Group Theory

One or more correct

• A$(\text{F}, \odot)$ is an Abelian group.
• B$(\text{F}, \odot)$ is an Abelian monoid.
• C$(\text{F}, \odot)$ is a non-Abelian group.
• D$(\text{F}, \odot)$ is a non-Abelian monoid.

One or more correct

• AOperator $\lozenge$ obeys the associative law
• BOperator $\square$ obeys the associative law
• COperator $\lozenge$ over the operator $\square$ obeys the distributive law
• DOperator $\square$ over the operator $\lozenge$ obeys the distributive law

Numerical answer

One or more correct

• A$H$ is a group.
• BEvery element in $H$ has an inverse, but $H$ is NOT a group.
• CFor every $A \in 2^X$, the inverse of $A$ is the complement of $A$.
• DFor every $A \in 2^X$, the inverse of $A$ is $A$.

Answer key coming soon

One or more correct

• AIf for all $x, y \in G$, $(xy)^2 = x^2y^2$, then $G$ is commutative.
• BIf for all $x \in G$, $x^2 = 1$, then $G$ is commutative. Here, 1 is the identity element of $G$.
• CIf the order of $G$ is 2, then $G$ is commutative.
• DIf $G$ is commutative, then a subgroup of $G$ need not be commutative.

Answer key coming soon

Single correct

• ABoth $G$ and $H$ are always cyclic.
• B$G$ may not be cyclic, but $H$ is always cyclic.
• C$G$ is always cyclic, but $H$ may not be cyclic.
• DBoth $G$ and $H$ may not be cyclic.

Answer key coming soon

Numerical answer

Answer key coming soon

Single correct

• A$R_1$ and $R_2$
• B$R_1$ only
• C$R_2$ only
• DNeither $R_1$ nor $R_2$

Numerical answer

Numerical answer

Numerical answer

Single correct

• ACommutative but not associative
• BBoth commutative and associative
• CAssociative but not commutative
• DNeither commutative nor associative

Answer key coming soon