I am starting to study group algebras and I am stuck on the following problem. The first part is easy, but I copy it in case it helps to prove the second part. This exercise is taken from Representations of groups by Lux and Pahlings.

Suppose $K$ is a field with char $\neq 2$ containing a primitive $4$th root of unity $i$, and let $\langle g \rangle = C_4$. Put $$a = \frac{1+i}{2}g+\frac{1-i}{2}g^3 \in KC_4$$ and $$b = \frac{1-i}{2}g+\frac{1+i}{2}g^3 \in KC_4.$$

(a) Show that $\{1,g^2,a, b\} \subseteq KC_4$ is a subgroup of the unit group of $KC_4$ isomorphic to $V_4 \cong C_2 \times C_2$. (easy part)

(b) Show that $KC_4 \cong KV_4$ as algebras over $K$.

Here is my problem: I can't find an isomorphism between $KC_4$ and $KV_4$. Actually, I don't understand a priori how they could be isomorphic since $C_4 \ncong V_4$ as groups.

As a natural second question, - since I guess that there should actually exist an isomorphism after all - I wonder: are the hypotheses "char$K\neq 2$ and there exists a primitive 4th root of unity in $K$" necessary in order to obtain part (b)?

I think my problem is partly conceptual: I don't really have any intuition on how to work with group algebras. Could you give me hints or help me to get a clearer view on this topic?


This is one of those cases where you should use part (a) to prove part (b). Note that the four elements you've written down are really a basis of $KC_4$, and let $\phi$ be the isomorphism between the subgroup of the unit group of $KC_4$ and $V_4$. You can define a map from $KC_4$ to $KV_4$ by linear extension of $\phi$, and it will be multiplicative and bijective (by dimensions).

If your intuition tells you that if $kG\cong kH$ implies $G\cong H$, you need to modify your intuition :) Take two abelian groups of the same finite order and $k=$ complex numbers: then both group algebras are isomorphic to $\mathbb{C}^{|G|}$ as algebras. This can be proved with the Wedderburn theorem on the structure of semisimple algebras. There are non-isomorphic groups whose group algebras are isomorphic over every field (there's a paper by Dade called Deux groupes finis distincts ayant la même algèbre de groupe sur tout corps, MR0280610). If $G$ and $H$ are $p$-groups, it is an unsolved problem to determine whether $\mathbb{F}_pG \cong \mathbb{F}_pH$ implies $G \cong H$. This is called the "modular isomorphism problem", a large literature exists.

When the characteristic is two, the group algebras really are different. One is isomorphic to $k[x]/x^4$, the other to $k[x,y]/(x^2, y^2)$. The second of these doesn't have a nilpotent element with third power non-zero while the first does.


The answer by mt covers the important points, but let me expand a bit. Over the complex field (or any algebraically closed field of characteristic zero, or even of characteristic coprime to the group order), the group algebra of a finite group $G$ is particularly easy to understand as an algebra. We have $\mathbb{C}G \cong M_{n_1}(\mathbb{C}) \oplus \ldots \oplus M_{n_k}(\mathbb{C}),$ where the degrees of the complex irreducible characters of $G$ are $n_1,n_2, \ldots,n_k$ (multiplicities included). Hence for finite groups $G$ and $H$, we have $\mathbb{C}G \cong \mathbb{C}H$ if and only if the degrees of the complex irreducible characters of $G$ and $H$ are the same. Now an irreducible complex character of a finite Abelian group has degree $1$ (by Schur's Lemma, for example). Hence it follows that if $G$ and $H$ are finite Abelian groups of the same order, then $\mathbb{C}G \cong \mathbb{C}H$. Another example of groups with equal irreducible complex character degrees occurs when $G = D_{8}$ and $H = Q_{8}$ (respectively, the dihedral group of order $8$ and the quaternion group of order $8$). These each have four irreducible characters of degree $1$, and one irreducible character of degree $2$. Hence we do have $\mathbb{C}G \cong \mathbb{C}H$ in this case too. I have chosen to work over the complex field for ease of exposition, but the theory is much the same over any field of characteristic coprime to the order $n$ of the groups in question which contains (say) a primitive $n$-th root of unity.


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