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Consider the set G = {0,{1},{2},{1,2}}. Does the set operation intersection de fine a binary operation on G? Does the set operation union de fine a binary operation on G? Is < G,(union) > a group? Explain. Is < G, (intersection) > a group? Explain.

So I know that a binary operation takes all possible ordered pairs of elements of G and outputs an element of the set G. However, I do not understand how to apply this in this context... I also know that in order to be a group it is necessary that: the group is closed under a binary operation associative: (a*b)c= a(b*c) identity: contains the identity element e inverse: for every a in G, a^-1 is also in G

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  • $\begingroup$ By $0$ do you mean the empty set? $\endgroup$ – Tobias Kildetoft Dec 19 '14 at 8:57
  • $\begingroup$ Yes, 0 is the empty set. $\endgroup$ – Lydia Dec 19 '14 at 9:14
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if $(G,(\text{intersection}))$ be a group, $\varnothing=\varnothing\cap \{1\}=\varnothing \cap \{2\}$ implies $\{1\}=\{2\}$. contradiction.

In $(G,(\text{union}))$, $\varnothing$ is identity. associative law and commutative law are satisfied. but $\{1\}$ have not inverse i.e. $$ \{1\}\cup A\ne \varnothing$$ for any $A\in G$.

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  • $\begingroup$ But how do we know if the set operation union and intersection define a binary operation? $\endgroup$ – Lydia Dec 19 '14 at 9:15
  • $\begingroup$ binary operator is just a function from $G\times G$ to $G$. $G$ is closed under union and intersection. $\endgroup$ – Mirin Dec 19 '14 at 9:23
  • $\begingroup$ Yes, but could you please explain/show me why in this specific case? $\endgroup$ – Lydia Dec 19 '14 at 9:26
  • $\begingroup$ $\varnothing=\varnothing\cap \{1\}=\varnothing\cap\{2\}=\varnothing\cap \{1,2\}, \{1\}\cap \{2\}=\varnothing, \{1\}\cap \{1,2\}=\{1\}, \{2\}\cap \{1,2\}=\{2\}$ $\endgroup$ – Mirin Dec 19 '14 at 9:44

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