4
$\begingroup$

In this post all groups are finite, and all representations are complex linear finite dimensional representations.

If a group $G$ is abelian, then all of its irreducible representations are of one dimension. The converse is also true: if all of its irreducible representations are of one dimension, then that group is abelian.

What interests me is to what extent can we detect nonabelianity from this information. To be more precise, I'd like to argue that nonabelian groups do not have $1$-dimensional representations other than the trivial one. However, this is a false statement because the permutation group $S_3$ over $3$ elements has a nontrivial yet $1$-dimensional representations.

But that representation comes from induction from its normal subgroup $A_3$! Therefore my guess should be modify so that the group is simple. The simplest nonabelian simple group is $A_5$. I also check $A_6$ and $A_7$ on Groupprop. Neither of them has nontrivial $1$-dimensional representations. So is my guess true:

Any nonabelian simple group has no nontrivial $1$-dimensional representation?

$\endgroup$
3
  • 3
    $\begingroup$ 1D reps of G factor through 1D reps of $G/G'$... $\endgroup$ Mar 13, 2020 at 13:20
  • $\begingroup$ Yeah.. I always forgot this. This makes my question too trivial to deserve an answer :( $\endgroup$
    – Student
    Mar 13, 2020 at 14:46
  • $\begingroup$ I shouldn't extend my question here, but is there a meaning for the minimal dimension ($>1$) among all irreducible representations of a simple group? $\endgroup$
    – Student
    Mar 13, 2020 at 14:53

1 Answer 1

2
$\begingroup$

Hint: kernels of characters are normal subgroups and $G$ is simple if and only if $ker(\chi)=1$ for all non-principle irreducible characters $\chi$ of $G$. Otherwise stated, a group is simple if and only if all its non-principle characters are faithful.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.