Understanding Factor Groups of Infinite Groups: Examples and Solutions

[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.

 

[Newtime]

Noone?

 

[Jose27]

8.- Let (a,b,c)+H be a coset induced by H:=&lt;(1,1,1)&gt; \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 &lt;(1,1,1)&gt; \cong \mathbb{Z} \times 0 \times 0 , &lt;(2,2,2)&gt; \cong 2\mathbb{Z} \times 0 and in general &lt;(k,...,k)&gt; \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}

 

[Newtime]

8.- Let (a,b,c)+H be a coset induced by H:=&lt;(1,1,1)&gt; \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 &lt;(1,1,1)&gt; \cong \mathbb{Z} \times 0 \times 0 , &lt;(2,2,2)&gt; \cong 2\mathbb{Z} \times 0 and in general &lt;(k,...,k)&gt; \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}

Thanks, this was quite helpful.