# Does taking quotients commute with taking the group algebra?

I ll preface this by excusing myself because I am not quite sure how to ask my question. Basically I was messing around with group algebras $$\mathbb{K}[G]$$, I found it interesting that sub algebras correspond to sub groups, and morphisms between group algebras correspond to group homomorphisms (for finite groups at least).

And then I was lead to wondering if $$\mathbb{K}[G/H]\simeq\mathbb{K}[G]/\mathbb{K}[H]$$.

I tried proving this "directly" but couldn't figure out how to take quotients of algebras well enough. And then I tried looking up if functors (in this case the functor which sends a group to its algebra) preserve quotients, but my category theory isn't good enough for me to figure it out alone.

So my questions are:

Is $$\mathbb{K}[G/H]\simeq\mathbb{K}[G]/\mathbb{K}[H]$$ even true?

If not, is there another "similar" results which holds?

And if it does hold, how do you prove it?

Thanks enough, hopefully my formulation isn't too bad

• Is $\mathbb{K}[H]$ even an ideal in $\mathbb{K}[G]$? Commented Sep 1, 2022 at 12:35
• Sub-algebras do not correspond to subgroups. There are subalgebras that are not a group algebra for any group. Commented Sep 1, 2022 at 13:02

So I guess the first observation is that the quotient you wrote down on the algebra side isn't doing the right thing. When we take a group quotient $$G/H$$ we are setting $$h = 1$$ for all $$h \in H$$, whereas when you take the quotient $$\mathbb{K}[G] / \mathbb{K}[H]$$ you are setting $$h = 0$$ for all $$h \in H$$.
If we want to set $$h =1$$ for all $$H$$ on the algebra side the way to do it is to quotient by the 2-sided ideal $$I$$ generated by all expressions of the form $$h-1$$ for $$h \in H$$. If we do that then indeed $$\mathbb{K}[G]/I \cong \mathbb{K}[G/H]$$ provided $$H$$ is normal, which I'll leave as an exercise.
$$\mathbb{K}[H]$$ contains $$1$$, so unless $$H=G$$, $$\mathbb{K}[H]$$ is not an ideal of $$\mathbb{K}[G]$$ and the quotient algebra $$\mathbb{K}[G]/\mathbb{K}[H]$$ is not defined. We can take the quotient vector space, but then the dimension of $$\mathbb{K}[G]/\mathbb{K}[H]$$ is $$|G|-|H|$$, while the dimension of $$\mathbb{K}[G/H]$$ would be $$|G|/|H|$$.