From: mckay@cs.concordia.ca (MCKAY john)
Subject: Re: Question of the concept isomorphism of group theory
Date: 22 Dec 1999 11:13:51 GMT
Newsgroups: sci.math
Keywords: Brauer pair (groups indistinguishable by character theory)
In article <83q5n9$9lk$1@nnrp1.deja.com> jrbao@my-deja.com writes:
>I have this basic and not too difficult question, but I haven't very
>clear idea. I hope someone can give me an explaination.
>
>We know, if two group A and B are isomorphic, they have the same
>character table. My question is: How about the inverse case? i.e. if two
>groups have the same character table, are they isomorphic?
>
NO - it is not. Even the stronger property that the char tables AND the
power maps (induced on classes by the maps g -> g^k, g in G) can
correspond and yet the groups are not pairwise isomorphic.
Such groups are called a Brauer pair.
There is a Brauer pair of order 2^8. They are subgroups of a common
index 2 supergroup. They have a common index 2 subgroup.
They have a faithful permutation representation of degree 32.
I do not know whether a minimal such pair is necessarily a p-group.
I believe the smallest order pair is of order 2^8.
John McKay
>I have this question because of two groups of order 8: Quaternion group
>Q8 and Dihedral group D8 (sometimes writing as D4). As I know they are
>nonisomorphic (is that right?), and group with order 8 has 5
>nonisomorphic groups, Q8 and D8 are the 2 cases of non-abelian groups.
>But they have the same character table, are they isomorphic?? Q8 is one
>important group because of hamiltonian groups, which are non-abelian but
>have only normal subgroups. of course, Q8 has only normal subgroups. But
>it seems D8 has only normal subgroups too. Therefore, all hamiltonian
>groups are consist of Q8*A*B is not right ?!? Here A, B are two special
>abelian groups.
>
>I hope someone can give me some explainations of above confusion.
>Thank you.
>Bao
>
>
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