Does representation irreducibility ensure non-zero determinant?

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If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



Thank you very much!



Cheers,
Collin



P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!






matrices group-theory finite-groups determinant







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    up vote
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    down vote

    favorite












    If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



    Thank you very much!



    Cheers,
    Collin



    P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!






    matrices group-theory finite-groups determinant







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



      Thank you very much!



      Cheers,
      Collin



      P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!






      matrices group-theory finite-groups determinant







      share|cite|improve this question













      If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



      Thank you very much!



      Cheers,
      Collin



      P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!






      matrices group-theory finite-groups determinant




      matrices group-theory finite-groups determinant






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      share|cite|improve this question











      share|cite|improve this question




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      asked 2 hours ago









      Collin Ren

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          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.






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          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago










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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









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          oldest

          votes






          active

          oldest

          votes








          up vote
          5
          down vote



          accepted










          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.






          share|cite|improve this answer




















          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago














          up vote
          5
          down vote



          accepted










          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.






          share|cite|improve this answer




















          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago












          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.






          share|cite|improve this answer












          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Daniel Mroz

          1,364414




          1,364414











          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago
















          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago















          Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
          – Collin Ren
          1 hour ago




          Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
          – Collin Ren
          1 hour ago

















           

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