※ 引述《kyod ( )》之銘言：
> (1)
> Is it true that a semigroup which has a left identity element
> and in which every element has a right inverse is a group ?
> 這題我的想法是想要找一個binary operation 然後有left identity
> and right inverse ，再利用群的性質(left identity = right identity)
> (left inverse = right inverse)
> 製造出矛盾。
符號不好打，只回答 (1):
Let G be the set consisting all real 2 by 2 matrix A
s.t. the leftup entry of A is nonzero and both the leftdown
and rightdown entries of A are 0.
Under usual matrix multiplication, G is a semigroup.
Then any left identity has the form [ 1 k ] and
[ 0 0 ]
for any [ a b ] in G, [ 1/a 0 ] is a right inverse.
[ 0 0 ] [ 0 0 ]
If [ x y ] is a left inverse of [ 1 0 ],
[ 0 0 ] [ 0 0 ]
then [ 1 k ] = [ x y ] [ 1 0 ] = [ x 0 ]
[ 0 0 ] [ 0 0 ] [ 0 0 ] [ 0 0 ]
So k must be 0.
If [ x y ] is a left inverse of [ 1 1 ],
[ 0 0 ] [ 0 0 ]
then [ 1 0 ] = [ x y ] [ 1 1 ] = [ x x ]
[ 0 0 ] [ 0 0 ] [ 0 0 ] [ 0 0 ]
This leads to a contradiction.
Thus G is not a group.

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