Determine if it is a group or not for each operation $\ast$ defined on a set. Determine if it is a group or not for each operation defined on a set.


*

*Define $\ast$ on $\Bbb Z$ by $a\ast b = |ab|$.

*Define $\ast$ on $\Bbb Z$ by $a\ast b = \max \{a,b\}$.

*Define $\ast$ on $\Bbb Z$ by $a\ast b = a + b +1$.

*Define $\ast$ on $\Bbb Z$ by $a\ast b = ab +b$.

*Define $\ast$ on $\Bbb Z$ by $a\ast b = a + ab +b$.


For 1, $a\ast b = |ab|$.
Identity: $a*e = |a \cdot  e| = a $
There is no such e. Thus, it is not a group.
For 2, $a\ast b = \max \{a,b\}$.
There is no identity element such that $a \ast e = \max \{a, e\} = a$
Thus, it is not a group.
For 3, $a\ast b = a + b +1$.
Identity: $a\ast e = a + e + 1$, with $e = -1$, $a *e = a - 1 +1 = a$
Inverse: There is no $x \in \Bbb Z$ such that $a * x = a +x+1 = e = -1$.
For 4, $a\ast b = ab +b$
$(a *b)*c =  (ab + b) * c = (ab + b)c +c = abc +bc +c$
$a*(b*c) = a *(bc +c) = a(bc +c) + bc + c = abc + ac +bc+c$
Thus, $(a *b)*c \neq a*(b*c)$
Therefore, it is not a group.
For 5, $a\ast b = a + ab +b$
Identity: $a*e = a + ae + e = a$ with $e =0$, $a*e=a + 0 + 0 = a$
Inverse: There is no $x \in \Bbb Z$ such that $a*x = a +ax + x = e$
Thus, it is not a group.
Those are my solutions. I am not quite sure if I got it though.
 A: You need to be a little more rigorous with your proofs. For $1$, you can say, let $a=-1$. Then for all $b \in \mathbb{Z}$, $|ab|$ is nonnegative, so $a*b=|ab|\neq a$, so there is no identity.
For $2$, assume there is an identity $e$. Then $e*(e-1)=\max\{e,e-1\}=e\neq e-1$. (We know that the difference between integers is an integer, so $e-1 \in \mathbb{Z}$). This contradicts $e$ being the identity, so it is not a group.
Try going through these again and go through what the statement says- if you go through it directly, it should be easy to give a rigorous proof.
A: For Q3 consider $x=-a-2$ for each $a \in \mathbb{Z}$ and you'll see that there is in fact an inverse. Now prove associativity.
A: For $3$ note that $x\mapsto x-1$ is a group isomorphism $(\Bbb Z,+)\to (\Bbb Z,*)$, hence this is a group.
For 4 and 5 (and other cases) I suggest to exhibit explicit counterexamples. The fact that two different expressions "look" different may not be sufficient (for example $|a|$ and $\sqrt{x^2}$ look different but are the same; how can you tell that $abc+bc+ac+c$ is not the same as $abc+ac+c$? Letting $a=2$, $b=1$, $c=0$, I get the same results!): Such as: For $a=b=c=1$ we get $abc+bc+ac+c=4\ne 3=abc+ac+c$. Similarly in case 5, for $a=1$, $a*x=e=0$ would imply $2x=-1$, which is not possible for $x\in \Bbb Z$. (Note however, that $a*b=a+ab+b$ does define a group structure on another, more suitable set such as the set of rational numbers $>1$)
A: You have the concepts down and no what to do but I'm slightly less than satisfied. (only slightly).
1) and 2) You simply claim there is no such element such that $a*e = a$.  True, that is the reason but how can you verify these claims? I could claim there is no $e \in \mathbb Z$ such that $a + e = a$ but without proving this (wrong) claim there is no reason for the reader to take my word for it.
3) you showed if $e = -1$ then $a*e = a$ but you also have to show $e*a = a*e = a$.
You claim there is no $x$ such that $a + x +1 = e = -1$.  Of course there is.  $x = -a -2$ is such a number.   What's more $a*x = x*a = e$ (which also had to be shown).  So inverse does exist for every $a$.
Just need to show it is associative.  $(a*b)*c = (a+b +1)+c + 1 = a + (b+c + 1) +1 = a*(b*c)$ so it is a group. 
3 is wrong.
4) I would have said this was very good... and then I read Hagen von Eitzen's answer and he/she is right.  The inequality only exists if $ac \ne 0$.  You should give a (trivial) counter example.  (Although, in my opinion I'm being quite picky).
[Also, I'm curious why you didn't simply claim "there is no such element e where $ae + e = a $" as you did in your others.  Is it not as obvious?]
5) same problems I had with 1) and 2)  You need to show $e*a = a*e = a$ and you need to say why there is no inverse.  ($a*x = a + a*x + x = 0$ so $x = -a/(1+a)$ why isn't that an inverse?)
