Given the presentation of a finitely generated Abelian group, is there a straight forward way to write it as a direct sum of cyclic groups?
The particular presentation I am looking at is
$[a,b,c,d | a^3b^6c^{12}d^9, a^2b^6c^{10}d^8, a^2b^4c^8d^6]$
Given the presentation of a finitely generated Abelian group, is there a straight forward way to write it as a direct sum of cyclic groups?
The particular presentation I am looking at is
$[a,b,c,d | a^3b^6c^{12}d^9, a^2b^6c^{10}d^8, a^2b^4c^8d^6]$
$a = a \cdot a^3 b^6 c^{12}d^9 \cdot a^{-2} b^{-6} c^{-10} d^{-8} \cdot a^{-2} b^{-4} c^{-8} d^{-6} = b^{-4} c^{-6} d^{-5}$ so we don't need the generator $a$, let's remove it to get the presentation $[b,c,d|b^6c^6d^6,b^2c^2d^2,b^4c^4d^4]$ only the middle relation is needed, let us change the basis to $[bcd,c,d|(bcd)^2]$ so we find the group is $C_2 \oplus \mathbb Z \oplus \mathbb Z$.
more generally, if you write the relations as a matrix you can compute the Smith Normal Form of that matrix
? matsnf(mattranspose([3,6,12,9; 2,6,10,8; 2,4,8,6; 0,0,0,0])) %1 = [0, 0, 2, 1]
which tells us that the group is isomorphic to Z/(0) x Z/(0) x Z/(2) x Z/(1) = Z x Z x C_2
.
The smith normal form comes from diagonalizing the relation matrix using a change of basis matrix, this is what we were doing in the calculation when we went from b,c,d to bcd,c,d.