# Group algebra for quaternion group

I'm trying to understand Hopf Galois Theory, and I decided to try studying some example of a non commutative ring extension. The papers I've studied tell me that, for a strongly $$G$$-graded algebra $$A$$ over a fixed commutative ring $$K$$, the corresponding Hopf algebra $$H$$ (that will verify that $$A$$ is a right $$H$$-comodule algebra) is the group algebra $$K[G]$$.

So I tried to put the theory into an example where $$G$$ is the quaternion group $$\{\pm1,\pm i,\pm j,\pm k\}$$.

My question: It's about how are these $$A$$ and $$H$$. $$A$$ is said to be a strongly $$G$$-graded algebra over $$K$$ ($$K$$ being a field). That means $$A=\bigoplus_{g\in G}A_g$$, where each $$A_g$$ is a $$K$$-subspace of $$A$$; and that $$A_gA_h\subseteq A_{gh}$$, for all $$g,h\in G$$. I guess $$A$$ can be expressed like this: $$A = A_1\oplus A_i\oplus A_j \oplus A_k = K\oplus Ki\oplus Kj \oplus Kk \ \cong \ Ke_1\oplus Ke_i\oplus Ke_j\oplus Ke_k,$$ where $$e_1=(1,0,0,0), e_i=(0,1,0,0), e_j=(0,0,1,0), e_k=(0,0,0,1)$$ (I prefer these vector space notation). I think I need just $$4$$ subalgebras instead of $$8$$ because $$-g=-1_K\cdot g$$. On the other hand, I believe $$H=K[G]$$ can be represented the exact same way. Is this correct?

I think I need just $$4$$ subalgebras instead of $$8$$ because $$-g=-1_K\cdot g$$. On the other hand, I believe $$H=K[G]$$ can be represented the exact same way. Is this correct?

No: you have to be careful not to confuse the "-" in the quaternion group with the "-" in $$K$$.

Explicitly, $$(1_K)(-g)$$ and $$(-1_K)(g)$$ are distinct elements of $$K[G]$$, and furthermore $$(1_K)(g)$$ and $$(-1_K)(-g)$$ are distinct from each other and from the last two.

$$K[G]$$ is necessarily a free $$K$$-module of rank $$|G|$$, one basis element per group element.

The same would be said of $$A=\bigoplus_{g\in G}A_g$$: the fact that $$g$$ and $$-g$$ "look" like scalar multiples of each other is just a red herring. They are distinct, and produce separate coordinates in the sum. This will be of rank $$|G|$$ too. The "-" in front of $$g$$ is just a notation from the group $$G$$, and not from the additive group of the module you're making.

• Ok, so that applies to $K[G]$. Is my $A$ correct? Thank you so much for answering Apr 13, 2022 at 15:16
• @Cafeinicola Sorry, I missed that you wrote $H=K[G]$. I thought you meant $A=K[G]$. I do not know what you mean by $A$ in this case... Apr 13, 2022 at 15:17
• $A$ is a strongly $G$-graded algebra over $K$, not the same as the group algebra in general. Apr 13, 2022 at 15:20
• I thought you chose $A$ and computed $H$. Without knowing anything about $A$, how do you know $H=K[G]$? Apr 13, 2022 at 15:24
• Because it's known that, given a strongly $G$-graded algebra over $K$, it always has right $K[G]$-comodule algebra structure (all the papers I've checked state the same) and I did in fact prove that $A$ has that structure (the wrong $H$ I used made no difference in the properties it had to verify). What happends is that I thought $A/K$ would be a right $H$-Galois extension, but considering $H$ is not as I thought then it's just a right $H$-extension. Apr 13, 2022 at 15:35