# Group theory problem

1. Jan 22, 2008

### ehrenfest

[SOLVED] group theory problem

1. The problem statement, all variables and given/known data
Classify the factor group (Z_4 cross Z_4 cross Z_8)/<(1,2,4)> according to the fundamental theorem of finitely generated abelian groups.

2. Relevant equations

3. The attempt at a solution
<(1,2,4)> has order 4 so the factor group has order 32, so there are seven possibilities:

(Z_2)^5
Z_32
(Z_2)^3 cross Z_4
Z_16 cross Z_2
Z_8 cross Z_2 cross Z_2
Z_2 cross Z_4 cross Z_4
Z_8 cross Z_4

Anyone have any ideas about how to do this without doing a lot of tedious calculations?

2. Jan 22, 2008

### Hurkyl

Staff Emeritus
It seems evident to me: your relation says:

(1, 0, 0) + (0, 2, 0) + (0, 0, 4) = (0, 0, 0)

*shrug*

Maybe rewriting it as a quotient of Z³ would help?

3. Jan 22, 2008

### ehrenfest

What is "it"?

4. Jan 22, 2008

### masnevets

Let a,b,c be the generators of Z/4, Z/4, and Z/8, respectively. Then modding out by (1,2,4) is the same as saying that a=-2b-4c. So this means that a is completely determined by b and c, and hence doesn't matter anymore. Now the question is whether b and c can be whatever they want (between 0 and 3 and 0 and 7, respectively) and give different elements. I'll leave the rest to you.

5. Jan 22, 2008

### ehrenfest

I am kind of confused about this. Does Z/4 mean the same thing as Z_4 i.e. the integers mod 4? If so, then a=b=c=1, and 1 is not equal to -6.

6. Jan 22, 2008

### ehrenfest

Can someone please elaborate on what masnevets is saying?

7. Jan 22, 2008

### ehrenfest

anyone? this problem is killing me!

8. Jan 23, 2008

### ehrenfest

9. Jan 23, 2008

### masnevets

Yes, Z/4 is the integers modulo 4. Yes, 1 is not equal to -6, but you're confusing what I mean now. a, b, and c mean (1,0,0), (0,1,0), and (0,0,1), respectively.

10. Feb 2, 2008

### ehrenfest

So, you're just saying that the class of (1,0,0) is in the same as the class of (0,-2,-4) ? I agree. Those two elements are clearly in the same coset. But how does that help you classify the quotient group according to the fundamental theorem of finitely generated abelian groups?!

11. Feb 2, 2008

### NateTG

The natural homomorphism from the group to the qotient is going to be onto, so the image of a set of generators is a set of generators.

12. Feb 2, 2008

### ehrenfest

I am not sure I have seen that proof...

But you're saying that the (1,0,0)+<(1,2,4)>,(0,1,0)+<(1,2,4)>,(0,0,1)+<(1,2,4)> will generate the quotient group?

Can someone just give me a concrete instruction so that I can make some progress classifying this group!

Maybe I need to use the Fundamental Homomorphism Theorem...to use this I need to find a group G' and a homomorphism phi such that ker(phi)=<(1,2,4)>. How would I figure out what G' is and what phi is though...

Last edited: Feb 2, 2008
13. Feb 3, 2008

### ehrenfest

anyone?

14. Feb 3, 2008

anyone?