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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?

2 Answers 2

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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.

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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.