Homeworks and Comments

This page is a collection of the homework questions and my comments about the mistakes that students had made. Different homeworks are separated by a horizontal line.

§7 Ex. 1(k), 1(l)

Determine whether the following sets build groups with respect to the operations given. In each case, state which group axioms are satisfied.

(k) {f, g} under the composition of mappings, where f : x --> x and g : x --> 1/(1 - x) are functions from R \ {1} into R \ {1}.

(l) {f, g, h} under the composition of mappings, where f : x --> x and g : x --> 1/(1 - x) and h : x --> (x - 1)/x are functions from R \ {0, 1} into R \ {0, 1}.

• Some students thought associativity hold on a set with two elements, on the ground that an equation like (a b) c = a (b c) requires three distinct entities. However, in such an equation, a, b, c, need not be distinct.
• Some students checked that f o g and g o f is in the set {f, g}, but forgot f o f and g o g.
• Many people wrote someting like (x)(g o g) = (x - 1)/x e/ {f, g}. But if (x - 1)/x stands for a real number and f, g are functions, (x - 1)/x can never be an element of {f, g}. The relevant thing that is not an element of {f, g} is g o g, not the value of g o g at x.
• Some people wrote x o x, where they should have written f o f. Some people wrote meaningless things like (a)(1/(x-1) o f). Make sure that you distinguish clearly between a function and the value of that function at a point.
• One student showed that f is a left identity and that every element of {f, g, h} has a right inverse. This argument does not prove that {f, g, h} is a group under o.
• Some people had the strange idea that R \ {1} or R \ {1, 0} could be closed under the composition of functions. One student said that {f, g} was not closed under o because f o g is not a member of R \ {1}.
• One student believed that there should be an element in a group that is the inverse of every element.
• Some people wrote the functions on the left side.

Although the following are not mathematical mistakes, they are mistakes.

• Many students have not read the question! The question asks which group axioms are satisfied, not merely whether certain objects are groups.
• A large number of students verified, or attemped to verify, the equation (a b) c = a (b c) in 8 or 27 cases in order to show that the operations are associative. Reference to Theorem 3.10 would be enough.
• Almost everybody made long calculations involving f, although they could appeal to the known (!) properties of the identity function on an arbitrary set.
• Some students computed g o f and found it to be g when checking the closure of {f, g, h}, repeated the same computation when checking that f is a right identity. They computed g o h and found it to be f when checking the closure of {f, g, h}, repeated the same computation when checking that g has a right inverse. Previous computations could be referred to.

§11 Ex. 2 Find all subgroups of a cyclic group of order 8, of a cyclic group of order 10, of a cyclic group of order 12.

§11 Ex. 4 Let G be a group and a an element of G. Let n, k be natural numbers and let m be their least common multiple. Prove that the intersection of <a^k> and <aⁿ> is equal to <a^m>.

• Many people, while trying to prove that the intersection of < a^k > and < aⁿ > is a subset of < a^m >, took an element x from the intersection, wrote x = a^kr = a^ns with integers r, s, and jumped to the conclusion kr = ns. This conclusion is valid only if we know or assume that o(a) is infinity. In case o(a) is finite, all we can concude from a^kr = a^ns is kr is congruent to ns (mod o(a)).
• Some students found the subgroups of the stated groups, but failed to invoke Lagrange's theorem and Lemma 11.10 to establish that their list is complete.
• Some students unnecessarily repeated the same reasoning three times, once for a cyclic group of order 8, for a second time for a cyclic group of order 10, then once again for a cyclic group of order 12.
• A few students listed, for example < a¹ >, < a² >, < a³ >, < a⁴ >, < a⁵ >, < a⁶ >, < a⁷ >, < a⁸ >, < a⁹ >, < a¹⁰ > and said that they are all the subgroups of a cyclic group of order 10. What about redundancies in this list?
• Some students tried to prove < a^m > is a subset of the intersection of < a^k > and < aⁿ > by taking an arbitrary element a^ms from < a^m > and showing that it belongs to the intersection. Although this is viable, it would be simpler and shorter to show that a^m is in the intersection. Then it would follow immediately that < a^m > is a subset of the intersection.
• One student said that if G is a finite group and n | |G|, then there is a unique subgroup H of G having order |H| = n. This is true for cyclic groups only. In general, if n | |G|, then there may or may not exist a subgroup of G of order n, and if there does exist a subgroup of order n, the number of subgroups of order n may be more than one.
• One student examined almost every subset and checked whether it could be a subgroup. For example, he/she argued in a long way that {1, a, a⁴, a⁷} or {1, a³ a⁴, a⁵} cannot be subgroups of a cyclic group < a > of order 8. No wonder his/her paper was very lengthy.
• One student forgot to record {1} and G among the subgroups of G.
• One student solved an unassigned problem instead of an assigned one.
• One student wrote consistently N in place of Z. Either he/she was not aware that certain integers in his/her answer could be negative, or he/she was not aware that N consists of positive integers only.
• Some students write ``p means q'' in place of ``p implies q'' or ``p yields q'' or ``p gives q'', etc. Notice that ``p means q'' means that p and q are logically equivalent, whereas ``p implies q'' means ``if p, then q''. Don't write ``p means q'' unless ``if p then q'' and ``if q then p'' are both true.
• Some students are still in the habit of mixing words and logical signs in sentences.

§13 Ex 2. Let m and n be two distinct lines intersecting at a point P. Show that s_ms_n is a rotation about P. Through which angle?

§13 Ex 3. Let and n be parallel lines. Show that s_ms_n is a translation.

Many people, while trying to prove that the composition of two reflections in intersecting lines is a rotation, took a point Q and showed that d(P, Q) = d(P, Qs_ms_n), and said that there is a rotation about P that maps Q to Qs_ms_n. This is true. But you cannot conclude from this that s_ms_n is a rotation! It is equally true that there is a translation that maps Q to Qs_ms_n, but this does make s_ms_n into any translation.

The rotation about P that maps Q to Qs_ms_n is indeed a rotation, but through an angle q that may depend on Q. Your argument shows that, for every Q, we can find an angle q in such a way that the rotation about P through q maps Q to Qs_ms_n. But what you should prove was this: we can find an angle q in such a way that, for all Q, the rotation about P through q maps Q to Qs_ms_n. Do you see the different order of the quantifiers?

You could show that the angle in question is twice the angle between the lines m and n, so this angle depends on m and n but not on Q, so it is the same for all points Q. This would prove that s_ms_n is a rotation.

The same remark applies, with appropriate modifications, to the second question.

§14 Ex 10. Find the symmetry group of a rectangle that is not a square.

§14 Ex 11. With the notation of definition 14.7, what is |D_2n: <a>|?

§15 Ex. 8. Show that V₄ := {id, (12)(34), (13)(24), (14)(23)} is a subgroup of S₄.

§15 Ex. 13. For s in S₄, we put sV₄ = {s \pi : \pi ranges over V₄} and V₄s = {\pis : \pi ranges over V₄}. Find sV₄ and V₄s when s = id, s = (12), s = (123), s = (12)(34), s = (1234).

§15 Ex. 14.%%%%%%%%

• Apparently, some students believe that rectangle is a regular polygon. It is not.
  The regular 4-gon is the square. The symmetry group of a rectangle that is not a square is different from the symmetry group of a square. The former has four elements, the latter eight.
• Some students speak of reflections in the x- and y-axes, without choosing any coordinate system at all. They say that these reflections are in the symmetry group of a rectangle. This is not always true. It depends on the location of the rectangle relative to the axes. For example, these reflections do not fix the the rectangle with corners (0, 0), (2, 0), (2, 1), (0, 1).
• Some students found four isometries in the symmetry group of our rectangle, and claimed that these four isometries make up the whole symmetry group. They had to show, in addition, that there are no other isometries in the symmetry group.
  In particular, some of these students repeated the argument on page 137 and concluded, as on page 137, that the symmetry group consists of the four isometries they found. However, the conclusion on page 137 rests on a previous result on page 136, namely that the symmetry group there has at most 2n elements. If you haven't proved before that the symmetry group of the rectangle has at most 4 elements, then finding 4 isometries in it does not enable you to infer that this symmetry group is exactly the set of these 4 isometries.
• Two students showed that V₄ contains the identity and that its is closed under the forming of inverses, and said that this proves V₄ is a subgroup of S₄. They should read the subgroup criterion more carefully.
• Since V₄ is a finite subset of S₄, in order to prove that V₄ is a subgroup of S₄, it is sufficient to show that it is closed under multiplication. You didn't have to, and you shouldn't, check that it is closed under the forming of inverses.
• In order to show that V₄ is a subgroup of S₄, you have to show that it is closed under multiplication. Some students checked some but not all the possible products to be in V₄.
• One student said (s)he would check ``closeness'' and ``inverses''. What you should check is ``closure (of the subset under composition)'' and ``existence of inverses (of elements of the subset in the same subset)''.
  ``Closure'' means being closed, being non-open, ``closeness'' means being close, being near, being not far away.
• One student said that V₄ is a subgroup of S₄ because V₄ is a subset of S₄ (?!).
• Three students found the elements of (12)V₄ (four things) but not (12)V₄ itself. Although this looks like a minor issue, you should be in a position to distinguish between a set and its elements!
• You could find (12)(34)V₄ and V₄(12)(34) without making any computation at all. Since (12)(34) is an element of V₄, the cosets (12)(34)V₄ and V₄(12)(34) are both equal to V₄ by Lemma 10.2(2).
• What does the set {(1234) (42) (1432) (13)} mean? The singleton whose only elements is the product (1234) (42) (1432) (13)?
• For some reason unknown to me, one student converted permutations in cycle notation to the double row notation, multiplied them in double notation, and then converted the product into cycle notation, instead of having the ease of multiplying them in cycle notation. Some students wrote (1)(2)(34) in place of (34), (1)(234) in place of (234).
  These are not wrong, but wasting time.
• One student thought that each of sV₄ and V₄s is a set with 15 (!) elements.
• One student forgot to simplify the products like (14)(21)(32)(43).
• One student wrote s = (12) ===> (12)V₄ = {...}. This is true. It is equally true that s ≠ (12) ===> (12)V₄ = {...}. What is the point of writing ``s = (12) ===>''? [Here and below I use the sign ===> to designate implication.]%%%%%%%
• Two students found (12)V<sub>4</sub> and stated without any justification that V<sub>4</sub>(12) is equal to (12)V<sub>4</sub>.
• In order to show that (123)H<sub>1</sub> ≠ H<sub>1</sub>(123), for example, it was enough to find one element in H<sub>1</sub> (123) which does not belong to (123)H<sub>1</sub>. This would eliminate almost half of your work in the last question, which does not ask you to find some cosets of H<sub>1</sub> and H<sub>2</sub>, but only to determine whether they are equal.
  It is always a good idea to read a question before trying to answer it!
• Some people wrote
  (123)H<sub>2</sub> = {(123), (23), (1243), (243)} ≠ H<sub>2</sub>(123) = {(123), (13), (1234), (134)}
  or
  (12)(34)H<sub>2</sub> = {(12)(34), (34), (12), i} = H<sub>2</sub>(12)(34) = {(12)(34), (34), (12), i}
  but the second (non)equality is far from clear. In such cases, you'd better write
  (123)H<sub>2</sub> = {(123), (23), (1243), (243)} ≠ {(123), (13), (1234), (134)} = H<sub>2</sub>(123)
  or
  (12)(34)H<sub>2</sub> = {(12)(34), (34), (12), i} = {(12)(34), (34), (12), i} = H<sub>2</sub>(12)(34).
• Somebody wrote ``for all s in {(12), (13), (123), (12)(34), (1234)}, s H₁ ≠ H₁ s and s H₂ ≠ H₂s''. This is false. For example, (12)(34) H₁ = H₁ (12)(34).
• One student concluded from (12)(13) ≠ (13)(12) that (12)H₁ ≠ H₁(12). This conclusion is not valid, for (12)(13) may very well be an element of H₁(12) and (13)(12) may very well be an element of (12)H₁. An equation like (12)H = H(12) does not mean that (12) commutes with all elements of H.
• Many answers were terribly long.

§17 Ex. 3. Write down the multiplication table of GL(2, Z₂). Compare it eventually after reordering the rows and columns) with the multiplication table of S₃.

§17 Ex. 6. Let K be a field and A =

be a matrix with elements from K. When a = 0 and b = 0, we have det A = 0. In case (a, b) ≠ (0, 0), prove that det A = 0 if and only if there is an element k in K such that c = ka, d = kb. Use this result and show that |GL(2, Z_p)| = (p² - 1)(p² - p).

§17 Ex. 15.%%%%%%%

§18 Ex. 5.%%%%%%%%

§18 Ex. 11. Let H be a normal subgroup of G and let Ha be an element of G/H. Show that o(Ha) = n (n is a natural number) if and only if n is the smallest natural number such that aⁿ belongs to H.

• Some students built the multiplication tables of GL(2, Z₂) and S₃, but forgot to compare them.
• A lot of students wrote the condition ad - bc = 0 as d/a = c/b. First a mistake of carelessness: we haven't defined the symbol d/a for all fields, for example d/a is meaningless when the field is Z₅. But it is clear what these students mean, and let's accept their notation. Now a more serious mistake: d/a = c/b makes sense only when a ≠ 0 and b ≠ 0. The argument of these students ignores the case where a or b is equal to 0.
  From d/a = c/b you can deduce ad - bc = 0, but not vice versa. From ad - bc = 0, you can obtain d/a = c/b only if a ≠ 0 and b ≠ 0. Many derivations in mathematics are not reversible.
  When we wish to show that a given set G is a group under a given operation o, we must find a right identity element and, for this purpose, we must solve the equation a e = a for e. Remember, after finding e, we had to verify that e does satisfy a e = a for all a. It is exactly for avoiding wrong conclusions (like the equivalence of ad - bc = 0 with d/a = c/b) that we make that verification.
• One of the problems was to show that |GL(2, Z_p)| = (p² - 1)(p² - p). So you had to count the number of elements of GL(2, Z_p) and find (p² - 1)(p² - p). Some students invented arguments that yield (p² - 1)(p² - p) but have nothing (or little) to do with the number of elements in GL(2, Z_p)!
  I am thinking of asking you to derive a wrong formula and see how many students are not incapable of obtaining it. Shall I do that?
• What on earth can this sentence ever mean: ``we have to show that a ≠ 0 and b ≠ 0 and there is k in K which c = ka and, d = kb det A = 0"?
• Some people have the impression that there are just two possibilities for a and b, namely a = b = 0 or a ≠ b ≠ 0. Stated differently, they think that the negation of ``a = 0 and b = 0'' is ``b ≠ 0 and a ≠ b''.
• I used to believe firmly that 1 is less than 6 and 7, that writing one symbol takes less time and less space than six or seven symbols. Having seen that a good many students preferred to write ``('', ``0'', ``-1'', ``1'', ``0'', ``)'' instead of ``j'' in a multiplication table, I am no longer sure about it!
• If H is a set with three elements, in order to check that H is closed under
  multiplication, you have to consider nine cases ``ab \in H'',
  where a takes on the three values from H and a takes on the three values from H. An astonishingly large number of students failed to list these cases properly.
• One student, deeply interested in the theory of permutations, wished to study the effect of the permutaton (68) in S₁₀ and labelled sixth, seventh and eighth pages of is homework by the numbers 8, 7, 6 instead of 6, 7, 8.
• A student who is very sensitive about privacy refused to write his/her name on the homework sheets.
• It is amazing to observe how many students, working independently, happened to have had just the same fruitless idea and give me the same wrong answer!

(a) Show that Z₁₉^x is a cyclic group by finding a generator of Z₁₉^x.

(b) Find all subgroups of Z₁₉^x and draw their Hasse diagrams.

(c) Find all subgroups of GL(2, Z₁₉) containing SL(2, Z₁₉) and draw their Hasse diagrams. Which of them are normal in GL(2, Z₁₉)?

• Applying Theorem 21.1 would not be justified unless you mention that det : GL(2, Z₁₉) --> Z₁₉^x is a surjective homomorphism. Very few students did so.
• There are people who believe that det is a one to one function.
• Some people said ``f preserves inclusion, hence it is an isomorphism.''
• Some people believe that there is a one-to-one correspondence between the subgroups of GL(2, Z₁₉) and the subgroups of Z₁₉^x. No. There is a one-to-one correspondence between the subgroups of Z₁₉^x and the subgroups of GL(2, Z₁₉) which contain SL(2, Z₁₉).
• The book describes Hasse diagrams so: Two points (subgroups) are joined by a line segment if and only if the lower subgroup is contained in the upper one. Some students drew horizontal diagrams like this, as if just to put subsets and oversets at the same level.

• Some students found the number of generators of Z₁₉^x. I don't know why.
• Somebody said that the subset {1, 2, ... 18} of Z₁₉ is a field.
• I'm quoting from a homework: ``By theorem of 21.1 there is an one to one correspondence between the set of all subgroups of GL(2, Z₁₉^*).''

(1) Let R be a ring, a, b, c, d elements of R. Evaluate (a + b)(c + d).

(2) Let R be a ring. If a² = a for all elements of R, show that R is commutative.

(3) Let R be a ring with identity. If (ab)² = a² b² for all a, b in R, show that R is commutative.

(4) Find a noncommutative ring R such that (ab)² = a² b² for all a, b in R.

No comment here.

§32 Ex. 3, 7, 10, 13

Neither here.

If R is a ring, prove that (Mat₂(R))[x] is isomorphic to Mat₂(R[x]).

• Some students examined the set of matrices with image

  1	0
  0	1

  to find the kernel of a ring homomorphism, whereas they should have examined those matrices with image

  0	0
  0	0

• Some people wrote functions on the left side.
• Some people still don't know when they ought to study whether a ``function'' is well-defined and when this question does not arise.
• If f, g are two arbitrary polynomials in R[x], you cannot always have f = S _{k = 0}ⁿ a_kx^k and g = S _{k = 0}ⁿ b_kx^k with the same n, but you can adjust notation and write these polynomials with the same number of terms.
• Most people used unbelievably long arguments to establish the isomorphism stated in the problem. Some failed to subsume the three cases deg f < deg g, deg f = deg g, deg f > deg g under a single notation. Some wrote polynomials as sequences and made their lives hard. Some wrote matrices like

  aij1	aij2
  aij3	aij4

  
  and had to go through a mess of arguments.
• If these people had spent five minutes in reading the proof of Lemma 33.9, they would have reduced the proof that a function preserves products of polynomials to the proof that it preserves products of monomials, saved themselves an hour's time, an hour's work, a lot of headache and five sheets of paper.
• Only one student gave a correct and concise answer. Five more points to him.

§35 Ex. 1, Ex. 2 (by Lagrange's interpolation formula).

• Some students found f = (x - a)q, but did not indicate which ring q lies in; some said that f is irreducible, but did not indicate whether f is irreducible in Q[x] or in Z[x].
• One student wrote that any rational number can be expressed as d/c, where d and c are natural numbers; (s)he should have said that d and c are (positive or negative) integers.
• Some students are not aware that, for a polynomial with coefficients from R, `irreducible over R' means the same thing as `irreducible in R[x]'. Don't say `irreducible over R[x]' when you mean `irreducible in R[x]'.
• Some students said that a polynomial f is irreducible in D[x] if and only if f is irreducible in D. But this is false. This is true for constant polynomials only, i.e., for elements of D only.
• Let f be a polynomial in Q[x]. Some students argued that if f is reducible over Q, then f must have a root in Q. This is false in general. Indeed, f is reducible over Q, then f has a nontrivial factor in Q[x]; wheras having a root in Q is equivalent to having a linear factor in Q[x]. Thus a biquadratic polynomial may be reducible over Q and yet may have no root in Q, for example if f is a product of two quadratic irreducible polynomials over Q.
  If f is reducible over Q and if f is of degree 2 or 3, then one of the nontrivial factors of f must be of degree 1, and then f will have a root in Q.
• The second problem was to be answered by Lagrange's interpolation formula.
• Some students made numerical errors while using that formula.

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