# Thread: Calculating Rank and Torsion Coefficients

1. ## Calculating Rank and Torsion Coefficients

How do i calculate the rank and torsion coefficients of the following:

B=Z3 x Z14 x Z24 x Z126 x Z
C = <a,b,c,d: 168b=0, 42c=0, 18d=0>

Am i right in thinking that B can be decomposed to:

Z3=Z3
Z14=Z2 x Z7
Z24 = Z3 x Z8
Z126 = Z2 x Z9 x Z7

Of so how, do I combine to get the torsion and rank coefficient?

Also, for C, am I right in thinking that this is Z0 x Z168 x Z42 x Z18

If, so can someone help me decompose this and recombine so i can calculate rank and torsion

2. Originally Posted by moolimanj
How do i calculate the rank and torsion coefficients of the following:

B=Z3 x Z14 x Z24 x Z126 x Z
C = <a,b,c,d: 168b=0, 42c=0, 18d=0>

Am i right in thinking that B can be decomposed to:

Z3=Z3
Z14=Z2 x Z7
Z24 = Z3 x Z8
Z126 = Z2 x Z9 x Z7

Of so how, do I combine to get the torsion and rank coefficient?
If you have the group $G=\mathbb{Z}_{q_1}\times ... \times \mathbb{Z}_{q_k} \times \mathbb{Z}^n$ then the torsion subgroup is $\mathbb{Z}_{q_1}\times ... \times \mathbb{Z}_{q_k}$. Where the $q_i$ str prime powers, not necessarily distinct. If we form the torsion subgroup $T\times \{ 0\}^n$ and we mod it out, i.e. compute $G/T \simeq \mathbb{Z}^n$. Thus, $G/T$ is a free abelian group. The "rank" is the # of elements in a basis for $G$. Since all basis has the same number of elements we see that $\{ (1,0,...,0),(0,1,...,0),...(0,0,...,1)\}$. And so the rank is $n$.