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

  • $\begingroup$ 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

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 $A\in G$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.