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So, basically I'm taking an intro into proofs class, and we're given homework to prove something in abstract algebra. Being one that hasn't yet taken an abstract algebra course I really don't know if what I'm doing is correct here.

Prove: The set $\mathbb{R} \backslash \left\{ 1 \right\}$ is a group under the operation $*$, where: $$a * b = a + b - ab, \quad \forall \,\, a,b \in \mathbb{R} \backslash \left\{ 1 \right\} .$$

My proof structure:

After reading about abstract algebra for a while, it looks like what I need to show is that if this set is a group, it has to satisfy associativity with the operation, and the existence of an identity and inverse element.

So what I did was that I assumed that there exists an element in the set $\mathbb{R} \backslash \left\{ 1 \right\}$ such that it satisfies the identity property for the set and another element that satisfies the inverse property for all the elements in the set. However I'm having trouble trying to show that the operation is indeed associative through algebra since

$$\begin{align} a(b * c) & = a(b+c) - abc \\ & \ne (a+b)c - abc = (a*b)c \\ \end{align}$$

So in short, I want to ask if it's correct to assume that an element for the set exists that would satisfy the identity and inverse property for the group. Also, is this even a group at all since the operation doesn't seem to satisfy the associativity requirements.

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    $\begingroup$ As a wise man once told me: "The easiest way to prove something exists is to go out there, find it, bring it back, and say: Look! Here it is." $\endgroup$ – vrugtehagel Mar 13 '17 at 13:06
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    $\begingroup$ For associativity, you need to show $a \ast (b \ast c) = (a \ast b) \ast c$. Multiplication should always be given by the operation $\ast$ and not "normal" multiplication. $\endgroup$ – Matt B Mar 13 '17 at 13:13
  • $\begingroup$ Thanks @MattB, gonna look at it to see if it works. $\endgroup$ – Aldon Mar 13 '17 at 13:14
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The statement of the associativity condition is wrong; it should be $$(a \ast b) \ast c = a \ast (b \ast c) .$$

Expanding the l.h.s. gives \begin{align}(a \ast b) \ast c &= (a + b - ab) \ast c \\ &= (a + b - ab) + c - (a + b - ab) c \\ &= a + b + c - bc - ca - ab + abc .\end{align}

Now, do the same thing for the r.h.s. and verify that the expressions agree.


We cannot simply assume the existence of an identity and inverse. There are plenty of associative binary operations which lack one or both!

We can pick out the identity element using the definition: We must have, for example, that $$a \ast e = a$$ for all $a$, and expanding gives $$a + e - ae = a.$$ Can you find which $e$ makes this true for all $a$?

Likewise, to show that the operation admits inverses, it's enough to produce a formula for the inverse $a^{-1}$ of an arbitrary element $a$, that is, an element that satisfies $a \ast a^{-1} = e = a^{-1} \ast a$. As with the identity, use the definition of $\ast$ and solve for $a^{-1}$ in terms of $a$.


Remark If we glance at the triple product, which by dint of associativity we may as well write $a \ast b \ast c$, we can see the occurrence of the elementary symmetric polynomials in $a, b, c$, which motivates writing that product as $(a - 1) (b - 1) (c - 1) + 1$. Then, glancing back we can see that we can similarly write $a \ast b = (a - 1) (b - 1) + 1,$ which is just the conjugation of the usual multiplication (on $\Bbb R - \{0\}$) by the shift $s : x \mapsto x + 1$, that is, $a \ast b := s(s^{-1}(a) s^{-1}(b))$. Since multiplication defines a group structure on $\Bbb R - \{0\}$, that $\ast$ defines a group structure follows by unwinding definitions. For example, to show associativity, we have $$(a \ast b) \ast c = s(s^{-1}(s(s^{-1}(a) s^{-1}(b))) s^{-1}(c)) = s(s^{-1}(a) s^{-1}(b) s^{-1}(c)) = s(s^{-1}(a)s^{-1}(s(s^{-1}(b) s^{-1}(c))) = a \ast (b \ast c) .$$ (Note that in writing the third expression without parentheses we have implicitly used the associativity of the usual multiplication.) You can just as well use this characterization to determine the group identity $e$ and the formula for the inverse of an element under $\ast$.

Indeed, this together with analogous arguments for the other group axioms shows that conjugating any group operation by a set bijection defines an operation on the other set.

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An important thing to prove here too is that for every pair $a,b\in\mathbb{R}\backslash\{1\}$, we need $a*b\in\mathbb{R}\backslash\{1\}$.

This is true, since for $a\neq1\neq b$ to have $a+b-ab=1$, we'd need $a(1-b)=1-b$ meaning either $a=1$ or $b=1$.


We have an identity, since (let's denote the identity with $I$, so $a*I=a$) for $a+I-aI=a$ we only need $I(1-a)=0$ so $I=0$. So $0$ is the identity, since, $a*0=a=0*a$. Note that, even in formal proofs, it's not necessary to include how you've found it; simply presenting it and showing it is the identity is sufficient.


This operation is associative, since: \begin{align} a*(b*c)&=a+(b+c-bc)-a(b+c-bc)\\ &=a+b+c-bc-ab-ac+abc\\ &=(a+b-ab)+c-(a+b-ab)c\\ &=(a*b)*c \end{align}


A group also needs inverse elements; elements $a^{-1}$ for every $a$ that satisfy $a*a^{-1}=0$ (because $0$ was the identity). For $a*b=0$, we need $a+b-ab=0$, so $b=\frac{a}{a-1}$ (and we won't be dividing by $0$ since $a\neq 1$). And so now: $a*\frac{a}{1-a}=0$, which is what we wanted.


Now we've checked every property a group must have, and it has it all, thus, it's a group. An important, useful fact to notice in this group is the fact that it is commutative, meaning $a*b=b*a$. This property often makes a lot of calculations and/or proofs a lot easier, so you should always look out for it.

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  • $\begingroup$ Looks good, thanks! I'm confused tho, why does the identity $a*b = a$? $\endgroup$ – Aldon Mar 13 '17 at 13:18
  • $\begingroup$ That's how the identity is defined. It is the number $I$ such that $a*I=a=I*a$. I edited my answer to hopefully make that a little clearer. $\endgroup$ – vrugtehagel Mar 13 '17 at 13:20
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Something that you always need to check is if the set is closed under the operation $*$. Let's assume that it is not closed and there is $a$ and $b$ in the set, such that

$a*b=1$

Then, we can deduce that

$a*b=a+b-ab=1 \implies a(1-b)=1-b \implies a=1 \hspace{0.1cm} or \hspace{0.1cm}b=1$

which is not possible, as both $a$ and $b$ are taken from $\mathbb{R} \backslash \left\{ 1 \right\}$.

In a group, you only have one operation defined, i.e. the operation $*$. So, for associativity, you have to show

$a*(b*c)=(a*b)*c$

rather than

$a(b*c)=(a*b)c$

where you have assumed the regular multiplication as an extra operation.

For the identity element, you may make a guess first. By definition, an element $x$ is the identity element of a group, if

$a*x=x*a=a \hspace{0.2cm} \forall a\in \mathbb{R} \backslash \left\{ 1 \right\}$

Then

$a*x=a+x-ax=a \implies x=0$

So, $x=0$ seems to be a valid identity element and then you need to proof, using the definitions, that it is actually the identity element.

Note that you need to find the identity element before you deal with the inverse of each element.

Finally, the inverse of an element $a$ is $b$ if and only if

$a*b=0$

Note that $0$ is used as the identity element.

Again make a guess

$a*b=0 \implies a+b-ab=0 \implies a(1-b)=-b \implies a=\frac{b}{b-1}$

and finish the job.

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Associativity is satisfied:

$$a*(b*c)=\\a*(b+c-bc)=\\a+b+c-bc-a(b+c-bc)=\\a+b+c-ab-bc-ac+abc$$

and $$(a*b)*c=\\(a+b-ab)*c=\\a+b-ab+c-(a+b-ab)c=\\a+b+c-ab-bc-ac+abc$$

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