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

  • 1
    $\begingroup$ Yes, this seems correct. $\endgroup$ – idok Apr 22 '18 at 12:51
  • $\begingroup$ @idok thanks.. :) $\endgroup$ – viru Apr 22 '18 at 13:34

Yes, this is correct.

For similar phenomena, consider studying semigroup theory, where there can exist an identity but no inverses.

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