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How do I prove that every non-zero ring is not a group under multiplication?

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3 Answers 3

up vote 1 down vote accepted

Hint: Note that by ring axioms, a ring should have an element $0$, which is the additive identity. Can you show that $0$ has no inverse?

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For any group $G$, left or right multiplication by any element $g \in G$ induces a map $G \to G$ which is both injective and surjective. Injective, since if $gh_1 = gh_2$ or $h_1g = h_2g$ we have, after multiplication by $g^{-1}$ on the left or right, respectively, $h_1 = h_2$. Surjective, since for any $h \in G$ we have $g(g^{-1}h) = (hg^{-1})g = h$. In a non-zero ring $R$, multiplication by $0$, either on the left or right, is neither injective nor surjective, since $0r = r0 = 0$ for all $r \in R$. Thus there can be no group structure consistent with the multiplication in $R$. QED.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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+1 because you avoid talking about $1\in R$ –  Hagen von Eitzen Jan 21 at 7:31
    
@ Hagen von Eitzen: you mean as in $0 \times 0^{-1} = 1$? ;-) . . . and speaking of "$1$", thanks for one! –  Robert Lewis Jan 21 at 7:47

$0 \times 0^{-1}=1$ and $0 \times x = 0$ for all $x$

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