Homework Solutions

Here you'll find solutions to the homeworks. When possible, we'll put up solutions from your peers. Give us time; the solutions will go up gradually.

HW 1
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  (from Ellen Li). Some common issues:
  1. Very few people bothered to show associativity for the group with 2 elements; but once you exhibit a bijection with Z/2Z (for example) that respects the group operation, you can conclude associativity (because you know that addition for Z/2Z is associative).
  2. We almost never use the adjective "homomorphic" in math. So for instance, we don't say "a function is homomorphic," we say "a function is a homomorphism." We also never say two groups are homomorphic, whatever one might mean by that. (This may be related: Can you prove that any group admits a homomorphism to another group?)
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  (from a student). There are two main points to this problem:
  1. In this context, power series are purely formal and do not need to correspond to functions on the real numbers, so there is no need to worry about Taylor expansions or convergence issues.
  2. The inverse power series can be constructed as long as the first coefficient is nonzero since we need to divide by it, and there's an explicit, recursive way to generate the coefficients of an inverse power series in terms of the original coefficients.
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  (from a student). This one was very tricky so don't worry at all if you didn't get it. The key points here are that S_n is generated by transpositions/swaps, and that since H is abelian, phi must send all transpositions to the same element in H.
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  (from Luke Melas-Kyriazi). For (a), lots of people gave finite groups for G which doesn't work, because for H = Z, there would only be one group homomorphism G to Z, namely the zero homomorphism (this is a good exercise), while Z is infinite. For (b), this is a very challenging problem given what we know. This kind of group is not like anything we have seen in this course so far: it's called the free group on two generators. If you want to learn more about free groups, you can check this Links to an external site. out.
HW 2
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  (from Dan Fu).
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  (from a student)
  1. By far the most common issue was failing to prove in part (c) that the function G/G_x-->O_x given by sending gG_x --> gx is well-defined. A coset can be written as gG_x in multiple ways, and you need to prove that they all give you the same answer. (Not proving injectivity was also common but less so.)
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HW 3
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  (from Ellen Li)
  1. One of the more common problems here was attempting to show that there was no isomorphism in part (d) by taking a particular homomorphism (or a homomorphism only of a specific form) and showing it is not a bijection, or taking a particular bijection and showing it is not a homomorphism. You need to show that there are no functions that are both bijections and homomorphisms, not that just some particular functions don't work.
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HW 4
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  (from Caleb Miller)
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HW 5 (20 Questions): See here.

Midterm One
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4. Traces and Group Actions on {1, ..., n}. Solutions Download Solutions
  (from Natalia Pacheco-Tallaj)
  1. Some people neglected to show rho indeed took values in GL_n(R).
  2. Some people gave a new name to the homomorphism (like "phi") when "rho" was already the name. (Nothing was deducted, just pointing out.)
  3. In (c), "this is the average number of fixed points" was a little too obvious an interpretation. A surprising interpretation is that this average is actually the number of orbits of the G action!
    1. How would you have come up with this? If you experiment with an element g in S_n, you would have found that this average is actually the number of cycles in the cycle notation for g---but that's precisely the number of orbits in the action of <g> on the set {1, .. , n}.
  4. Some people had a hard-time articulating that matrix multiplication "is" function composition. We'll soon see that there's a ring homomorphism from M_{nxn}(R) to End(R); this is one way of articulating the sense in which matrix mult. is compatible with composition of functions.
  5. There seems also to be a confusion about the term "well-defined." Ironically, "well-defined" is a common term in math whose meaning is not always clear. When, in general, does one have to show "well-definedness" of a function f? When f involves choices.
    1. As an example, in the function m: G/H x G/H --> G/H, where do we send two cosets I and J? Well, we first choose an element g1 in I, and g2 in J, so we can write I =g1H and J = g2H. After choosing such elements, we then declared m(I,J) to equal the coset (g1g2)H. Because this m(I,J) involved choices g1 and g2, we need to show that m doesn't depend on these choices---if we choose g1' and g2' to be elements in I and J, respectively, we have to show that (g1'g2')H = (g1g2)H to show that m is indeed "well-defined"--i.e., choice-independent.
  6. Points: (a), (b) worth 5 points. (c) worth 10 points.
5. Counting Homomorphisms. Solutions Download Solutions
  (from Hiro) and a student's Solutions Download a student's Solutions
  (from Emily Saunders).
  1. To show that a function f is a bijection, some people tried to show f is an injection, and a surjection. Keep in mind that sometimes, it's much easier to exhibit a two-sided inverse to f---that is, some function g such that gf = id and fg = id.
  2. In hindsight, part (f) led you through an inefficient solution; that A(mn) = A(m)A(n) is most easily seen by exhibiting an bijection between A(m) and the set of numbers coprime to m. I apologize for the length of the road some of you had to walk down for (f). But you can take a look at the solutions (from Hiro) above to see a not-too-bad way of proving (f).
  3. A homomorphism from G to G need not be an automorphism. In particular, A(m) is not the same thing as |hom(Z/mZ,Z/mZ)|.
  4. Some submissions still used the adjective "homomorphic." We almost never use this term in the math community--this is just how the culture has turned out, and I have no wisdom to share except that this adjective sounds a bit jarring to read because I am not used it.
  5. A lot of people try to enumerate elements of G: "Let G be a group, and consider its elements g_1, g_2 , ...." Unless a choice of bijection from G to this indexing set {1, 2, ... } is *necessary* for your problem, you should avoid this. If you later need to choose an element of g, just say "Let g be an element of G." You don't need to enumerate. And if you need to relate an element of G to an element of H, just choose a function phi: G --> H and write, "Let h = phi(g)."
    1. One reason to avoid choosing an index for elements of G: For instance, what if G is uncountable? Then notation like {g_1, g_2 , ...} is misleading. Moreover, G indexes itself, so you are most likely using cluttering notation unnecessarily.
  6. Please don't ever call the trivial homomorphism the "identity homomorphism." The identity homomorphism is the identity function id: G --> G sending g to g. This is not trivial unless G has only one element.
  7. Z is cyclic. (It can be generated by one element, namely 1 or -1.)
  8. Not every finite group is cyclic. (See: S_3, S_4, S_5, ... , D_{2n}, Q_8, ...)
  9. A lot of people said "These have the same size, therefore there is a bijection" without ever showing two sets have the same size. In general, when you have no a priori information on the sizes of your sets, you have to show a bijection to conclude two sets have have the same cardinality.
  10. Points: (a)-(e) worth 3 points. (f) worth 5 points.
6. (Optional.) More mattresses. Solutions Download Solutions
  (from Hiro)
  1. Points: (a),(b),(c) worth 2 points. (d) worth 4 points.
7. (Optional.) Hypermattresses. Solutions Download Solutions
  (from Hiro)
  1. Points: (a),(b), and (c) worth 3, 3, and 4 points, respectively.

HW 6
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  (from Brian Sapozhnikov)
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HW 7
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  (from Anne Carlstein)
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HW 8
HW 9

Midterm Two Solutions Download Solutions
(from Hiro). Here are some common mistakes.
1. For part (a), several people said that homomorphisms must preserve the order of elements in the group. This is false. (It's true for isomorphisms though.) Also, a number of people said that $LaTeX: G/\ker\phi$ $G/\ker\phi$ is a subgroup of G. It's not; it's a quotient group of G. (This is a bit orthogonal to the problem itself, but you should know this.)
2. For part (a), it was commonly claimed that there's a map $LaTeX: R/\phi^{-1}(I)\to S$ $R/\phi^{-1}(I)\to S$ by the universal property of quotient groups. This is not necessarily true: the universal property only gives us this map if $LaTeX: \phi^{-1}(I)\subset\ker\phi$ $\phi^{-1}(I)\subset\ker\phi$ . Also, several people were under the impression that a coset of I in S is of the form $sI$ . It's actually of the form $s+I$ . Remember, the group operation on a ring is addition! (Similarly, the cosets of $LaTeX: \phi^{-1}(I)$ $\phi^{-1}(I)$ in R are of the form $LaTeX: r+\phi^{-1}(I).$ $r+\phi^{-1}(I).$ For part (b), by far the most common mistake (which also happened in many cases in part (a)) was to treat $LaTeX: \phi^{-1}$ $\phi^{-1}$ as a function $LaTeX: S\to R$ $S\to R$ . It's not--this notation doesn't actually mean the inverse of $LaTeX: \phi$ $\phi$ , it just takes an element x and gives the *set* of elements that map to x. So saying something like $LaTeX: \phi^{-1}(x)\phi^{-1}(y)\in \phi^{-1}(I)$ $\phi^{-1}(x)\phi^{-1}(y)\in \phi^{-1}(I)$ is not a statement that makes sense, as the objects being "multiplied" on the left are actually sets.
3. Part (a): The point here is that the only extra condition you need to satisfy (beyond being a subgroup) is closure under scaling--that is, showing that for all subgroups $LaTeX: M'\subset M$ $M'\subset M$ and elements $LaTeX: r\in\mathbb{Z}/p\mathbb{Z}$ $r\in\mathbb{Z}/p\mathbb{Z}$ , $LaTeX: \forall m\in M',rm\in M'.$ $\forall m\in M',rm\in M'.$ By far the most common mistake was not realizing that this is necessary to show. It's nontrivial, however--if you replace $LaTeX: \mathbb{Z}/p\mathbb{Z}$ $\mathbb{Z}/p\mathbb{Z}$ by, say, $LaTeX: \mathbb{R}$ $\mathbb{R}$ , this part is false! For part (b), an extremely common mistake was to claim that all subgroups of A are given by choosing $LaTeX: \mathbb{Z}/p\mathbb{Z}$ $\mathbb{Z}/p\mathbb{Z}$ or 0 in each component. This is false: Consider the subgroup generated by $LaTeX: (1,1,\ldots,1).$ $(1,1,\ldots,1).$ This subgroup is not 0 on any component, but it's definitely not all of A. (It only contains elements with all components equal.)

HW 10