[Fraleigh, p.133, ex. 13.7] Let $f_i: G_i \rightarrow G_1 \times G_2 \times \dots \times G_r$ be given by $f_i(g_i) = (e_1, e_2, ..., g_i, ..., e_r),$ where $g_i \in G_i$ and $e_j$ is the identity element of $G_j$.
Injection map. Compare with Example 13.8. $$\begin{align} f(ab) & = (e_1, e_2, ..., ab, ..., e_r) \\ & = (e_1, e_2, ..., a, ..., e_r)(e_1, e_2, ..., b, ..., e_r) \\ & = f(a)f(b) \end{align}$$

[Fraleigh, p.135, ex. 13.53] Let $h, k \in \text{group } G$ and let $\phi: \mathbb{Z \times Z} \rightarrow G$ be defined by $\phi(m, n) = h^m k^n$. Give a necessary and sufficient condition, involving $h$ and $k$, for $\phi$ to be a homomorphism. Prove your condition. enter image description here

Update Fev. 3 2014: (1.) I understand $\langle \mathbb{Z}, +\rangle$ is a group and $\langle \mathbb{Z}, \times\rangle$ isn't. I meant to ask why the question in (53.) said nothing about what it chose for the binary operation for $\mathbb{Z}$?. Shouldn't the question notify the reader about this? How do readers know the question chose integer addition? What if readers use some other binary operation that still induces $\mathbb{Z}$ to be a group?

(2.) Questions say nothing about binary operation in $G$.
Hence in (7), why are you authorized to flesh out $ab$ like that? In (53.), why vector addition?

(3.) In (53.), how do you envisage the necessary condition $hk = kh$ is the sufficient condition too?

Update: (4.) What's the proof strategy? How do you envisage and envision to start by calculating $\phi(1,0), \phi(0,1)$? Then you magically envisage to calculate $\phi(1,1)$ in 2 ways?
Then this magically induces the necessary condition for homomorphism.

(5.) Shouldn't $f$ be $f_i$? Question (7.) never defined $f$.

Update: (6.) (54.) is the next question. But isn't it just (53.)? (53.) proves the necessary and sufficient condition is $hk = kh$ for all arbitrary $h,k \in G$. But this $\iff G$ Abelian.

(54.) Find a necessary and sufficient condition on G such that the map in (53.) is a homomorphism for all choices of $h, k \in G.$

  • $\begingroup$ Sorry about the long post. I'm posting (7.) and (53.) together because I want to see the connection between binary operations for different homomorphisms. Why's the textbook authorized to choose them randomly? $\endgroup$
    – NNOX Apps
    Jan 26, 2014 at 9:26

1 Answer 1


(In the following, for the sake of simplicity I'm only going to discuss products of two groups. The generalization to more factors is immediate.)

Given groups $G$ and $H$, in order for it to be meaningful to talk about $G\times H$ as a group rather than simply as a set, you need to have a group structure defined. You can't talk about "the group" $G\times H$ until you have (or at least someone has) defined what group structure it's supposed to carry.

By convention, in the absence of any special indication in the surrounding context of what group structure you're meant to impose, the notation $G\times H$ indicates that you should take $(g_1, h_1)\cdot(g_2, h_2) := (g_1 g_2, h_1 h_2)$. When you take the set $G\times H$ and endow it with this particular group operation, the resulting group is what is referred to as the direct product of $G$ with $H$. (Amazon shows that, at least in the 6th edition of the book, direct products are discussed in Fraleigh section 2.4 (page 128).)

(1.) According to the above discussion, it's not the case that you're doing "vector addition" in $\mathbb{Z}\times\mathbb{Z}$ just out of the blue for no good reason. Rather, the group structure on $\mathbb{Z}$ is addition, so when we see the notation $\mathbb{Z}\times\mathbb{Z}$ without any further indication what group operation we should use, we understand that to mean we should impose the group operation coming from the direct product of $\mathbb{Z}$ with $\mathbb{Z}$, which in this case will correspond to coordinatewise addition.

Update to (1.) I will let you consider for yourself how addition can be interpreted as a group operation on the integers. It will be a good exercise for you in verifying that something satisfies the group axioms. Note that the fact that $\mathbb{Z}$ also happens to carry a multiplication is irrelevant. The group axioms don't say a group can't carry additional structure. They only specify certain minimal structure a set is required to possess to qualify as a group. Incidentally, algebraic objects that carry a commutative group structure along with a compatible (associative) multiplication that distributes over it in the same way multiplication distributes over addition have a special name. They're called rings. You'll find them discussed later on in Fraleigh. (Chapter 5 in the 6th edition.)

Here are some questions to guide your consideration of how Z endowed with addition can count as a group: First and foremost, group operations are supposed to be associative. Is integer addition associative? Once you've got that out of the way, what will play the role of the additive identity? Given a number $n$, what will play the role of its additive inverse?

Updated update to (1.) In modern mathematical discourse, especially in the context of abstract algebra, you will find that the notation $\mathbb{Z}$ is used pretty much exclusively to refer to the group (or ring) of integers with respect to the usual operations. This has been the case for decades, presumably influenced directly by Bourbaki's decision to use it in their text on algebra, Book II of their grand project. (To be more precise, I believe Chapter 1 of Bourbaki's Algèbre first appeared in 1942, and the integers are denoted there as $\mathbf{Z}$, which is, as far as I care to discuss the matter, essentially the same thing as $\mathbb{Z}$.) This, in turn, may have been influenced by Edmund Landau's use of $\overline{\mathfrak{Z}}$ for the integers in Grundlagen der Analysis in the 1930s. In any case, no matter where it came from, it is widely accepted that the Z stands for Zahl or Zahlen, which is German for number or numbers, respectively.

Why do the integers get to have their own dedicated notation? There is no mystery here. The integers constitute one of the most important and fundamental constructions in all of mathematics. If any algebraic object deserves its own notation, I would think the integers should.

Even apart from these broader considerations, however, just considering Fraleigh's text as a universe unto itself, if you scan through pages 1 through 135 I believe you will find that the first occurrence of the notation $\mathbb{Z}$ will describe that it means the group $\langle \mathbb{Z}, +\rangle$, and every subsequent occurrence will reinforce this interpretation. By page 135, further reminders to the reader that the notation still means what it has always meant up to that point would come off as redundant, superfluous, or unnecessarily repetitive. :-P

(2.) This is an immediate consequence of how the group operation is defined in a direct product.

(3.) The sufficiency of $hk = kh$ is explained in the section beginning "Conversely, suppose that $hk = kh \dots$" I could say more on this point, but it's probably not necessary. I suspect that once you've reviewed and gotten comfortable with the definition of direct products, the explanation you already wrote down will be clear enough.

(4.) $\phi(m,n)$, as a map of sets, is defined to be $\phi(m,n) = h^m k^n$. Therefore, by definition, $\phi(1,0) = h^1 k^0 = h$. Similarly, we find that $\phi(0,1) = k$. I'm not sure what else to say on this point. Could you say more about which part of that explanation/unpacking of the definition leaves you feeling dissatisfied?

Update to (4.) Every element in $\mathbb{Z}$ can be written as some multiple of the generator $1$. Similarly, the observation being made here is that every element in $\mathbb{Z}\times\mathbb{Z}$ can be written as some multiple of $(1,0)$ plus some multiple of $(0,1)$. Whenever you're confronted with a group about which it is practical think in terms of generators (in particular, this is the case if the group is finite, or, more generally, finitely generated), once you know what happens to the generators, you know everything there is to know: Every other element in the group is just going to be some (not necessarily unique) finite product of the generators and their inverses (that's what I mean by "generators"). So you think to examine what happens to (1,0) and (0,1) because they generate ZxZ. You think to look at the 2 different ways you can write (1,1) because you're leveraging a relation that holds in ZxZ, the domain of $\phi$, in order to force a constraint on what happens to the corresponding images of those elements over in $G$, and (1,1) is the simplest element in ZxZ that makes that possible.

This is one of the fundamental strategies in modern mathematics: You have some objects with some structure. You consider maps between two such objects that you require to respect that structure. Then you use the fact that the map is supposed to respect the structure to describe characteristics the map will be forced to exhibit.

(5.) I don't have a copy of Fraleigh handy. Supposing what you've transcribed is the whole story, it seems clear we're meant to understand from context that $f = f_i$ where $i$ could be any index from 1 to $r$. If I'm right about this, I wouldn't interpret the omitted subscript as a typo. Rather I would see it as an attempt by the author to make things clearer by simplifying notation, which maybe didn't work so well in this instance.

(6.) $h$ and $k$ weren't arbitrary in (53.). The work you transcribed for (53.) indicates quite clearly that $h$ and $k$ need to satisfy a certain condition in order for $\phi$ to be a homomorphism.

Update to (6.) In (53.), $\phi$ is defined in terms of a specific selection of $h$ and $k$ in $G$. Maybe it will be easier to process conceptually if I denote it as $\phi_{h,k}$. When it is first defined in (53.), $\phi_{h,k}$ is only guaranteed to be a map of sets; it is not yet necessarily a homomorphism of groups. Then the argument proceeds to show that $\phi_{h,k}$ is a homomorphism if and only if the $h$ and $k$ you picked commute with one another. If you pick $h$ and $k$ that do not commute with one another, you're out of luck, $\phi_{h,k}$ is not a homomorphism of groups, it is merely some map between sets.

(54.), in contrast, wants you to characterize groups that satisfy the condition that $\phi_{h,k}$ defined as in (53.) would be a homomorphism regardless of what $h$ and $k$ you might pick. I agree with your conclusion on what this says about $G$.

  • 1
    $\begingroup$ Thanks so much. Upvoted. Exquisite answer! I really appreciate the detail and time that you very nicely put in. Upvoted your other posts. $\endgroup$
    – NNOX Apps
    Feb 5, 2014 at 8:44

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