# Factor Group Computations

1. Nov 17, 2009

### Newtime

For those of you with "A First Course in Abstract Algebra 7th ed.," I'm on chapter 15 and in particular, the first 12 questions. Basically, I understand very well how to classify any factor groups of finite groups, but it's these factor groups of infinite groups that just seem so arbitrarily classified. Here's a few examples that exemplify my problem:

8. (Z x Z x Z) / <(1,1,1)>. That is, Z x Z x Z "over" the subgroups generated by (1,1,1). Now the answer according to the book is that this factor group is isomorphic to Z x Z since every coset of the factor group contains a unique element of the form (0,m,n) where m and n are integers. This reasoning makes sense but then why can't I use the same reasoning and say that each coset also has a unique element of the form (0,0,m) or (m,n,q) thus making the factor group isomorphic to Z or Z x Z x Z respectively?

11. (Z x Z) / <(2,2)>. According to the book, this factor group is isomorphic to Z2 x Z. If I'm given the answer I can come up with a reason why it's true: the factor group contains an element of order 2 and an element of infinite order... but not only is this logic flimsy at best, I also would have no idea how to come up with the answer in the first place.

12. (Z x Z x Z) / <(3,3,3)> The answer here is Z3 x Z x Z with much of the same logic/problems as number 11. above.

Any help is appreciated, these problems have been bugging me.

2. Nov 18, 2009

### Newtime

Noone?

3. Nov 18, 2009

### Jose27

8.- Let $$(a,b,c)+H$$ be a coset induced by $$H:=<(1,1,1)> \subset \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$$ we want $$(0,d,e) \in (a,b,c)+H$$ but this happens iff $$(0,d,e)-(a,b,c) \in H$$ iff $$(a,b-d,c-e) \in H$$ so we have $$b-d=a$$ and $$c-e=a$$ so taking $$d=b-a$$ and $$e=c-a$$ we get our element and the fact that it's unique is rather easy: Assume $$(0,x,y) +H=(0,z,w)+H$$ then $$x-z=0=y-w$$.

Now take $$(a,b,c)+H$$ we want an element $$(0,0,d) \in (a,b,c)+H$$ and reasoning as before we have $$a=b=0$$, $$d=c$$ which clearly isn't satisfied by all elements $$(a,b,c)$$ so not ALL cosets have an element of the form $$(0,0,d)$$. On the other hand if you want an element $$(d,e,f)$$ we get $$a-d=b-e=c-f$$ and we loose the fact that in each coset the element $$(d,e,f)$$ is unique.

Another way to tackle this problems is to note that $$<(1,1,1)> \cong \mathbb{Z} \times 0 \times 0$$ , $$<(2,2,2)> \cong 2\mathbb{Z} \times 0$$ and in general $$<(k,...,k)> \cong k\mathbb{Z} \times 0 \times ... \times 0$$ (where there are n k's and (n-1) 0's ) and with a little knowledge of direct products and quotient groups you get:

$$\frac{ \mathbb{Z} ^n }{ k_1\mathbb{Z} \times ... \times k_n\mathbb{Z} } \cong \mathbb{Z} _{k_1} \times ... \times \mathbb{Z} _{k_n}$$

4. Nov 19, 2009

### Newtime

Thanks, this was quite helpful.

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