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Given group $\mathbb{Z}\backslash\{1\}$ with operation $a \otimes b = a + b − a · b$. Here $+, ·$ denote the normal operations of addition and multiplication in $\mathbb{Z}$. Which of the group axioms are satisfied.

First request code for typing $\otimes$.

Next, there are four group axioms / properties.

  1. Closure. Results always in a number in set $\mathbb{Z}$.

  2. Identity: Let $e$ be identity, then for all elements $a$ in $\mathbb{Z}$, have $$a \otimes e = a + e − a · e = a$$ $$\implies e(1 − a) = 0$$ $$\implies e(1 − a) = e$$ This implies $a=1$. So, how to find identity?

If take $e\ne 0$, then: $$\implies e(1 − a) = 0$$ $$\implies e = \frac0{(1-a)}=0$$ Then get below:

  1. Inverse wrt operation $\otimes$: let, $b$. But, $$a \otimes (b) = a +b -a · b = e$$ $$a +b -a · b = e$$ Leads nowhere.

Edit

In view of need for changes to domain, as pointed below; want to add that it is taken varbatim from the Q.3 of Chapter 20 of book titled: Abstract algebra- an Inquiry-Based based, by Jonathan Hodge, Ted Sundstorm.

jiten
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    $-a$ is not the inverse with respect to $\otimes$. – Cheerful Parsnip May 26 '22 at 03:56
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    "First request code for typing $⊗$." We have a \otimes b: $a\otimes b$. – Arthur May 26 '22 at 03:57
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    When you got to $e(1-a)=0,$ it only hows that either $e=0$ or $(1-a)=0.$ Since it must hold for every $a$ you have $e=0.$ [And indeed $0$ is the identity operation for this definition of $\otimes.$] – coffeemath May 26 '22 at 03:57
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    $a+b-ab=0$ and $a\neq 1$ imply $b=-a/(1-a)$. So the inverse with respect to $\otimes$ is $-a/(1-a)$ when $a\neq 1$. – Cheerful Parsnip May 26 '22 at 04:03
  • @CheerfulParsnip If inverse of $a= \frac {-a}{1-a}$, then get $a+ \frac {-a}{1-a} -a. \frac {-a}{1-a} =0\implies \frac{(a -a-a^2+a^2)}{1-a}=0\implies 0=0$. But, this means need modified $ \mathbb{Z^*}= \mathbb{Z}-1$. – jiten May 26 '22 at 04:14
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    @jiten Your observation is really just a check that Cheerful's inverse is correct, given the identity is $0.$ – coffeemath May 26 '22 at 04:17
  • @coffeemath But, the question needs modified $ \mathbb{Z^*}= \mathbb{Z}-1$. – jiten May 26 '22 at 04:17
  • Ah, you are right. One must throuw out $1$ from $\mathbb Z.$ OP failed to say that. – coffeemath May 26 '22 at 04:18
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    You have received some great comments from other users above! One thing I wanted to add is as follows:

    You note in the final statement of your original post that your attempt to compute the inverse leads nowhere. This should be expected, as it is a bit unwieldy to attempt to compute an inverse without knowing what the identity is! If you do not know what the identity is, you will not be able to compute an inverse, and as such, any computations you attempt for an inverse should not contain an arbitrary element like e, but should be instead an element of the group that you already know about!

     – JJ Hoo May 26 '22 at 04:21
  • Everything follow immediately from transport of structure as explained in the linked dupes (and many more posts on such. – Bill Dubuque May 26 '22 at 05:20

1 Answers1

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Let us check each of the group axioms, in the order you indicate.

  1. Closure, as you mention, is inherited due to the closure of both + and $\cdot$ over $\mathbb{Z}$

  2. For the identity, your computation is correct, but your final line of the first set of implications is incorrect. What you have written does NOT imply that $a=1$. Keep in mind that you quantify $a$ using a universal quantifier. Thus, $1-a$ could be anything! Another way to look at this is that since $a$ is an arbitrary integer, we can instead look at an arbitrary integer $b=1-a$, and ask ourselves:

Given an arbitrary $b$, how do we solve $eb=e$? The implication is not that $b=1$, but instead that since this holds true for all $b$, $e$ must be 0. Your next line does provide the contradiction that shows indeed that $e$ must be 0, albeit in a slightly roundabout manner. Please let me know if this clears up your confusion regarding 0 being the identity!

Note that we need not split this into cases where we consider $e\neq 0$.

  1. Now that we know that $0\in\mathbb{Z}$ is the identity element of this group. If $a\neq 1$, we have:

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

However, we have here an issue, if $1$ is included in this group. $$1+b-1\cdot b = 0\implies 1+b-b=0\implies 1=0$$ There is clearly an issue here, so if $1$ is an element in the set to which we are applying this $\otimes$ operation, then this is not a group!

It might perhaps be useful to define your set along with your operation before asking if it's indeed a group, but this should be your thought process as you work out the problem!

P.S. Don't forget to check associativity! It might be as easy as claiming inheritance from other operations, as you do with closure in this case, but it should still be stated!

JJ Hoo
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    I think this will not be a group even if we remove $1$ from $\mathbb Z.$ Because then $3$ will not have in integer inverse, since $3^{-1}=\frac{3}{3-1}=\frac{3}{2}$ is not in $\mathbb Z$ with $1$ removed. I think things work OK if we use $\mathbb Q \setminus {1}$ as the underlying set of the group. – coffeemath May 26 '22 at 04:33
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    I think you might be correct here. I was unsure what the set actually was prior to the comments made to the original post, and thus wanted to leave my answer as just stating that the problem as originally stated is NOT in fact a group. However, I think you are right as to $(\mathbb{Q}\backslash {1},\otimes)$ being a group!

    Edit: I say that this is a group simply because, as coffeemath implies above,, $\frac{a}{a-1}$ will always be in $\mathbb{Q}\backslash{1}$, and so we can always construct an inverse element.

     – JJ Hoo May 26 '22 at 04:47
  • What is this group? – Kolja May 26 '22 at 04:56
  • I don't know if this is a canonical or well-known group, but it's certainly a group. In the two comments above, coffeemath suggests that when we consider the set $\mathbb{Q}\backslash{1}$, we indeed have that $(\mathbb{Q}\backslash{1},\otimes)$ is a group, which is certainly true, and probably the smallest "extension" of $\mathbb{Z}\backslash{1}$ that allows satisfaction of group axioms. However, we can go further to say that $(\mathbb{R}\backslash{1},\otimes)$ is a group as well! – JJ Hoo May 26 '22 at 05:05
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    Well, $1-a\otimes b=(1-a)\times(1-b)$ tells you it's isomorphic (via $x\leftrightarrow 1-x$) to $\Bbb Q$ or $\Bbb R$ or whatever with the usual multiplication operation. Very common trick for these problems. It's an example of "transport of structure": conjugate a group's group operation by any other permutation of the group's underlying set and you get a potentially wonky-looking, but isomorphic, group operation on the same set. – anon May 26 '22 at 05:08
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque May 26 '22 at 05:21
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    @Kolja This is just the usual multiplicative group on either the nonzero rationals or nonzero reals, but acted on by the self-inverse map $h(t)=1-t$ as follows: To compute $a \circ b$ we calculate $h(h(a)\cdot h(b)),$ where the $\cdot$ in the last is ordinary multiplication on the nonzero rationals or reals. In particular $h(0)=1$ must be removed to get the underlying set of the transformed group. – coffeemath May 26 '22 at 05:45