I'm going over the construction of Tensor Products using modules as in Dummit and Foote. So they have explained Tensor Products as motivate by a desire to "extend" scalar multiplication to a module which another is embedded in. Specifically they take $R$-module $N_R$ and look to embed it in $L_S$ an $S$-module with $R \leq S$.They then take an $S$-module homomorphism $\phi:F[S \times N_R]/H \rightarrow L_S$ where $F[S \times N_R]/H$ turns out to be the Tensor Product itself and $\phi$ turns out to be the unique bi-linear map (which I saw in the universal property for Tensor Products).
I noticed that the construction seems to work for $H$ a subgroup because the relations that define $H$ are expressed as elements of $F[S \times N_R]$ being added to each other. So when you conjugate any $(a,b)$ on $H$ it will certainly be normal as in $(a,b) + H - (a,b) = H$. If this is the case does this imply that any subgroup defined by relations which explicitly use addition in a similar way will define something besides a Tensor Product when modding out $F[S \times N_R]$?
Clarifying the Question: Let $R, S$ be rings such that $R\leq S$ and let $N_R$ be an $R$-module. Let $L_S$ be an $S$-module. Then the Tensor Product is defined by the module homomorphism
$\phi: F[S \times N_R]/H \rightarrow L_S$
where $H$ is the relation$\langle(s,n_1 + n_2) - (s, n_1) + (s, n_2), (s_1 + s_2,n) - (s_1,n) + (s_2,n), (sr, n) - (s, rn)\rangle$
But suppose there was another relation $K$ which also relied totally on addition to define the relation. Would $\psi: F[S \times N_R]/K \rightarrow L_S$ also define a Tensor Product or maybe something different?