Let $A$ be a subgroup of $\mathbb{Z}^n$ of index $3$. Prove that there exists a basis $v_1,...,v_n$ of $\mathbb{Z}^n$ such that $A$ is generated by $v_1,...,v_{n-1},3v_n$.

My attempt: Let $B_1 = \{v_1,...,v_n\}$ and $B_2 = \{v_1,...,v_{n-1},3v_n\}$. $B_1,B_2$ are $n\times n$ matrices. I think I should somehow form a matrix using $B_1$ and $B_2$ and expect its Smith normal form to be $\begin{pmatrix} I_{n-1} & 0 \\ 0 & 0 \\ \end{pmatrix}$ because $v_1,...,v_{n-1}$ are the invariants going from one basis to another? Also, I'm not sure where I would use index of $A$ equals $3$. Thanks and appreciate a hint!


First I'd like to point out something very confusing you have done. You are meant to prove the existence of a basis $v_1,\ldots,v_n$ with $A=\langle v_1,\ldots,v_{n-1},3v_n\rangle$, but you have started with a (presumably arbitrary) basis $v_1,\ldots,v_n$ and seem to expect $A=\langle v_1,\ldots,v_{n-1},3v_n\rangle$ which clearly won't be true in general. I'm probably mistaken about your approach, but in any case it is not clear.

So we want to construct this basis. Despite my comments above I will start with an arbitrary basis $u_1,\ldots,u_n$ of $\mathbb{Z}^n$. Reordering if necessary, we may assume $u_n\notin A$.

Step 1: Show that $3u_n\in A$ (you may be able to assume this). Let $v_n=u_n$.

Step 2: For $i=1,\ldots,n-1$, if $u_i\in A$ then great, let $v_i=u_i$. If $u_i\notin A$ find $v_i\in A$ such that $v_1,\ldots,v_i,u_{i+1}\ldots,u_n$ is a basis.

Step 3: Explain that the resulting $v_1,\ldots,v_n$ is an appropriate basis.

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  • $\begingroup$ Thanks. May I know where we would use the index of $A = 3$, presumably in Step 1 but am not sure how. Also, I did this thing in my approach too, but how do we know apriori that rank of $A$ is $n$? $\endgroup$ – manifolded Mar 19 '19 at 18:53
  • $\begingroup$ You do use $[G:A]=3$ in step 1. Consider the image of $3u_n$ in $G/A$. I don't use anywhere that the rank of $A$ is $n$. $\endgroup$ – Robert Chamberlain Mar 19 '19 at 21:23
  • $\begingroup$ Sorry, not clear yet. Let $\phi : G\rightarrow G/A$ s.t. $\phi(g) = gA$, $3u_n\in G \implies \phi(3u_n)=3u_n A$ and for $3u_n\in A$ to hold true $\phi(3u_n) = A$ because $[G:A] = 3$? Thanks. $\endgroup$ – manifolded Mar 19 '19 at 22:37
  • $\begingroup$ Because we are using additive notation, $\phi(g)=g+A$. So $\phi(3u_n)=3\phi(u_n)$. Can you see that whatever $\phi(u_n)$ is, we must have $3\phi(u_n)=A$? $\endgroup$ – Robert Chamberlain Mar 19 '19 at 22:46
  • $\begingroup$ Got it, thanks. $\endgroup$ – manifolded Mar 19 '19 at 22:52

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