We know that every abelian group $H$ has a presentation of the form $\langle S |R\rangle$, where $S$ are the generators and $R$ are the relations.
Intuitively there should be some connection between $T$- the torsion subgroup of $H$ and $R$. Is it true that $R$ generates $T$? If not, is there any well known relationship between them?
Assuming that $ S,R $ are both finite (i.e. $H$ is finitely presented), and that the set of relations $R$ does not contain any "redundancy" (see the remark below), there is a very interesting relationship between $T$ and $R$. I know one way of seeing this that arises in homological algebra. It may not be the simplest way (actually it is quite sophisticated), but its generalizations have several deep applications (e.g. in the universal coefficient theorem in algebraic topology). I will try to make my answer as self-contained as possible.
Any finite presentation of the abelian group $H$ gives rise to a free resolution of $H$, i.e. a sequence of abelian groups with homomorphisms of the following form: $$ 0 \rightarrow F_2 \xrightarrow{\phi_2} F_1 \xrightarrow{\phi_1} H \rightarrow 0 $$ where:
Remark: for this construction to work and satisfy these properties, we need the relations $R$ to be "non-redundant" / linearly independent, in the following sense: denoting the set of generators by $ S = \left\{ a_1, \ldots, a_m \right\} $, and writing each relation $ r_i $ as a linear combination $ r_i = \sum_j n_{ij} a_j $ (with integer coefficients $ n_{ij} $) that vanishes in $H$, we require that there does not exist any non-trivial linear combination of the $ r_i $ that vanishes in $ F_1 $. That is, for any set of integer coefficients $ \left\{ m_i \right\} $, we have that $ \sum_i m_i \sum_j n_{ij} a_j = 0 $ in $ F_1 $ (i.e. $ \sum_i m_i n_{ij} = 0 $ for all $j$) implies that $ m_i = 0 $ for all $i$. Note this is simply a condition on the integer matrix $ n_{ij} $; and even if this condition is not fulfilled, we can use the Hermite normal form of this matrix to find a new set of relations $R'$ that would fulfil the condition.
The properties 1-5 above imply that the chain of groups and homomorphisms is a short exact sequence, meaning that for every group $A$ in the sequence, the kernel of the homomorphism from $A$ equals the image of the homomorphism to $A$ (this is what "exactness" means; the "short" part says that there are five groups in the sequence, with the first and last being the zero group).
Now, we dualize this short exact sequence. This means we apply the functor $ \mathrm{Hom} \left( -, \mathbb{Z} \right) $ to the groups and homomorphisms. I.e., we replace each abelian group $A$ by the abelian group $ A^* := \mathrm{Hom} \left( A, \mathbb{Z} \right) $ of homomorphisms from $A$ to the integers; and we replace each homomorphism $\phi : A \rightarrow B $ from the original sequence by its dual, defined by precomposing with $ \phi$: $$ \phi^* : B^* \rightarrow A^* $$ $$ \phi^* : \sigma \mapsto \sigma \circ \phi $$ This way we obtain a dual sequence: $$ 0 \leftarrow {F_2}^* \xleftarrow{\phi_2^*} {F_1}^* \xleftarrow{\phi_1^*} H^* \leftarrow 0 $$ However, it turns out that this dualized sequence fails to be exact at $ {F_2}^* $ (and only there). This means that $ \phi_2^* $ is not surjective; and the cokernel of $ \phi_2^* $ turns out to be isomorphic to the torsion subgroup of $H$: $$ \mathrm{coker} ( \phi_2^* ) = {F_2}^* / \mathrm{Im} ( \phi_2^* ) \cong T $$ To see this, use the structure theorem for abelian groups to write: $$ H \cong \mathbb{Z}^r \oplus T , \quad T = \left( \mathbb{Z} / d_1 \mathbb{Z} \right) \oplus \ldots \oplus \left( \mathbb{Z} / d_n \mathbb{Z} \right) $$ then $ F_1 \cong \mathbb{Z}^{r+n} $ and $ F_2 \cong \mathbb{Z}^n $; and $ \phi_2 $ sends the generator of the $i$th component of $F_2$ to $ d_i $ in the $(r+i)$-th component of $ F_1 $. Now, since $ F_1, F_2 $ are free, it is not hard not see that they are naturally isomorphic with their duals. Moreover, the dual map $ \phi_2^* : \mathbb{Z}^{r+n} \rightarrow \mathbb{Z}^n $ acts trivially (vanishes) on the first $r$ components and maps the generator of the $(r+i)$-th to $ d_i $ in the $i$th component. Thus, we have: $$ \mathrm{Im} (\phi_2^*) = d_1 \mathbb{Z} \oplus \ldots \oplus d_n \mathbb{Z} $$ implying that: $$ \mathrm{coker} ( \phi_2^* ) = \mathbb{Z}^n / \left( d_1 \mathbb{Z} \oplus \ldots \oplus d_n \mathbb{Z} \right) \cong T $$ as required.