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Given a finite group $G$, one can define the group algebra $\mathbb{C}[G]$ as the algebra having the elements of $G$ as a basis, with the multiplication of $G$. Clearly, any group homomorphism induces an algebra homomorphism on the group algebras.

I'm wondering whether one can prove that any algebra homomorphism of two group algebras must always come from a homomorphism of groups.

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Perhaps jstor.org/discover/10.2307/… would help. –  Sanath Devalapurkar Mar 23 at 16:15
    
That's got to do with compact Lie groups, is it applicable to finite groups? –  Turion Mar 23 at 16:17
    
I don't think it should be difficult to give an example of an algebra homomorphism that doesn't come from a group homomorphism. –  Dustan Levenstein Mar 23 at 16:21
    
@Dustan, That's what I suspected, but I'm trying since half an hour. Maybe I'm looking at the wrong groups... –  Turion Mar 23 at 16:23
    
For instance, there's only one automorphism of $G = \mathbb Z/2\mathbb Z$ (the trivial one), but $\mathbb C[G] \simeq \mathbb C \oplus \mathbb C$ has at least one nontrivial automorphism, given by interchanging the summands. –  Dustan Levenstein Mar 23 at 16:23

2 Answers 2

up vote 5 down vote accepted

Let $G = \{e, \sigma\}$ be the group of order $2$. Then $\{e, \sigma\}$ is a basis for $\mathbb C[G]$. The linear map given by $e \mapsto e$ and $\sigma \mapsto -\sigma$ is an automorphism of $\mathbb C[G]$ that does not come from a group homomorphism.

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It is well known that the dihedral group of order $8$ and the quaternions have isomorphic group algebras. Taking an isomorphism and restricting it to the basis of the dihedral group given by the elements of the group does not give you the basis of the quaternions given by the elements of the group because if it did this would imply that the dihedral group and the quaternions were isomorphic as groups. So in this sense it does not "come from a group homomorphism".

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That sounds interesting! You say "it is well known", but when I search for dihedral group, group algebra and quaternions, I can't find anything about this algebra isomorphism. –  Turion Mar 23 at 16:37
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@Turion in general, if the set of irreducible representations have the same dimensions, then the group algebras are isomorphic. This is related to the remark I was making above, and is not a trivial result. –  Dustan Levenstein Mar 23 at 16:39
    
Right, so I could have also just used any two nonisomorphic abelian groups of the same order, for example. Also it should be intuitive that "most" algebra homomorphisms between group algebras are not induced by group homomorphisms. –  Seth Mar 23 at 16:46
    
Ah, so already $\mathbb{C}[\mathbb{Z}_4] \cong \mathbb{C}[\mathbb{Z_2} \times \mathbb{Z}_2]$? I wonder whether these kind of isomorphisms are also equivalences of representation categories. –  Turion Mar 23 at 16:51
    
Yes, for complex group algebras over finite groups their irreducible representations are the same iff the group algebras are isomorphic. This result takes some representation theory and structure theorems to prove. –  Seth Mar 23 at 16:53

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