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Let the following operation be defined on $T=\mathbb Z \times \mathbb Q$:

$$\begin{aligned}(a,b)\centerdot (c,d) = (-ac, b+d+2) \end{aligned}$$

in the commutative semigroup $(T, \centerdot)$, find the identity element, invertible elements and inverses.

I need to find the identity element, so I need an $(\alpha, \beta)$ so that

$$\begin{aligned}(a,b)\centerdot (\alpha,\beta) = (a,b) \end{aligned}$$

as $(T, \centerdot)$ is a commutative semigroup, I assume that $(\alpha,\beta)\centerdot (a,b) = (a,b)$ holds.

$$\begin{aligned}-a\alpha = a \Leftrightarrow \alpha = -1 \end{aligned}$$ $$\begin{aligned}b+\beta+2 = b \Leftrightarrow \beta = -2 \end{aligned}$$

so the identity element relative to this semigroup is $(-1,-2)$.

In order to find the invertible elements and inverses:

$$\begin{aligned}(a,b)\centerdot (a',b') = (-1,-2) \end{aligned}$$

so

$$\begin{aligned}-aa' = -1 \Leftrightarrow a'=-\frac{1}{a} \in \mathbb Z \Rightarrow a = 1 \end{aligned}$$

$$\begin{aligned}b+b'+2=-2 \Leftrightarrow b'=-(b+4) \end{aligned}$$

in conclusion only the elements $(1,b)$ are actually invertible and their inverse is $(1, -(b+4))$.

Does this whole thing hold? Am I wrong in any part of it?

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2  
In the last bit, why can't $a$ be $-1$? –  Gerry Myerson Jul 11 '12 at 9:37
    
@GerryMyerson because I need $(-aa') = -1$ if $a = -1$ then $(-aa') = 1$. –  haunted85 Jul 11 '12 at 9:42
1  
As @GerryMyerson pointed out $a$ cab be $-1\implies a'=-1\implies $ inverse of $(-1,b)$ also exists and is equal to $(-1,-(b+4)).$ –  Aang Jul 11 '12 at 9:46
1  
After equation $-aa'=-1$ $a'=\frac{-1}{a}$ is wrong, it should be $\frac{1}{a}$. –  Aang Jul 11 '12 at 9:49
    
@avatar thank you I've really overlooked that minus sign!! what a shame! –  haunted85 Jul 11 '12 at 9:59

1 Answer 1

up vote 2 down vote accepted

In all of these problems, you are taking a structure with operations that you "already know", and are trying to test to see whether a new operation, defined in terms of the ones you know, satisfies certain properties.

It's usually best to denote the "new" operation with a symbol that is unlikely to generate confusion. You've been doing this elsewhere, but here you are using $\cdot$, which can easily be confused with regular integer multiplication. So I'm going to replace it with $\odot$, if you don't mind.

Let's approach it systematically, and essentially "following your nose." I'm going to check a few things that you seem to take for granted, just to get familiarity with the operation.

Now, we have the set $T=\mathbb{Z}\times\mathbb{Q}$, and the operation $$(a,b)\odot (c,d) = (-ac, b+d+2).$$ First, the operation is commutative, since integer multiplication and addition of rationals is commutative. It is also associative: $$\begin{align*} \Bigl( (a,b)\odot (c,d)\Bigr)\odot (x,y) &= (-ac, b+d+2)\odot (x,y)\\ &=(-(-ac)x,(b+d+2)+y+2)\\ &= (acx, b+d+y+4)\\ (a,b)\odot \Bigl((c,d)\odot (x,y)\Bigr)&= (a,b)\odot (-cx, d+y+2)\\ &= (-a(-cx), b+(d+y+2)+2)\\ &= (acx, b+d+y+4). \end{align*}$$ So the operation is commutative and associative, and we do indeed have a semigroup.

Now we are trying to see whether this is a monoid (semigroup with identity). So we are trying to see whether there exist $\alpha\in\mathbb{Z}$ and $\beta\in\mathbb{Q}$ such that, for all $a\in\mathbb{Z}$ and $b\in\mathbb{Q}$, we have $$(\alpha,\beta)\odot(a,b) = (a,b)$$ (exactly as you did). This amounts to solving the equations you solved, which you did correctly, and so we discovered that $(T,\odot)$ is indeed a monoid, and that the identity element is $e_T = (-1,-2)$. Great!

To figure out what elements have inverses, you do set $(a,b)\odot (x,y) =e_T = (-1,-2)$, and figure out if you can determine necessary and sufficient conditions on $a$ and $b$ for $x$ and $y$ to exist. As you note, we have: $$(-1,-2) = (a,b)\odot(x,y) = (-ax, b+y+2).$$ This requires $-ax = 1$, which means that $a$ has to be an integer that has a multiplicative inverse in the integers (namely, $-x$); there are only two possibilities: $a=1$ or $a=-1$.

On the other hand, for any rational number $b$ we can always find a rational number $y$ such that $b+y+2=-2$, namely, as you note, $y = -(b+4)$. So it would seem that provided that $a=\pm 1$, then $(a,b)$ has an $\odot$-inverse in $T$. And, indeed, if $a=\pm 1$, then we have $$(\pm 1,b)\odot (\pm1,-b-4) = (-(\pm1)(\pm 1), b-b-4+2) = (-1-2).$$ So a necessary and sufficient condition for $(a,b)\in T$ to be $\odot$-invertible is that $a=\pm 1$.

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thank you so much for your patient, clear and most of all exhaustive answer, appreciated. –  haunted85 Jul 11 '12 at 19:23

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