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The groups to be determined are $$(\mathbb{R},*) \; \mathbb{defined \; by \;} a*b=a+b+ab\;\;\;\;\;\forall \;a,b\in\mathbb{R}$$ and $$\mathbb{Z\times Z\;defined \;by} \; (a,b)*(c,d)=(ad+bc,bd) \;\;\;\;\;\forall\;a,b,c,d\in\mathbb{Z}$$

I've made an attempt at the both, but am not sure if I am correct or even have sound arguments.

Denote our group to $G$. Here's an attempt

If $a,b \in G=(\Bbb R,*)$, then it follows that $a+b \in G$ and $ab \in G$. Thus, $a + b + ab \;\mathbb{must\;} \in G$ so we have closure. Now, suppose $c \in G$. Then, $$a*b*c=a+(b+c+abc) = (a +b+c)+abc$$ and so $G$ satisfies the associative property. Now, we need to find an identity. Let's denote our potential identity to be $e$ $\in G$. Then, for $a\in G$ $$a=a*e=a+e+ae=a+0+a(0)=a$$ since under addition, we have that $e=0$. Lastly, we need to see if there exists an inverse for $G$. We have to satisfy $a*f=e$, so we let f be our potential inverse. Hence, $a*f=a+f+af=a+a^{-1}+aa^{-1} \neq e = 0$. Under addition, $a^{-1}=-a$, but $a(-a)=-a^2$. This does not give us 0, unless $a=0$. Therefore, G is not a group. $$\\$$ $$\\$$ Now, if $a,b\in G=(\Bbb Z\times \Bbb Z,*)$, then it follows that $(a,b)*(c,d)=(ad+bc,bd) \;\mathbb{must}\; \in G$ so we have closure. Now, suppose $(q,r)\in G$. Then $$(a,b)*(c,d)*(q,r)=(ad+bc,bd)+(cr+dq,dr)$$ It is impossible to satisfy the associative property with our $*$ is defined, so $G$ is not a group.

So I'm sure there are many mistakes, as I began to get very confused with this problem. Could anyone guide me along the correct road?

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+1 for effort (: –  Ivo Terek Aug 3 at 3:23

4 Answers 4

up vote 4 down vote accepted

Closures are trivial. Let's list what we have to prove:

i) Associativity;

ii) Existence of identity;

iii) Existence of inverses.

For $(\Bbb R, \ast)$, you should see what $(a \ast b) \ast c$ and $a \ast (b \ast c)$ are. What you have done does not makes it clear that they're equal. We have: $$\begin{align} (a \ast b) \ast c &= (a + b + ab)\ast c = a+b+ab+c + ac+bc+abc \\ a \ast (b \ast c) &= a \ast (b + c + bc) = a+b+c+bc+ab+ac+abc\end{align}$$ hence they're equal. So we've checked i). For the identity, we want $e \in G$ such that $a \ast e = e \ast a = e$ for all $a \in G$. So we want: $$a+e+ae= a \implies e + ae = 0 \implies e(1+a) = 0$$ Now, $e = 0$ will work for every $a$, but if $a = -1$, we have a problem here, because, calling $a^{-1}$ a candidate for the inverse of $-1$, we get: $-1 \ast a^{-1} = -1 + a^{-1} + (-1)a^{-1} = 0$, and so $-1 = 0$ by cancelling $a^{-1}$, a contradiction. Hence $-1$ doesn't have an inverse, and $(\Bbb R, \ast)$ is not a group. Property iii) fails.

Now, let's see the other one.

Consider $(\Bbb Z^2, \ast)$. If we want to see that $$\left((a,b)\ast(c,d)\right) \ast (q,r) = (a,b) \ast \left((c,d) \ast (q,r)\right)$$ then we should see each side separately, and then compare. We have: $$\begin{align} \left((a,b)\ast(c,d)\right)\ast (q,r) &= (ad+bc, bd)\ast(q,r) = (adr+bcr + bdq,bdr) \\ (a,b) \ast \left((c,d) \ast (q,r)\right) &= (a,b) \ast (cr + dq, dr) = (adr + bcr + bdq, bdr)\end{align}$$ hence property i) holds. For ii), we want $(e_1, e_2)$ such that $(a,b) \ast (e_1, e_2) = (e_1, e_2) \ast (a,b)$, that is: $(ae_2 + be_1, be_2) = (a,b)$, which gives $e_2 = 1$. And that gives $a + be_1 = a$, so $e_1 = 0$. To check that $(0,1)$ really is the identity, see that $(0,1)\ast(a,b) = (0b + a, 1b) = (a,b)$. So, property ii) also holds. Now, let $(\bar{a}, \bar{b})$ be a candidate for the inverse of $(a,b)$. We want: $(a,b) \ast (\bar{a}, \bar{b}) = (0,1)$. That is: $$(a\bar{b} + b\bar{a}, b\bar{b}) = (0,1)$$ This already gives us problems, since if you choose $b = 0$, the inverse will not exist. Hence $(\Bbb Z^2, \ast)$ is not a group, because property iii) fails.

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Thank you, my friend. The way you expanded everything out: I figured that would be the way, but I was just too confused as a whole. –  user146925 Aug 3 at 3:32
    
One step at a time, and you can manage it. It is important to be patient and work out the definitions given carefully. Listing the things that you want to prove/check is useful not only for group theory, but in everything in maths, in general. Glad to help. Good studies (: –  Ivo Terek Aug 3 at 3:34
    
You should probably mention that the first product is $(a + 1)(b + 1) - 1$. –  Ryan Reich Aug 3 at 4:57
    
A question regarding $(R,*)$. Is the jump from $$a=a+e+ae$$ to $$0=e+ae$$ valid because $-a\in \mathbb{R}$? –  user146925 Aug 3 at 5:00
1  
@Ryan, I didn't see what product you refered to. user146925 Yes, the step is valid because the cancellation law holds in $(\Bbb R, +)$. –  Ivo Terek Aug 3 at 5:03

For the first, note the identity must be $0$. However, then $-1$ does not have an inverse.

For the second, the identity must be $(0,1)$. However, then $(0, 0)$ has no inverse.

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For the first problem, you should be careful about replacing $f$ with $a^{-1}$, since we don't even know if $a^{-1}$ even exists yet. Indeed, $a^{-1}$ is not the inverse of $a$ under addition; it's the inverse of $a$ under the operation that we're talking about, namely $*$. In fact, $(\mathbb R, *)$ is almost a group, if it weren't for the fact that $-1$ has no inverse.

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For $(\mathbb{R},*)$ everything you need to prove follows from homomorphism with multiplication in $\mathbb{R}$ by adding one. This could work by showing that using this homomorphism as an alternate definition of $*$ is equivalent with the original definition.

The alternate definition of $a*b$ would then require to add one to each element to get to $\mathbb{R}$, multiply, and then subtract one to get back. This would imply $a*b = (a+1)(b+1)-1$, which is obviously equivalent with the original definition.

The question then is, if $\mathbb{R}$ with multiplication is a group, which it only is, if you ignore $0$.

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