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I wanted to have example of group which can not be ring?

I think if we have non abelian group with some operation we can not proceed to ring . Is it correct or it required to more argument ?

Actually this question asked by my prof in class I answered same but he is not satisfied . so I was confused whether I am correct or not ?

Please give me suggestion .

Curious student
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    Did your professor explain why he was not satisfied with your answer? Also, what exactly do you mean by "being made into a ring"? – 5xum May 28 '19 at 12:38
  • That means using any other possible operation (multiplication) . NO he just kept question open . And today he did not talk about that question – Curious student May 28 '19 at 12:43
  • Well clearly, just by adding multiplication, there is no way to change a non-abelian group into a ring. Your answer is perfectly fine... under one assumption. That assumption is that you correctly understood the question. – 5xum May 28 '19 at 12:45
  • You can take any existing ring $R$ and create a group ring $R[G]$ which are formal combinations of elements of the form $rg$ with $r \in R$ and $g \in G$.https://en.wikipedia.org/wiki/Group_ring – CyclotomicField May 28 '19 at 12:46
  • Since the requirements for a ring include "abelian group with secondary operation that is associative" you need to specify that this group is also not associative, or find another non-qualifying property. The group in question could be non-abelian and then there could be some way for this to be the secondary operation for a ring. – abiessu May 28 '19 at 12:46
  • Obviously the underlying group has to be abelian, otherwise the answer is trivial yes. That's probably why the professor is not satisfied. If by "ring" you mean "unital ring", then the answer is also "yes, there are such groups" in abelian case, see this: https://math.stackexchange.com/questions/93409/does-every-abelian-group-admit-a-ring-structure Note that every abelian group can be turned into a nonunital ring via $xy:=0$. – freakish May 28 '19 at 13:08
  • Note that if you want a finite non-abelian group to act as the multiplicative group of non-zero elements of a ring, all non-zero elements are then invertible, we have a division ring and Wedderburn's Theorem tells us that the multiplication must be commutative. – Mark Bennet May 28 '19 at 14:09

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