# Is there a general way to simplify such group presentations (Free Abelian Group with Relations)?

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 $$n_1,n_2,n_3\in\mathbb{Z}$$.

Thanks.

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

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 $$\begin{bmatrix}n_{1}&-n_{2}&0\\0&-n_{2}&n_{3}\end{bmatrix} .$$