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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:

$1, x, y$

$x^2, xy, y^2$

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

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

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

$x^3y^3$

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

$1, x, y$

$xy, y^2, x^2$

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

$1, xy^3, yx^3$

$x, y$

$xy$

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.

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2 Answers 2

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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 $\overline{x}^2$.

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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 ${\bf Z}_4\times{\bf Z}_4$?

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