Let $G,G_1$ and $G_2$ are three abelian groups with group homomorphisms $\phi_i:G\to G_i$. This gives $k$-algebra homomorphisms $k[\phi_i]:k[G]\to k[G_i]$. So we can consider $k[G_i]'s$ as $k[G]$-module via the homomorphisms $k[\phi_i]$. We can consider the tensor product $k[G_1]\otimes_{k[G]}k[G_2]$ and this will be again $k$-algebras.
My question is there a simpler way to describe the $k$-algebra: $k[G_1]\otimes_{k[G]}k[G_2]$?
For example take $G=\{e\}$, the identity group; then $k[G]=k$ and hence $$k[G_1]\otimes_{k[G]}k[G_2]=k[G_1]\otimes_kk[G_2]\cong k[G_1\times G_2].$$
So I was wondering if there exists any simpler way to express $k[G_1]\otimes_{k[G]}k[G_2]$ like above.
Note that here the groups are abelian and hence the group algebras are commutative rings. Therefore the tensor product makes sense.
Thank you in advance.
Yes, it's just the group algebra of the pushout of $G_1$ and $G_2$ over $G$ (explicitly, the quotient of $G_1\oplus G_2$ by the image of $(\phi_1,-\phi_2):G\to G_1\oplus G_2$). This follows immediately from the fact that the group algebra functor from abelian groups to commutative $k$-algebras is left adjoint to the multiplicative group functor, and thus preserves colimits, so it turns the pushout of $G_1$ and $G_2$ over $G$ into the pushout of $k[G_1]$ and $k[G_2]$ over $k[G]$ which is just the tensor product.