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What would be the inverse element in this abelian group: $(P(M),\triangle)$? I know the neutral element is the empty set and I thought the inverse element would be $A^{c}$ for every $A$. Turns out $A\triangle A^{c}=P(M)$ but it has to return the empty set.

Arthur
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  • P(M) means power set of M – Arthur Feb 21 '15 at 01:57
  • A set $B$ with $A\Delta B=\emptyset$ must have every point in $A\cup B$ also in $A\cap B$. This would work, for instance, if $A\cup B=A\cap B=A...$ (in fact this is the only possibility, since $A\cap B\subset A\subset A\cup B$.) Can you think of any sets like that? – Kevin Carlson Feb 21 '15 at 02:24
  • also consider what is the inverse of the neutral element $\varnothing$ – David Holden Feb 21 '15 at 02:34

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Ok, I got it. The inverse element for every $A \in P(M)$ is A itself!

Because: $A \triangle A=(A\cup A)\backslash (A\cap A)=A\backslash A=\varnothing$

Arthur
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