Below are one student's answers to the "Concepts"
questions in the exercises. I've put my
responses below each of the student's answers.
Section
Right! (We usually write "exactly one" or "a unique" instead of "only one.")
Actually, it is true that any two groups of order three are isomorphic. You might try proving that---it would be a good exercise.
I don't know what "linear equation" means for a group. I think what they're fishing for is the statement of Thm. 4.16.
What you've written is true; what's written in the book is true, too. You might try proving it---it would be a good exercise.
Actually, it does always have a unique solution. Namely, x = a^(-1) * c * b^(-1).
Nope. A group must contain an identity element.
A binary algebraic structure is just a set with a binary operation. (p. 29) So, yes, every group is a binary algebraic structure.
Section
Right.
Right.
Right.
Right.
A better way to phrase this is ". . . it is not necessarily true that every element is a generator." (In fact, there is only one cyclic group G for which it is true that every element is a generator of G. Can you figure out what G is?)
Right.
A better way to phrase this is "Not every set of numbers that is a group under addition is also a group under multiplication." Your example is a good one. (In fact, if G is a set of numbers that is a group under addition and G has more than one element, then G cannot be a group under multiplication. Why not? Hint: Find three numbers a, b, and c for which the cancellation law fails.)
I'm not sure what they're getting at with this one. Maybe they just want you to write the definition of a subgroup.
Rught.
A better way to phrase this is "Not every subset of every group is a subgroup under the induced operation." Another true statement would be " Every subgroup of every group is a subgroup under the induced operation."
Section
Right.
Right.
This is false. You might try proving that Q is not cyclic---it would be a good exercise.
Right.
I'm not sure what your parenthetical remark means.
Not true. The Klein 4-group V is not cyclic. (p. 59)
Right.
Subtle point: G and G' might be disjoint. You can fix this by writing "If G and G' are subgroups of a group H, then G intersect G' is a group."
Right.
Right. You might try proving this statement---it would be a good exercise.
Section
Right.
Right.
False. Consider a constant function.
Change "may not" to "is." This is Cayley's Theorem.
Right.
A better way to phrase this is ". . . it is not necessarily true that every element is a generator."
Right.
False. (It's not abelian.)
False. See Section 8 Ex. #46 and Thm. 6.1.
Right.