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

  • 4
    $\begingroup$ Is $\mathbb{K}[H]$ even an ideal in $\mathbb{K}[G]$? $\endgroup$
    – Randall
    Commented Sep 1, 2022 at 12:35
  • $\begingroup$ Sub-algebras do not correspond to subgroups. There are subalgebras that are not a group algebra for any group. $\endgroup$ Commented Sep 1, 2022 at 13:02

2 Answers 2


As others have pointed out, what you have written isn't quite right, but it can be salvaged to an extent.

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.


I'm afraid the statement is not even false.

$\mathbb{K}[H]$ 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 $|G|/|H|$.


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