Haskell class constraints for group definition

Question

I came up with a typeclass definition for Group , but I found a counterexample type that is not really a group.

Here is a class definition and an instance that is a group ℤ₂:

class Group g where
  iden :: g
  op :: g -> g -> g
  inv :: g -> g

data Z2T = Z0 | Z1

instance Group Z2T where
  iden = Z0
  Z0 `op` Z0 = Z0
  Z0 `op` Z1 = Z1
  Z1 `op` Z0 = Z1
  Z1 `op` Z1 = Z0
  inv Z0 = Z1
  inv Z1 = Z0

However, these type signatures for are Groupnecessary but not sufficient for the type to really be a group, here is my counterexample that compiles:

data NotAGroup = N0 | N1

instance Group NotAGroup where
  iden = N0
  N0 `op` N0 = N0
  N1 `op` N0 = N0
  N0 `op` N1 = N0
  N1 `op` N1 = N0
  inv N0 = N0
  inv N1 = N0

How can I code enough rules for a type that should be a group in a class Groupin Haskell?

Answer 1

, . Monoid, , , Monoid:

-- | The class of monoids (types with an associative binary operation that
-- has an identity).  Instances should satisfy the following laws:
--
--  * @mappend mempty x = x@
--
--  * @mappend x mempty = x@
--
--  * @mappend x (mappend y z) = mappend (mappend x y) z@
--
--  * @mconcat = 'foldr' mappend mempty@

- . Group Monoid; , .

class Monoid g => Group g where
  ginverse :: g -> g
  -- ginv must obey the following laws
  -- x `gappend` (ginverse x) == gempty
  -- (ginverse x) `gappend` x == gempty

-- g-prefixed synonym for mappend
gappend :: Group g => g -> g -> g
gappend = mappend

-- g-prefixed synonym for mempty
gempty :: Group g => g
gempty = mempty

Answer 2

, Group, , .

, , Monad, . Monad, Group.

, Haskell, , . , , , , Haskell .

, , , , .