# Group theory problems

1. Jun 17, 2005

### Oxymoron

Question 1 a

Find the order of

$$(\mathbb{Z}_4 \times \mathbb{Z}_2) \backslash \langle (2,1) \rangle$$.

Question 1 b

Find the order of

$$(\mathbb{Z}_2 \times \mathbb{Z}_4) \backslash \langle (1,1) \rangle$$.

2. Jun 17, 2005

### Oxymoron

I was attempting these two questions from Section 14 of "A First Course in Abstract Algebra" by John Fraleigh (where I get all my questions from in fact).

Although the two questions were slightly different, I was unable to distinguish them due to my lack of insight.

For (a), I can write out $\mathbb{Z}_4\times\mathbb{Z}_2[/tex] as $$\mathbb{Z}_4\times\mathbb{Z}_2 = \{ (0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1) \}$$ Since I am asked to compute the factor group, I want to be able to picture a factor group in the same way as 'picking' bits out. In other words we have the 8 elements in $\mathbb{Z}_4\times\mathbb{Z}_2[/tex] and I want to factor out [itex]\langle (2,1) \rangle$. Now $\langle (2,1) \rangle$ is the subgroup generated by $(2,1)$. So $$\langle (2,1) \rangle = \{ (2,1),(4,2),(6,3),(8,4),\dots\}$$ But $(4,2) \equiv (0,0)$, $(6,3) \equiv (2,0)$, $(8,4) \equiv (0,0)$, $(10,5) \equiv (2,1)$, etc. So $$\langle (2,1) \rangle = \{ (0,0),(2,1),(2,0) \}$$ What am I missing? Shouldn't this have 2 elements? EDIT: The set $\mathbb{Z}_4\times\mathbb{Z}_2$ is now ammended. Last edited: Jun 17, 2005 3. Jun 17, 2005 ### Muzza That looks more like Z_2 * Z_4 ;) The computation of <(2, 1)> is indeed faulty. You've written that (6, 3) = (2, 0), which is wrong. It should be (6, 3) = (2, 1), so that <(2, 1)> consists of (0, 0) and (2, 1). And btw, you don't need to explicitly calculate the factor groups. Last edited: Jun 17, 2005 4. Jun 17, 2005 ### Oxymoron AHA!! Geez I make some pretty stupid mistakes! Ok so now I have $$\langle (2,1) \rangle =\{ (0,0),(2,1) \}$$ So $$|\langle (2,1) \rangle | = 2$$ Therefore the order of the factor group is $$\frac{|\mathbb{Z}_4 \times \mathbb{Z}_2|}{|\langle (2,1) \rangle|} = \frac{8}{2} = 4$$ 5. Jun 17, 2005 ### Oxymoron For part b we now have $$\mathbb{Z}_2 \times \mathbb{Z}_4 = \{ (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)\}$$ Which is obviously slightly different to $\mathbb{Z}_4 \times \mathbb{Z}_2$. And now $$\langle (1,1) \rangle = \{ (1,1),(2,2),(3,3),\dots\}$$ $$= \{ (0,0),(1,1),(2,0),(3,1)\}$$ So the orde of the factor group is 2. The purpose of this exercise was to illustrate that factor groups of finite groups can be done by 'picking' out what you dont want. 6. Jun 17, 2005 ### Oxymoron Ive shown you the way that I do it - it is rather long and exhausting (particularly with more complicated examples), could you show me exactly how you would do these problems? (just one will suffice). 7. Jun 17, 2005 ### AKG The questions you've been given have simply been to find the order of the factor groups. It should seem obvious to you that for a finite group G, |G/H| = |G|/|H|. So in your case, you just need to find the order |H|. You're given the generator of H in your problems, like (1,1). It's not hard to see that this has order 4, and that G has order 4 x 2 = 8, so you know |G/H| = |G|/|H| = 8/4 = 2. 8. Jun 17, 2005 ### Oxymoron Yes I do see that G has order 8. But it is not easy for me, at this stage, to see that something like (1,1) has order 4, or (2,1) has order 2. THIS is what I lack. Some of my friends, and you, for instance, can just look at that and know its order...so how do you do it? 9. Jun 17, 2005 ### AKG I don't know, maybe do a bunch of problems and it will start to come naturally two you. Well, I can give you the ideas but it's up to you to practice with them so they become natural. First, groups like $\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \dots \times \mathbb{Z}_{n_k}$ are abelian. Secondly, any element $(a_1, \dots , a_k)$ is just $(a_1, 0, \dots , 0) + (0, a_2, 0, \dots , 0) + \dots + (0, \dots , 0, a_k )$. Finally, if you have an element $x_1x_2\dots x_k$, or, using additive notation, $x_1 + x_2 + \dots + x_k$, the order of this element is the least common multiple of the order of each of the $x_i$. So, for example: (1,1) = (1,0)(0,1) [OR if you like additive notation, (1,0) + (0,1)] The order of (1,0) in $\mathbb{Z}_2 \times \mathbb{Z}_4$ is just the order of 1 in $\mathbb{Z}_2$, and the order of (0,1) in $\mathbb{Z}_2 \times \mathbb{Z}_4$ is just the order of 1 in $\mathbb{Z}_4$. Those numbers are, respectively, 2 and 4, whose least common multiple is 4. If we're looking at (1,2), then we have (1,2) = (1,0) + (0,2) and the order of (1,0) in $\mathbb{Z}_2 \times \mathbb{Z}_4$ is the order of 1 in $\mathbb{Z}_2$, which is 2, and the order of 2 in $\mathbb{Z}_4$ is also 2. The least common multiple of 2 and 2 is 2. So it depends on how quickly you can find the order of a single number in a single cyclic group, and how quickly you can find the least common multiple a pair (or more) of numbers. It's pretty easy to do the first thing. If you want to find the order of x in $\mathbb{Z}_n$, just compute n/gcd(n,x). Most of the time, though, you can just look at it and tell. You can tell easily that 2 has order 2 in $\mathbb{Z}_4$. Finding the lcm of some numbers is also generally not too tough. If you need a precise method for finding it, you can ask, or look it up at mathworld, but in most cases, it's something you can tell by looking at it, again, like the lcm of 2 and 4. 10. Jun 18, 2005 ### Oxymoron Perfect, that is exactly what I needed!!! One more question concerning factor groups. Say we factor the group $H$ from the group $G$. So we have $$G\backslash H$ Now, how come the order of $H$ always seems to be a divisor of $G$? I mean, if it wasn't, then the factor group would be impossible - because you cant have half an element. This lead me to believe that $H$ must have a special property. That is, the order of $H$ must be a divisor of $G$. But how can we guarantee that in all these factoring problems, that our $|H|$ will divide $|G|$ with no remainder? Would you care to explain this phenomenon? Do we only allow $H$ to be normal? As in the left and right cosets coincide? 11. Jun 18, 2005 ### AKG This is Lagrange's Theorem. A subgroup of a finite group G has order that divides |G|. The proof uses the idea of cosets. Suppose you have a subgroup H. Now take some g in G that isn't in H. gH must have the same size as H, obviously, and the two must be disjoint. Suppose they weren't disjoint, then for some h in H, and some h' in H, we'd have: gh = h' this being the element in the intersection of gH and H. But this gives us: $g = h'h^{-1}$ and since H is a subgroup, $h'h^{-1} \in H$, which means g is in H, contradicting our choice of g. Continuing this process, we must eventually fill out all of G with distinct disjoint cosets of H. It should be clear from here why any subgroup H has order that divides |G|. 12. Jun 18, 2005 ### Oxymoron Thankyou AKG, I will look into that. You're explanations are worth their weight in gold! (especilly since I have finals coming up). Question 2 Find all prime ideals of $\mathbb{Z}_6$ To find a prime ideal $N$ of $\mathbb{Z}_6$ I would begin by finding ideals such that by finding factor rings $\mathbb{Z}_6/N$ that are integral domains. In my opinion, all prime ideals will contain 0. Because by factoring out 0 from $\mathbb{Z}_6$ is a good start to assuring $\mathbb{Z}_6/N$ is an integral domain - because an integral domain is a commutative ring with unit $1\neq 0$ containing no divisors of 0. Now Im not sure how to continue. Last edited: Jun 18, 2005 13. Jun 18, 2005 ### Oxymoron A prime ideal of $\mathbb{Z}_6$ is [tex]N_1 = \{ 0,2,4 \}$$

Since (note that elements of $(\mathbb{Z}_6)$ commute under multiplication)

$$0\times 2 = 0\times 4 = 0 \in N_1$$

$$2\times 2 = 4 \in N_1$$

$$2\times 4 = 8 \equiv 2 \in N_1$$

$$4\times 4 = 16 \equiv 4 \in N_1$$

Therefore $N_1$ is a prime ideal. The set

$$N_2 = \{ 0,3 \}$$

is also a prime ideal for exactly the same reasons. My question now becomes: How do you find these prime ideals without trial-and-error? I mean, it took me a while to work out which elements belong to the prime ideal because I calculated every one of them - surely there is an easy way. And also, I found 2 prime ideals - is there a way of knowing that you found them all?

Last edited: Jun 18, 2005
14. Jun 18, 2005

### Oxymoron

For a illustration as how this method become tedious. Consider $\mathbb{Z}_{12}$. A prime ideal is

$$N_1 = \{ 0,2,4,8 \}$$

How do I know it is maximal? How do I know there is or isn't any more? If $N_1$ is a prime ideal, then isn't

$$N_2 = \{ 0,2,4 \}$$

also a prime ideal? Surely there must be an easier way that to going through all different multiples of two numbers and checking to see if they belong in the prime ideal or not.

15. Jun 18, 2005

### Hurkyl

Staff Emeritus
N1 isn't even an ideal, let alone a prime ideal.

Remember the definition of ideal -- adding any two things in an ideal is again in an ideal. Also, multiplying a number in the ideal by a number of the ring is again in the ideal!

16. Jun 18, 2005

### AKG

This is the first time I'm seeing prime ideals, but I'll try to answer your question. An ideal is a subset of your ring that forms an additive group, so we want to look at subgroups of $\mathbb{Z}_6$ where we treat $\mathbb{Z}_6$ as the additive group. These subgroups are just, <0>, <1>, <2>, <3>. Next, we want it so that whenever x is in $\mathbb{Z}_6$ and y is in our ideal, both xy and yx are in the ideal. But this is true for all those subgroups, so they're all ideals. For prime ideals, it must be that whenever xy is in the ideal, then either x or y are. <0> doesn't work. If x = 2, y = 3, then xy = 0, but neither of x nor y are in <0>. <1> works because it's all of $\mathbb{Z}_6$. Since 6 is even, if xy is in <2>, then x*y must be even, so one of x or y must be even, so one of them must be in <2>, so <2> is a prime ideal. Note I'm using x*y to denote normal multiplication, and xy to denote ring multiplication, which is multiplication modulo 6. Finally, if xy is in <3>, then x*y is a multiple of 3. If x*y = 0, then one of x or y is zero, so one of them is in <3> since zero is in <3>. Otherwise, x*y is a positive multiple of 3. Since 3 is prime, then one of x or y must be a multiple of 3, and since x and y are elements of {0,1,...,5}, then one of them must be 3 itself, and hence in <3>.

Look here:

http://mathworld.wolfram.com/PrimeIdeal.html

The second sentence is:

For example, in the integers, the ideal $\mathfrak{a} = \left \langle{p}\right\rangle$ is prime whenever p is a prime number.

Here, we weren't dealing with $\mathbb{Z}$ but $\mathbb{Z}_6$, but I'd guess that a similar result holds.