Show, using presentations, that $\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{{\rm lcm}(m,n)}\times \Bbb Z_{\gcd(m,n)} .$

Question

Note: This is an alternative-proof question and thus is not a duplicate.

The Question:

Show, using group presentations, that $$\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{{\rm lcm}(m,n)}\times \Bbb Z_{\gcd(m,n)}.$$

Motivation:

I was trying to answer this question in particular . . .

$\Bbb Z_m \times \Bbb Z_n$ isomorphic to $\Bbb Z_{\operatorname{lcm}(m,n)}\times \Bbb Z_{\gcd(m,n)}$

. . . using only Tietze transformations. (Why not, eh?) But I got stuck.

Thoughts:

Following the initial setup in the question, I began with

By this standard result, we have

$$\begin{align} \Bbb Z_m\times \Bbb Z_n &\cong\langle z,w\mid z^m, w^n, zw=wz\rangle\\ &\cong \langle x,y,z,w\mid x=zw, y=z^{m/d}w^{n/d}, z^m, w^n, zw=wz\rangle\\ &\cong\langle x,y,w\mid y=(xw^{-1})^{m/d}w^{n/d}, (xw^{-1})^m, w^n, (xw^{-1})w=w(xw^{-1})\rangle\\ &\cong\langle x,y,w\mid y=(xw^{-1})^{m/d}w^{n/d}, (xw^{-1})^m, w^n, xw=wx\rangle\\ &\cong\langle x,y,w\mid y=x^{m/d}w^{(n-m)/d}, x^m=w^m, w^n, xw=wx, xy=yx, yw=wy\rangle\\ &\cong\langle x,y,w\mid w^{{(n-m)/d}}=x^{-m/d}y, x^m=w^m, w^{n}, xw=wx, xy=yx, yw=wy\rangle\\ &\cong\langle x,y,w\mid w^{n-m}=x^{-m}y^d, w^{{(n-m)/d}}=x^{-m/d}y, x^m=w^m, w^{n}, xw=wx, xy=yx, yw=wy\rangle\\ &\cong\langle x,y,w\mid w^n=y^d, w^{{(n-m)/d}}=x^{-m/d}y, x^m=w^m, w^{n}, xw=wx, xy=yx, yw=wy\rangle\\ &\cong\langle x,y,w\mid w^n=y^d, w^mw^{n/d}=w^{m/d}x^{-m/d}y, x^m=w^m, w^{n}, xw=wx, xy=yx, yw=wy\rangle \end{align}$$

That's about it. How ugly!

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Surely there's a way to manipulate either this answer to the linked question (in the Motivation section) . . .

Fix $u,v\in\Bbb Z$ with $un+vm=d$ (Bezout). The map $$\Bbb Z_{\operatorname{lcm}(n,m)}\times\Bbb Z_{\gcd(n,m)} \to\Bbb Z_m\times\Bbb Z_n$$ $$ (a+\operatorname{lcm}(n,m)\Bbb Z,b+\gcd(n,m)\Bbb Z)\mapsto(ua+\tfrac mdb+m\Bbb Z,va-\tfrac ndb+n\Bbb Z)$$ is well-defined(!) and clearly a group homomorphism. For the element on the left to be in the kernel, $ua+\tfrac mdb$ must be a multiple of $m$ and $va-\tfrac ndb$ a multiple of $n$. But then $$\frac nd\left(ua+\frac mdb\right)+\frac md\left(va-\frac ndb\right) =\frac{nu+vm}{d}a=a$$ is a multiple of $\frac{nm}d=\operatorname{lcm}(n,m)$, i.e., we may as well assume that $a=0$. Then $\frac mdb$ must be a multiple of $m$, i.e., $b$ a multiple of $d$, i.e. $b\equiv 0$. We conclude that the kernel is trivial and our homomorphism injective. As both groups are finite of same order, the homomoprhism must be an isomorphism.

. . . or this answer to a slightly different question:

Write $m=dm', n=dn', d=mu+nv$. Then $l=m'n=mn'$.

These row and columns operations prove that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l$: $$ A=\pmatrix{ m & 0 \\ 0 & n} \to \pmatrix{ m & mu \\ 0 & n} \to \pmatrix{ m & mu+nv \\ 0 & n} = \pmatrix{ m & d \\ 0 & n}\\ \to \pmatrix{ 0 & d \\ -m'n & n} = \pmatrix{ 0 & d \\ -l & n} \to \pmatrix{ 0 & d \\ -l & 0} \to \pmatrix{ d & 0 \\ 0 & l}=B $$

An explicit isomorphism can be written by collecting the row and columns operations into two matrices $P,Q$ so that $B=PAQ$: $$ P = \pmatrix{ 1 & 0 \\ -n' & 1} \pmatrix{ 1 & v \\ 0 & 1} =\pmatrix{1 & v \\ -n' & 1 - v n'} \\ Q = \pmatrix{ 1 & u \\ 0 & 1} \pmatrix{ 1 & 0 \\ -m' & 1} \pmatrix{ 0 & -1 \\ 1 & 0} = \pmatrix{u & -1 + u m' \\ 1 & m'} $$

If $e_1, e_2$ is the canonical basis for $\mathbb Z^2$, then the basis $f_1, f_2$ given by $F=Q^{-1}E$ is such that this diagram commutes: $$ \matrix { \mathbb Z^2 , \{ e_1, e_2\} & \to & \mathbb Z^2, \{ f_1, f_2\} \\ \downarrow & & \downarrow \\ \mathbb{Z}_m\oplus \mathbb{Z}_n & \to & \mathbb{Z}_d\oplus \mathbb{Z}_l } $$

This isomorphism does not use prime factorizations nor explicitly the Chinese Remainder Theorem.

Please help :)

Answer 1

One of the arguments that you linked can in fact be translated into an argument using nothing but Tietze transformations.

For clarity, let's work with a special case, say $n=6, m=8$. I'll say a word about how to pass to the general case at the end, but it should be clear.

We want to pass from $\langle a, b \mid a^n, b^m, [a,b] \rangle$ to $\langle x,y \mid x^{n \land m}, y^{n \lor m}, [x,y] \rangle$. Here I'm writing $\gcd(n,m) = n \land m$ and $\text{lcm}(n,m) = n \lor m$ for conciseness. Let's also write $n \land m = kn + lm$ by the extended euclidean algorithm.

One of your included arguments builds the isomorphism

$$ \begin{align} \langle x,y \mid x^{n \land m}, y^{n \lor m}, [x,y] \rangle &\to \langle a, b \mid a^n, b^m, [a,b] \rangle \\ x &\mapsto a^{\frac{n}{n \land m}} b^{\frac{-m}{n \land m}} \\ y &\mapsto a^k b^l \end{align} $$

Indeed, notice $x^{n \land m}$ gets sent to $a^n b^{-m} = 1$, and $y^{n \lor m}$ gets sent to $a^{\frac{knm}{n \land m}} b^{\frac{lnm}{n \land m}} = \left ( a^n \right )^{\frac{km}{n \land m}} \left ( b^m \right )^{\frac{ln}{n \land m}} = 1$.

In our special case $n=6, m=8$ this amounts to

$$ \begin{align} \langle x,y \mid x^{2}, y^{24}, [x,y] \rangle &\to \langle a, b \mid a^6, b^8, [a,b] \rangle \\ x &\mapsto a^{3} b^{-4} \\ y &\mapsto a^{-1} b^{1} \end{align} $$

where it's now much easier to verify that $x^2$ gets sent to $a^6 b^{-8} = 1$ and $y^{24}$ is sent to $a^{-24} b^{24} = 1$.

Now that we have this isomorphism in hand, there's a standard way of tietze-ifying the argument. We can add new generators, based on the image of this homomorphism, then add in relations saying the new generators work how we expect, next we add relations to get our old generators in terms of our new ones, and lastly we remove the old generators.

Let's see it in action:

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We start with

$$\langle a, b \mid a^6, b^8, [a,b] \rangle.$$

Then we add the new generators to get

$$\langle a, b, x, y \mid a^6, b^8, [a,b], x = a^3 b^{-4}, y = a^{-1} b^1 \rangle.$$

Notice the identities for these new generators came from our isomorphism. Now we can add in the other relations we care about between our new generators

$$\langle a, b, x, y \mid a^6, b^8, [a,b], x = a^3 b^{-4}, y = a^{-1} b^1, x^{2}, y^{24}, [x,y] \rangle$$

write our old generators in terms of the new generators

$$\langle a, b, x, y \mid a^6, b^8, [a,b], x = a^3 b^{-4}, y = a^{-1} b^1, x^{2}, y^{24}, [x,y], a = x^{-1} y^{-4}, b = x^{-1}y^{-3} \rangle$$

and finally kill our old generators (and all relations involving them)

$$\langle x, y \mid x^{2}, y^{24}, [x,y] \rangle.$$

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This maneuver will work to Teitze-ify any isomorphism of groups, provided you can compute the image of each set of generators (note we needed the image of $x$ and $y$ to add them, and the image of $a$ and $b$ to remove them!)

In particular, it will work for our desired isomorphism

$$ \begin{align} \langle x,y \mid x^{n \land m}, y^{n \lor m}, [x,y] \rangle &\to \langle a, b \mid a^n, b^m, [a,b] \rangle \\ x &\mapsto a^{\frac{n}{n \land m}} b^{\frac{-m}{n \land m}} \\ y &\mapsto a^k b^l \end{align} $$

All we have to do is compute $a$ and $b$ in terms of $x$ and $y$ (which in any specific case of $m$ and $n$ is easy to do, and is probably not so hard to do in the abstract, though it looks really annoying) and then play the same game we just played.

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I hope this helps ^_^

Answer 2

The idea of the Smith normal form transformations from the second proof you cited is simply that the isomorphism class of the cokernel of $\begin{bmatrix} a & b \\ c & d \end{bmatrix} : \mathbb{Z}^2 \to \mathbb{Z}^2$ remains invariant under elementary row and column operations. This argument is easy enough to cast in terms of Tietze transformations, where the obvious presentation of the cokernel is $\langle x, y \mid x^a y^c, x^b y^d, xy = yx \rangle$.

The proof for a column operation is very simple; for example, for adding $n$ times the first column to the second column, we have $$\langle x, y \mid x^a y^c, x^b y^d, xy = yx \rangle \simeq \\ \langle x, y \mid x^a y^c, x^b y^d, x^{b+na} y^{d+nc}, xy = yx \rangle \simeq \\ \langle x, y \mid x^a y^c, x^{b+na} y^{d+nc}, xy = yx \rangle. $$

The proof for a row operation takes just a bit more work, because it involves replacing a generator. For example, for adding $n$ times the first row to the second row: $$\langle x, y \mid x^a y^c, x^b y^d, xy = yx \rangle \simeq \\ \langle x, x', y \mid x^a y^c, x^b y^d, xy = yx, x' = x y^{-n} \rangle \simeq \\ \langle x, x', y \mid x^a y^c, x^b y^d, (x')^a y^{c+na}, (x')^b y^{d+nb}, xy = yx, x' y = y x', x' = x y^{-n}, x = x' y^n \rangle \simeq \\ \langle x, x', y \mid (x')^a y^{c+na}, (x')^b y^{d+nb}, x' y = y x', x = x' y^n \rangle \simeq \\ \langle x', y \mid (x')^a y^{c+na}, (x')^b y^{d+nb}, x' y = y x' \rangle. $$

I will leave the proofs for swapping rows or columns, or negating a row or column (i.e. multiplication by the only nontrivial unit of $\mathbb{Z}$), as exercises for the reader.

Now, all you need to do is follow the steps of the matrix manipulation using the lemmas proved above, and get $\langle x, y \mid x^m, y^n, xy = yx \rangle \simeq \langle x, y \mid x^m, x^{mu} y^n, xy = yx \rangle \simeq \cdots \simeq \langle x, y \mid x^d, y^l, xy=yx \rangle$.