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If I have a set endowed with an addition operation (say a general group) and I know it contains 1, can I always define a multiplication operation so that it distributes over the addition? By this I don't require that the multiplication to even be associative but I wish for it to satisfy the condition that 1*a=a*1=a. Does it matter if addition was defined to be commutative?

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  • $\begingroup$ If you have addition, then there is $0$. I guess that then $a\cdot b=0$ (trivial multiplication) is distributive. I know, zou are asking for non-trivila multiplication. $\endgroup$ – Janko Bracic Jan 23 '15 at 6:48
  • $\begingroup$ well I didn't think of that, but I am still interested in nontrivial ones, maybe a proof that nontrivial ones exist? $\endgroup$ – davik Jan 23 '15 at 6:49
  • $\begingroup$ wait what if I required the identity property to hold $\endgroup$ – davik Jan 23 '15 at 6:51
  • $\begingroup$ You mean, there is $1\ne 0$ in the additive group, such that $1\cdot a=a$ for all $a$? Then it will be even harder to find non-trivial multiplication. $\endgroup$ – Janko Bracic Jan 23 '15 at 6:53
  • $\begingroup$ I have restated the question this way, but I wouldn't mind if you gave a proof of a nontrivial multiplication that didn't satisfy this, I just wanted a way to get rid of the trivial case :) $\endgroup$ – davik Jan 23 '15 at 6:58
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My idea is to steal a multiplication from an algebraic structure which already has distributive multiplication. More precisely, let our additive group be $(G,+)$. Assume that there exist a ring $(R,+,\cdot)$ with a nontrivial multiplication and an injective additive map $\phi: G \to R$. Then one can define $\phi^{-1}:\phi(G) \to G$ and this is again an additive map. If $\phi(G)$ is closed for the multiplication in $R$, then one can define a multiplication on $G$ by $$ a\cdot b:=\phi^{-1}(\phi(a)\phi(b))\quad (a, b\in G).$$ This will be a distributive multiplication. If $R$ has an identity $1$ and $1\in \phi(G)$, then $\phi^{-1}(1)$ is the identity for the multiplication in $G$.

Of course, in the above reasoning, it is not necessary that we have a ring, it is enough that $(R,+)$ is an additive group with additional non-trivial binary operation which is distributive.

Of course, the above idea is not a solution of the posed problem. It is just a reformulation of the question.

One more comment, not related to the above thinking. If the multiplication in $(G,+)$ is defined by $a\cdot b=a$, then we have distributivity from one side: $$ (x+y)\cdot z=x+y=x\cdot z+y\cdot z\quad (x,y,z\in G). $$

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For non-commutative multiplication, you cannot. Every element's order must divide the order of $1$: if the order of $1$ is $n$, then $$ \underbrace{a+\dots+a}_{n\text{ times}} = a\cdot (\underbrace{1+\dots+1}_{n\text{ times}}) = a\cdot 0 = 0. $$ (And $a\cdot 0$ has to be $0$ since $a\cdot 0 = a\cdot (0+0) = a\cdot 0 + a\cdot 0$.)

Now consider $S_3$: its elements have orders $1,2$ and $3$, so whatever element is chosen to be $1$, there would be elements such that their order does not divide its order.

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