Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is problem 21 in chapter 3, section 1 of Dummit and Foote. I'm having a lot of trouble with it, so I was hoping anyone would help me out. There are two particular parts I'm struggling with:

Let $G=\langle x,y \mid x^4=1=y^4, yx=xy\rangle$ and consider $\overline{G}=G/\langle x^2y^2\rangle$

Exhibit each element of $\overline{G}$ in the form $\overline{x}^a\overline{y}^b$ for some integers $a, b$.

Prove that $\bar{G}\cong Z_4\times Z_2$.

So to do this I listed all 16 elements of $G$ and took each element and multiplied it by $x^2y^2$ and reduced according to the relations and saw what equivalences I had. I'll make a table:

$1, x, y$

$x^2, xy, y^2$

$x^3, y^3, xy^2, yx^2$

$x^2y^2, yx^3, xy^3$

$x^3y^2, x^2y^3$


Now, right multiplying by $x^2y^2$, I get

$1, x, y$

$xy, y^2, x^2$

$xy^2, yx^2, x^3, y^3$

$1, xy^3, yx^3$

$x, y$


Clearly I don't know or understand what I'm doing. I know that since $G$ is Abelian and $\langle x^2y^2\rangle$ is a (claimed) subgroup I don't know what all the elements of this subgroup are. Are they $\{1,x^2y^2\}$?, this subgroup must be normal and hence $G/\langle x^2y^2\rangle$ is well defined and forms a quotient group whose order must be $16/2=8$.

I don't know how to properly list the elements in the quotient group, especially in terms of $\overline{x}^a\overline{y}^b$. I attempted to do so in the table after I did this nonsense, but since I have 12 distinct "elements", I stopped bothering.

After all this is said and done, I still don't know how I'd go about showing it's isomorphic to $Z_4\times Z_2$. I see that the order is 8. I don't understand what is $Z_4\times Z_2$. It is the cartesian product of two groups, but it must be more than that since I am only dealing with one "component" in $G/\langle x^2y^2\rangle$. In fact, the problem starts by saying let $G=Z_4\times Z_4$ but be given by the relation above. I understand the relation, but I don't understand how it's isomorphic to $Z_4\times Z_4$.

I am sorry to ask two questions in one "stack exchange question", but they are so related.

share|cite|improve this question
up vote 2 down vote accepted

Every element of $G$ can be written uniquely as $x^ay^b$ with $0\leq a\lt 4$ and $0\leq b\lt 4$, giving you the 16 elements of $G$. Multiplication is by addition of exponents, modulo $4$.

Of these, the only elements in $\langle x^2y^2\rangle$ are $x^0y^0 = e$, and $x^2y^2$ (since $(x^2y^2)^2 = x^4y^4 = x^0y^0$). Let's call this subgroup $N$.

What are the elements of $G/N$? They are congruence classes modulo $N$. When are two elements of $G$ congruent modulo $N$? $x^ay^bN = x^ry^sN$ if and only if eiether $x^ay^b=x^ry^s$, or $x^ay^b = (x^ry^s)(x^2y^2)$.

So, for example, $x^1y^0N = x^3y^2N$; $x^2y^0N = x^0y^2N$; $x^3y^0N = x^1y^2N$; and so on.

So what is important is to keep track of which element you multiply by $x^2y^2$ and what element you get.

In the end, what are the eight different cosets? You can take any element of $G$ and get a coset: and if you add two to the exponents of $x$ and $y$ (modulo $4$), you get the other "name" for the coset. So we have:

  1. $\overline{x}^0\overline{y}^0 = x^0y^0N = x^2y^2N = \overline{x}^2\overline{y}^2$.
  2. $\overline{x}^1\overline{y}^0 = x^1y^0N = x^3y^2N = \overline{x}^3\overline{y}^2$.
  3. $\overline{x}^2\overline{y}^0 = x^2y^0N = x^0y^2N = \overline{x}^0\overline{y}^2$.
  4. $\overline{x}^3\overline{y}^0 = x^3y^0N = x^1y^2N = \overline{x}^1\overline{y}^2$.
  5. $\overline{x}^0\overline{y}^1 = x^0y^1N = x^2y^3N = \overline{x}^2\overline{y}^3$.
  6. $\overline{x}^1\overline{y}^1 = x^1y^1N = x^3y^3N = \overline{x}^3\overline{y}^3$.
  7. $\overline{x}^2\overline{y}^1 = x^2y^1N = x^0y^3N = \overline{x}^0\overline{y}^3$.
  8. $\overline{x}^3\overline{y}^1 = x^3y^1N = x^1y^3N = \overline{x}^1\overline{y}^3$.

$Z_4\times Z_2$ is the set of all pairs of the form $(a,b)$ with $a\in Z_4$ and $b\in Z_2$.

Remember that isomorphism does not mean they are identical, it just means that there is a way of identifying the two groups so that the operations "match up". So it doesn't matter that $G/N$ has "one component" and $Z_4\times Z_2$ has "two components"; that's just how we are writing them, not what group they are.

To give you a hint about what the identification might be, you'll want to define a function that maps an element of the form $\overline{x}^a\overline{y}^b$ in $G/N$ to an element of the form $(r,s)$ in $Z_4\times Z_2$. This identification must be such that if $\overline{x}^a\overline{y}^b$ maps to $(r,s)$, and $\overline{x}^c\overline{y}^d$ maps to $(t,u)$, then the product $$\overline{x}^{a+c\bmod 4}\overline{y}^{b+s\bmod 4}$$ must map to $(r,s)+(t,u) = (r+t,s+u)$.

As a further hint, look at the first column of my list above. Notice that the exponent of $\overline{x}$ is always between $0$ and $3$ (just like elements of $Z_4$) and that the exponent of $y$ is always either $0$ or $1$ (just like the elements of $Z_2$). Might the obvious thing to try work?

Now, the above is fine, but it doesn't give much intuition about what is "going on" (and I apologize for that... my only excuse is that it was late and I was tired).

It should be clear that elements of $G$ can be written as $x^ay^b$ with $0\leq a,b\lt 4$, with the group operation being "add exponents modulo $4$", $$x^ay^b\cdot x^ry^s = x^{a+r\bmod 4} y^{r+s\bmod 4}.$$ What does taking the quotient modulo $\langle x^2y^2\rangle$ do?

You can think of taking the quotient modulo $x^2y^2$ as "forcing $x^2y^2=1$ and seeing what you get". If you make $x^2y^2=1$, then you make $y^2 = (x^2)^{-1}=x^2$. That means that any time you have an element of the form $\overline{x}^a\overline{y}^b$, if you have $b\geq 2$, you can replace $\overline{y}^2$ with $\overline{x}^2$ and still have the same element.

So this immediately tells you that you can write each element of $G/\langle x^2y^2\rangle$ as $\overline{x}^a\overline{y}^b$, and restrict $b$ to $b\in\{0,1\}$: because if you had a square, you can replace it with an $x^2$ instead, and if you had $\overline{y}^3$, you can replace it with $\overline{x}^2\overline{y}$. In essence, you never need more than one $\overline{y}$ for any element. So you can certainly "cut down" on the $\overline{y}$s.

However, notice that the "obvious thing" I suggest you try first above the line does not actually work: it doesn't work because if you map $\overline{x}^0\overline{y}^1$ to $(0,1)$ in $Z_4\times Z_2$, then you would need its square to be trivial; but $$\overline{x}^0\overline{y}^1\cdot\overline{x}^0\overline{y}^1 = \overline{x}^0\overline{y}^2 = \overline{x}^2\overline{y}^0$$ which is not trivial. Instead we need to be a bit more careful; mapping $\overline{x}^1\overline{y}^0$ to $(1,0)$ is okay as far as powers of $\overline{x}$ go. Now, $(2,0)$ in $Z_4\times Z_2$ is twice $(1,0)$, but it is also equal to twice $(1,1)$, twice $(3,0)$, and twice $(3,1)$. Since $\overline{x}^2\overline{y}^0$ is the square of $\overline{x}^1\overline{y}^0$ and $\overline{x}^0\overline{y}^1$, $\overline{x}^3\overline{y}^0$, and $\overline{x}^2\overline{y}^1$, this suggests that if we map $\overline{x}^1\overline{y}^0$ to $(1,0)$, we should map $\overline{x}^1\overline{y}^1$ to $(0,1)$, since then we get $\overline{x}^2\overline{y}^1$ corresponding to $(1,1)$, $\overline{x}^3\overline{y}^0$ corresponding to $(3,0)$, and $\overline{x}^0\overline{y}^1$ to $(3,1)$.

This makes more sense with the realization that $\overline{y}^2$ is not trivial, but rather equal to $\overline{x}^2$.

share|cite|improve this answer
Wow, thank you so much for the detailed reply; I really appreciate it... With your hints, I guess I would say that $\overline{x}^a\overline{y}^b\mapsto (a (mod 4),b (mod 2))$? Since $(a,b)$ will be between $(0,0)$ and $(3,1)$ then, the group operation mod 4 and 2 for the individual coordinates. I was worried about just sending it to (a,b) since the far right column involves powers of 3 and 4 for $y$. Is this map more precise? Looking at the left or right column is just picking two different representatives for the same coset, I think. Again, thank you so much. – mathmath8128 Oct 6 '11 at 5:21
@aengle: Yes: the far right column are just "different names" for the same elements, so you can pick one set of "names" and stick to it; but you need to make sure that this map is a group homomorphism, and that it is bijective. – Arturo Magidin Oct 6 '11 at 5:31
@aengle: Please see the addition. Sorry if I caused any confusion last night. – Arturo Magidin Oct 6 '11 at 13:28
I appreciate your help on the homework problem. It's been long due now, but I still don't know the proper explicit mapping to $Z_4\times Z_2$. I was wondering if you wouldn't mind letting me know exactly what it was. I'm studying for the math GRE and this problem is a good reinforcement of computing quotient groups like this. Thanks a lot :-) – mathmath8128 Nov 5 '11 at 3:33
@aengle: The penultimate paragraph gives an explicit mapping. The class of $x$ goes to $(1,0)$; the class of $xy$ goes to $(0,1)$; the class of $x^2y$ goes to $(1,1)$; the class of $x^3$ goes to $(3,0)$; the class of $y$ goes to $(3,1)$. With these you should be able to figure out everything. – Arturo Magidin Nov 5 '11 at 4:34

I'll only try to clear up a small part of the confusion. ${\bf Z}_4\times{\bf Z}_4$ is the group of all ordered pairs $(a,b)$ where $a$ is taken from ${\bf Z}_4$ and $b$ is taken from ${\bf Z}_4$. It has an element $(1,0)$, which I'll call $x$, and an element $(0,1)$, which I'll call $y$. Notice that $x^4=1$ (where I'm writing the addition operation on ${\bf Z}_4$ as if it were multiplication), and $y^4=1$, and $xy=yx$, exactly the relations defining $G$. Does that help you see how $G$ is isomorphic to ${\bf Z}_4\times{\bf Z}_4$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.