Let $G$ be a group and $g,h \in G$ two elements. I want to describe a set of generators for the left ideal $\mathbb{C}[G](1-g) \cap \mathbb{C}[G](1-h)$.
Suppose we have a relation $$g^{n_1}h^{m_1}g^{n_2}h^{m_2}\dots g^{n_r}h^{m_r}=1$$ in $\langle g, h \rangle$ and let $x_k = g^{n_1}h^{m_1}g^{n_2}h^{m_2}\dots g^{n_k} $ as well as $y_k = x_k h^{m_k}$.
We then have $x_k - y_k \in \mathbb{C}[G](1-h)$ for $1 \leq k \leq r $ as well as $x_{k+1} - y_{k}\in \mathbb{C}[G](1-g)$ for $1 \leq k < r $ and $x_1 - y_{r} = g^{n_1} - 1\in \mathbb{C}[G](1-g)$.
It follows that $$\sum_{k=1}^r (x_k - y_k) = (x_1 - y_r) + \sum_{i=1}^{r-1} (x_{k+1}-y_{k}) \in \mathbb{C}[G](1-h) \cap \mathbb{C}[G](1-g)$$
I want to show that elements obtained from relations like these generate $\mathbb{C}[G](1-h) \cap \mathbb{C}[G](1-g)$.
In fact, if $m = \sum_{x} a_x x \in \mathbb{C}[G](1-h) \cap \mathbb{C}[G](1-g)$, then $\sum_{i} a_{xg^i} = 0 = \sum_{i} a_{x h^i} $ for all $x \in G$. Let $\text{supp}(m) \subseteq G$ be the set of $x \in G$ such that $a_x \neq 0$ and let $x \in \text{supp}(m)$. It suffices to show that we can find an element $t$ so that $m -t$ has strictly smaller support than $m$ and $t$ is a $\mathbb{C}[G]$-linear combination of elements obtained from relations as above.
To do so, we can note that since $a_x \neq 0$ and $\sum_{i} a_{xg^{i}} = 0$, there exists $n_1$ such that $g^{n_1} \neq 1$ and $xg^{n_1} \in \text{supp}(m)$ and since $\sum_{i} a_{xg^{n_1}h^i} = 0$ we similarly find that $xg^{n_1}h^{m_1} \in \text{supp}(m)$ for some $m_1$ with $h^{m_1} \neq 1$. Proceeding like this we obtain sequences $(n_k)$ and $(m_k)$ with $xg^{n_1}h^{m_1} \dots g^{n_k} \in \text{supp}(m)$ as well as $xg^{n_1}h^{m_1} \dots g^{n_k}h^{m_k} \in \text{supp}(m)$ and since $\text{supp}(m)$ is finite we find a repetition $xg^{n_1}h^{m_1} \dots g^{n_k} = xg^{n_1}h^{m_1} \dots g^{n_r}$ (or one or both terms ending on $h^{m_k}$ or $h^{m_r}$ respectively). After cancelling we obtain a relation in $\langle g,h \rangle$ and subtracting an adequate multiple of the element obtained from this relation we can reduce the support as desired.
Question: If this is all correct, it seems to me that if $\langle g,h \rangle$ is finitely presented, this might imply (with some extra work) that $ \mathbb{C}[G](1-h) \cap \mathbb{C}[G](1-g)$ is finitely generated in this case. Is this right and what about the converse?