Let R ∗ be group of non-zero real numbers where the operation is multiplication. The group is 2x2 matrices with non-zero determinant.Let f : G → R ∗ be a function given by f ( A ) = d e t ( A ).My proof to show f is homomorphorism is:We must show f ( a b ) = f ( a ) f ( b ). Here we can seeif A = ( a b c d ) and B = ( e f g h ) Then f ( A B ) = f ( C ) where C = A B = ( a e + b g a f + b h c e + d g c f + d h ) Note det ( C ) = f ( C ) = ( a e + b g ) ( c f + d h ) − ( a f + b h ) ( c e + d g ) = ( a e c f + b g c f ) + ( a e d h + b g d h ) − ( a f c e + b h c e ) − ( a f d g + b h d g ) = b g c f + a e d h − b h c e − a f d gAlso note that f ( A ) = a d − b c and f ( B ) = e h − f gthen f ( A ) f ( B ) = ( a d − b c ) ( e h − f g ) = a d e h − b c e h + a d f g − b c f gThus f ( A B ) = f ( A ) f ( B ) and as such f is a homomorphism.Now since we showed f is surjective, we need to show f is injective to prove f is a bijective mapping and thus isomorphic.One aspect of 2 × 2 determinant we can use is that it produce the area of parallelogram. If we think of the properties of rotation matrices, we know their determinants all produce 1. Thus f is not injective and bijective. Meaning f is not isomorphic.Is this the correct way to go about this proof or am I making wild mis-understandings?

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Let             R        ∗   be group of non-zero real numbers where the operation is multiplication. The group is 2x2 matrices with non-zero determinant.Let   f  :  G  →            R        ∗   be a function given by   f  (  A  )  =  d  e  t  (  A  ).My proof to show f is homomorphorism is:We must show   f  (  a  b  )  =  f  (  a  )  f  (  b  ). Here we can seeif   A  =      (                            a                          b                                      c                          d                      )   and   B  =      (                            e                          f                                      g                          h                      )  Then   f  (  A  B  )  =  f  (  C  ) where   C  =  A  B  =      (                            a          e          +          b          g                          a          f          +          b          h                                      c          e          +          d          g                          c          f          +          d          h                      )  Note   det  (  C  )  =  f  (  C  )  =  (  a  e  +  b  g  )  (  c  f  +  d  h  )  −  (  a  f  +  b  h  )  (  c  e  +  d  g  )  =  (  a  e  c  f  +  b  g  c  f  )  +  (  a  e  d  h  +  b  g  d  h  )  −  (  a  f  c  e  +  b  h  c  e  )  −  (  a  f  d  g  +  b  h  d  g  )  =  b  g  c  f  +  a  e  d  h  −  b  h  c  e  −  a  f  d  gAlso note that   f  (  A  )  =  a  d  −  b  c and   f  (  B  )  =  e  h  −  f  gthen   f  (  A  )  f  (  B  )  =  (  a  d  −  b  c  )  (  e  h  −  f  g  )  =  a  d  e  h  −  b  c  e  h  +  a  d  f  g  −  b  c  f  gThus   f  (  A  B  )  =  f  (  A  )  f  (  B  ) and as such   f is a homomorphism.Now since we showed   f is surjective, we need to show   f is injective to prove   f is a bijective mapping and thus isomorphic.One aspect of   2  ×  2 determinant we can use is that it produce the area of parallelogram. If we think of the properties of rotation matrices, we know their determinants all produce 1. Thus   f is not injective and bijective. Meaning   f is not isomorphic.Is this the correct way to go about this proof or am I making wild mis-understandings?

Mateo Carson · Accepted Answer

f is surjection, but not injection. It can be prooved very easily, giving a specific example.  ∀  x  ∈            R        ∗    ,  f      (                            x                          0                                      0                          1                      )    =  x  .  f  (  I  )  =  f  (  −  I  )The first example shows   f is surjective, and the second shows   f is not injective.Also, your approach to injectivity is right, but your approach to surjectivity does not show that   f is surjection.

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