# Finding Identity Element for given operation

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 seems correct. – idok Apr 22 '18 at 12:51
• @idok thanks.. :) – viru Apr 22 '18 at 13:34