thanks, Arturo! You've really kindled my interest in the subject. Looking forward to checking this stuff out. $\endgroup$ – user20682 Feb 10 '12 at 12:58
  • $\begingroup$ why does the determinant happen to be the index of B in Z^n? $\endgroup$ – Marco Flores Nov 17 '16 at 1:15

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