I'm beginner in Group Theory and learning it by following 'Pinter algebra'
I just want to verify if my thought process is correct in following exercise
$(a, b)\times(c, d) = (a \cdot c, b \cdot c + d)$, on the set R * R does this form group ? If not, Give reason why it doesn't form a Group
ans: NO
reason: some elements don't have inverse element and not because it doesn't have identity element
My question is does above one have identity element w.r.t to $\times$?
Checking for Right Identity
$(a,b) \times (x,y)=( a\ \cdot x, b \cdot x+y )=(a,b)$
$a \cdot x = a \rightarrow a(1-x)=0: a=0$ or $x=1$
but for this to be true for all '$a$', so $x=1$
$b \cdot x+y =b$
substituting $x=1$
$b+y=b \rightarrow y=0$
Right Identity is $(x,y)=(1,0)$
Checking for Left Identity
$(x,y) \times (a,b)=(x \cdot a, y \cdot a+b)=(a,b)$
$x \cdot a=a \rightarrow a(1-x)=0 \rightarrow a=0$ or $x=1$
but for this to be true for all 'a' $x=1$
$y \cdot a+b=b \rightarrow y \cdot a=0 \rightarrow y=0$ or $a=0$
but for this to be true all '$a$', so $y=0$
Left Identity is $(x,y)=(1,0)$
Therefore it has $(1,0)$ as identity element
Yes, this is correct.
For similar phenomena, consider studying semigroup theory, where there can exist an identity but no inverses.