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Since addition requires 0 as it's identity and 0 has no inverse under multiplication this would seem to suggest that it is impossible but I am unable to prove it or find an example. Perhaps the rules are different enough under complex numbers, quaternions, or octonions to allow such a set to be possible.

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  • $\begingroup$ It is impossible, the closest that you can come is a division ring which is an additive group and the nonzero elements form a group. $\endgroup$ Apr 9, 2015 at 20:41
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    $\begingroup$ We should let the OP weigh in, since instinctually I assumed the object to be a field, forcing $0 = 1$ (but obviously Josh B. did not, which is valid; it's not explicitly stated): Do addition and multiplication have to "respect each other" via distributive properties? $\endgroup$
    – pjs36
    Apr 9, 2015 at 20:56
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    $\begingroup$ I have not studied fields much but I was not assuming distributivity would be given since it is not part of the definition of a group. $\endgroup$
    – Math Man
    Apr 9, 2015 at 21:02
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    $\begingroup$ If you are not assuming distributivity, you should ask yourself what it means to "be" addition and multiplication. The fact you cite that $0$ (the additive neutral element) has no inverse under multiplication depends on a supposed relation between addition and multiplication, and that relation is precisely distributivity. If you don't care about distributivity, nothing prevents you from for instance calling the same operation both addition and multiplication. It can even be natural: a cyclic group is written with addition or multiplication depending on your mood. But not both simultaneously. $\endgroup$ Apr 10, 2015 at 3:36
  • $\begingroup$ A division ring is an abelian group under addition, and its units ($R \backslash \{0\}$) form a group under multiplication. $\endgroup$
    – Myridium
    Apr 10, 2015 at 11:57

4 Answers 4

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The only way this can happen if your set also has the distributive property is if the multiplicative identity equals the additive identity, i.e. $1=0$. But this is the trivial group $\{0\}$ with $0+0=0$ and $0\cdot 0=0$.

If you do not have the distributive property, many more options are available.

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    $\begingroup$ Proof that $0 = 1$, assuming that multiplication distributes over addition: let $a$ be a left inverse of zero. Then $1+1 = a(0+0) = a(0) = 1$; because the identity element of a group is unique, we must have $0 = 1$. If there is no relationship like distributivity relating multiplication and addition, this is not true; just pick two random group structures on your set. $\endgroup$
    – user98602
    Apr 9, 2015 at 20:50
  • $\begingroup$ Why are we proving that 0=1? Are we redefining 1 and 0 for the purposes of identities? $\endgroup$
    – Math Man
    Apr 9, 2015 at 20:53
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    $\begingroup$ Agh! I saw the question title and this was literally going to be my answer. Well done! +1 $\endgroup$ Apr 9, 2015 at 20:54
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    $\begingroup$ @MathMan: Standard notation is to use $0$ for the identity in an additive group, and $1$ for the identity in a multiplicative group. This is just notation. $\endgroup$
    – user98602
    Apr 9, 2015 at 20:57
  • $\begingroup$ @mike: That makes sense. $\endgroup$
    – Math Man
    Apr 9, 2015 at 21:03
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You can just define the operations on a set to work. \begin{array}{c|cc} + & a & b \\\hline a & a & b \\ b & b & a \\ \end{array} \begin{array}{c|cc} * & a & b \\\hline a & b & a \\ b & a & b \\ \end{array} Here, $a$ is the additive identity and $b$ is the multiplicative identity. You do lose the distributive property though:

$a*(a+a)=a*(a)=b$

$a*a+a*a=b+b=a$

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    $\begingroup$ If we simply redefine the operations then it is no longer addition and multiplication is it? Isn't that like saying that If I want 2+2=5 then I redefine+ as ( * ) for some ( * ) where a( * )b=a+b+1 $\endgroup$
    – Math Man
    Apr 9, 2015 at 21:07
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    $\begingroup$ I guess it comes down to how you want to define addition and multiplication. When is a group operation addition, when is it multiplication, and when is it neither? $\endgroup$
    – Josh B.
    Apr 9, 2015 at 21:49
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    $\begingroup$ A ring has two operations, addition and multiplication on a set which satisfy certain axioms including distributivity. Certainly in that framework the other answer is more applicable. If you let go of the distributive axiom you don't have a ring. $\endgroup$
    – Josh B.
    Apr 9, 2015 at 21:51
  • $\begingroup$ that makes sense. I guess it is just a matter of what assumptions you start with. $\endgroup$
    – Math Man
    Apr 9, 2015 at 22:23
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    $\begingroup$ @MathMan: Correct. If you want to define two group operations then it doesn't really make sense to choose to call them "addition" and "multiplication" unless they have a distributive relationship. Better to call them "OpA" and "OpB". In fact, "0 has no inverse under multiplication" is already defining a relationship between the two operations. And if the operations can be anything then just define them the same, or as Josh has done define them isomorphically. His groups are both the cyclic group order 2, with the elements assigned the other way around. $\endgroup$ Apr 9, 2015 at 23:15
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The set $\{0\}$ forms a trivial group under both of the operations $+$ and $\times$ given their ordinary meaning on the integers.

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  • $\begingroup$ Well, there's really only one binary operation over a singleton. :-) $\endgroup$
    – Asaf Karagila
    Apr 10, 2015 at 14:42
  • $\begingroup$ @AsafKaragila Absolutely. I'm not claiming anything of any significance; just that the ordinary definitions of addition and multiplication both correspond to the (unique) trivial group when restricted to $\{0\}$. This isn't toooooootally trivial, since it doesn't work for, e.g., $\{1\}$. $\endgroup$ Apr 10, 2015 at 14:47
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I think the straight forward answer is no, because a group can only have 1 operation (addition or multiplication for your case). If you have 2 operations (addition and multiplication for your case) then it is considered a ring.

Edit (for Math Man): "In mathematics and abstract algebra, group theory studies the algebraic structures known as groups. The concept of a group is central to abstract algebra: other well-known algebraic structures, such as rings, fields, and vector spaces, can all be seen as groups endowed with additional operations and axioms."

http://en.wikipedia.org/wiki/Group_theory http://en.wikipedia.org/wiki/Group_%28mathematics%29

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  • $\begingroup$ "a group can only have 1 operation" could you link to a source on that? Because the definition of group In my college textbook doesn't include that as a requirement for membership in a group. $\endgroup$
    – Math Man
    Apr 10, 2015 at 1:16
  • $\begingroup$ I read it on wikipedia. I'll add the link right now. $\endgroup$
    – Neil
    Apr 10, 2015 at 1:17
  • $\begingroup$ I think you are missing the point of the question. But it's a good thing to add to supplement the current answers perhaps. $\endgroup$ Apr 10, 2015 at 1:29
  • $\begingroup$ Thanks for the link. The definition of a group (en.wikipedia.org/wiki/Group_%28mathematics%29#Definition) does not say that it cannot have other operations as well. And (en.wikipedia.org/wiki/Group_theory#Applications_of_group_theory) mentions that rings are simply abelian groups with additional operations. But groups nontheless. $\endgroup$
    – Math Man
    Apr 10, 2015 at 1:31
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    $\begingroup$ @MathMan A group has a single operation. When people say things like "A ring is a group with extra properties", it's a shorthand for "A ring is a set $S$ with operations $+$ and $\times$ such that $(S,+)$ is a group and...". It's not that a ring is a two-operation group; rather, it's that the ring restricted to one of its operations is a group. $\endgroup$ Apr 10, 2015 at 11:52

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