If I want to simplify the group presentation (free abelian group with relations)

$$\langle a,b,c\mid 2a=b=2c\rangle,$$

I can simplify it as $$\langle a,c\mid 2a=2c\rangle\cong\langle a,a-c\mid 2(a-c)\rangle\cong\mathbb{Z}\oplus\mathbb{Z}_2.$$

How about for more complicated presentations such as $$\langle a,b,c\mid 2a=3b=5c\rangle?$$

How do I simplify it, if possible?

In general, I am interested in simplifying $$\langle a,b,c\mid n_1a=n_2b=n_3c\rangle,$$ where $n_1,n_2,n_3\in\mathbb{Z}$.


  • 3
    $\begingroup$ The general method for this problem is the Smith Normal Form diagonlization algorithm for matrices over the integers. $\endgroup$ – Derek Holt Feb 5 '18 at 9:46

Here is an outline so that this question has an answer. As Derek Holt said in the comments, the general method involves the Smith Normal Form.

In your first example we have $2a-b=0$ and $2c-b=0$. Thus the matrix of relations is $$\begin{bmatrix}2&-1&0\\0&-1&2\end{bmatrix} $$ we compute the Smith Normal Form of this (computations omitted - I just used a computer) to obtain $$\begin{bmatrix}1&0&0\\0&2&0\end{bmatrix} $$.

Thus the group is isomorphic to $$\mathbb{Z}_{1} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}$$ which is the same as $$\mathbb{Z}_2 \oplus \mathbb{Z}$$.

For an explanation of what is going on, see the answers here, but basically we have to add a copy of $\mathbb{Z}$ since we get only two cyclic factors. Also note that $\mathbb{Z}_{1}$ is just the trivial group.

For the second example you posed you would look at $2a-3b=0$ and $5c-3b=0$, so find the Smith Normal Form of $$\begin{bmatrix}2&-3&0\\0&-3&5\end{bmatrix} $$.

For your final more general example, you would look at $$\begin{bmatrix}n_{1}&-n_{2}&0\\0&-n_{2}&n_{3}\end{bmatrix} .$$


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