You've already turned this exam in?This task was on my exam
Hi, lately i started posting a lot in this part of the forum (group theory mainly), cause i had this course on faculty and im not that good at abstract maths. but im trying to tackle some basics down so i can get my self into it more.., now let's get down to the task i wanted to ask you folks here:
This task was on my exam and i partially solve it but i need to know if the rest of it im doing ok:
Task: Let Z[i] = {a+bi | a,b E Z} is set of complex numbers with the normal addition and multiplication defined for complex numbers, check the following:
1)Is Z[i] a ring?
2)Does Z[i] posses divisors of zero?
3)Is Z[i] a field?
4)Is (x-y)(x+y)=x^2-y^2 for every element of Z[i] (give explanation)
5)Does Z[i] posses finite cyclic subgroup?
My explanation:
First as the task states the set is discrete since the real and imaginary part belongs to Z.
1)For (Z[i], +, *) to be ring (+ addition, * multiplication defined standard for complex numbers) by definition
a. Z[i], + must be abelian group
b. Z[i], * must be semigroup (* posses associativity law on the elements of Z[i])
c. Distribution law for * over + must hold in the structure
Z[i] is abelian group!
-first set closure: let x,y E Z[i] such that x=a+bi and y=c+di for some a,b,c,d E Z
now x+y=(a+bi)+(c+di) = (a+c)+i(b+d) , a+b E Z, b+d E Z => x+y E Z
- associativity: let x,y,z E Z[i] x and y defined as previously and z = e+fi, e,f E Z
now x+(y+z) = (a+bi)+((c+di)+(e+fi)) = (a+bi)+((c+e)+i(d+f)) = a+(c+e) + i(b+(d+f)) = (a+c)+e + i((b+d)+f)) = ((a+c)+i(b+d))+(e+if)=((a+ib)+(c+id))+(e+if)=(x+y) +z = > associativity holds.
- identity element in Z[i], + is 0, let x E Z[i] => x+0=(a+bi)+(0+i0)=(a+0)+i(b+0)=a+ib=x.
- inverse element to x is -x that is -(a+ib)=-a-bi, indeed: x+(-x) = (a+ib)+(-a-ib)=
(a+(-a))+i(b+(-b))=(a-a)+i(b-b)=0+i0=0 (the id in Z[i] according to +).
so from all above => Z[i] is group under op. + for complex numbers, now all left over is to prove that Z[i] is closed under * and distribution law for * over + holds..
Z[i] is closed under * indeed, let x,y E Z => (a+ib)*(c+id)=ac+adi+bci-bd=ac-bd+i(ad+bc), ac-bd E Z, as well as ad+bc => x*y E Z[i], thus is closed.
now distribution laws:
let, x,y,z E Z[i], then x*(y+z) = x*((c+id)+(e+if)) = x*((c+e)+i(d+f))=(a+ib)*((c+e)+i(d+f))=(a(c+e)+ia( d+f) + ib(c+e)-b(d+f)) = (a(c+e)-b(d+f))+i(a(d+f)+b(c+e))
now the other side of the identity:
x*y+x*z = (a+ib)*(c+id)+(a+ib)*(e+if)=ac-bd+i(ad+bc)+ae-bf+i(af+be)=ac+ae-bd-bf+iad+ibc+iaf+ibe=a(c+e)-b(d+f)+i(a(d+f)+b(c+e)) and this is equal to the result bit upper so distribution law holds, thus Z[i],+,* is ring.
Now that aside, 1) is complete i think this one i did ok, i just wrote the whole work cause i think i can use this fact that Z[i] is ring to solve part 4) which most concerns me and i didnt solve it on the exam cause i didnt even look at it dont know whu :P,
the second one divisors of zero are simply x*y=0 but x and y are not equal to 0.
(a+ib)*(c+id)=0+i0 => ac-bd = 0, ad+bc = 0 => ac=bd and ad=-bc, from where a=bd/c and b=ac/d => d=-bc/(bd/c)=> -bc^2/bd and similar you express c and find the values of all a,b,c,d such that x*y=0 for x,y different then 0.. nvm this one is not that hard to do i guess..
Now task 3) about field i know how to do it, now about task 4, i tought of it this way:
Since i prove that Z[i] is ring follows that Z[i],+ is abelian group thus there's inverse element to every element in Z[i] so the cancellation laws hold in Z[i] now, i can take it down this way (im not sure if this is the right approach):
(x-y)*(x+y)=x^2-y^2, now for x^2 -x^2 is inverse according to +, so i apply it on the left side for both sides of the identity => -x^2+(x-y)(x+y)=-x^2+x^2-y^2 =>
cancellation law kicks in => -x^2+(x-y)(x+y)=-y^2, now for -y^2 inverse is y^2 now i apply y^2 on right sides on both sides of the identity => -x^2+(x-y)(x+y)+y^2=e (the identity or 0), now -x^2+(x-y)(x+y)+y^2=-x^2+x^2+xy-yx-y^2+y^2, cancellation law now again kicks in thus => +xy-yx = e, now cause for normal multiplication commutativity law holds => +xy-xy=e => e=e, so indeed everything cancelled on the long run, but i dont know if this is consistent as proof, i need some help from you guys to assure me if im doing ok, thanks!!
And for the last one, can i just say that there's no generator in Z[i] thus there cant be finite cyclic subgroup..
for question number 5...have you thought about powers of -1?
for question number 4....by the distributive law: (x+y)(x-y) = (x+y)x - (x+y)y = xx + yx - xy + yy = x^2 + yx - xy + y^2.
so the question really is: does yx - xy = 0? can you answer that question?
what's all about "have you turned the exam".. i just wrote what is the task and i said i didnt do it all and i need to do it all as exercise, about the second replay: what you mean powers of -1 can you be more descriptive, the set is complex numbers i dont get it.., btw about 4) didnt i write the whole procedure and asked is my answer correct? cause at the end i get xy-yx = e, since Z[i] is ring there must be inverse of yx which is ofc another element of Z, so the inverse is -yx, vice versa yx is inverse of -yx so by the cancellation law you get xy-yx+yx=e+yx => xy=yx...
And if you replay me next be more descriptive, not everyone on this forum is guru
if this is homework, we are not allowed to do it for you. and you can get warned, infracted or banned for posting homework questions (according to the mods).
one has a (ring) isomorphism of Z into Z[i], given by n-->n+0i. so one can consider -1 as an element of Z[i]: -1+0i.
by powers of -1, i mean: -1+0i, (-1+0i)^2, (-1+0i)^3, etc. (multiplicative powers).
... it is not homework it is task that i had on my exam and now im doing exercises cause i want to take the exam again (that is abstract algebra course) so i resolve those task and some similar to those (anyway who is going to school/faculty in the summer??, there is just this exams in the summer that i can redo to get better grade on faculty etc..), and if you cant give me straight answer (evaluate my work) then no problem, btw i give my explanation you could just tell me if im doing ok about that what i did so far, i didnt ask for you to solve it for me -.-, after all the name of the forum is "math help forums", did i ever wrote "solve my problem", why you ppl just dont read what is written in the post i did? Anyway thanks for replays, and i was most concern about task 4) for which i was not sure if im doing ok!
Deveno: (EDIT: I meant goroner; very sorry) In all but your first post, your tone comes off as very aggressive and confrontational; people are just trying to help you and not get themselves in trouble. It seems like English is not your first language so I'm not going to be too critical, but I'm just letting you know that the tone you're sending is quite negative.
So the only question left is 5?
There seems to be a misunderstanding about the meaning of the question. "Does Z[i] possess a finite cyclic subgroup?" means "Is there a which is finite and cyclic?"
I'm sorry if i sounded rude, but you don't seem to read my posts well as i can see, i never asked for you to solve my tasks nor i want to, i want to get the topics down by my self cause i will use this in my life later on thus i must understand the essence of it.Anyway my mother language is not English but i think i'm not that bad with it, about my latter posts i never meant to be rude and i don't want people to think of me being rude, but why don't you just try to be more descriptive when you posting something, as i said in my posts not everyone is guru in maths and this forums is all about help, is it?, help doesn't mean solving ones tasks but evaluating ones work and guide one if one is trying to solve something but looking the wrong way and not help one that neither tried to start solving a task and just wait for others to solve it, and i think i'm not of that kind.Don't get me wrong here, i just try to explain it cause i don't want to be black sheep on the forum, and i read the rules, rule number 13 speaks it all...
About task 5, as it's stated one need to find if Z[i] posses finite cyclic subgroup that is G proper subset of Z[i], such that G = <a>, or more precisely G={a^n | n E Z}, as you guys wrote about -1, i don't get the point how come -1 generates Z[i], ye -1 is generator indeed but in Z, but Z[i] = {a+ib | a, b E Z}, so i can't say -1 is generator, Z[i] is not cyclic, Z[i] proper subset of C, and C is not cyclic, but as a theorem states every cyclic group has it's subgroups to be cyclic, if this is the case for Z[i] it would be easy to say that every possible subgroup of Z[i] is cyclic, but i can't tell that, so this kinda confuse me a bit..
Thanks for help!
it is not my intention to "be aggressive or confrontational". i have made two posts, one in which i gave some hints (and apparently was "ok in tone"), and one in which i endeavored to a) explain what i meant by powers of -1, and b) to give the original poster some information about the forum rules. i want to add that those rules do not necessarily reflect my personal feelings about what is, or is not appropriate; but since i am posting here, i have given my (tacit) acceptance to abide by them.
to the original poster: about number 4) i assure you that you can refer to "e" (the additive identity) as "0" (which is customary for a ring) without further explanation, because this identity is unique, and the traditional symbol for a ring additive identity. i will suggest that instead of starting with what you wish to prove:
(x+y)(x-y) = x^2 - y^2
and then deriving 0 = 0
you ought to "run the proof backwards" to show that 0 = 0 --> (x+y)(x-y) = x^2 - y^2.
in point of fact, this identity holds for any commutative ring, which is what yx - xy = 0 says (if x,y are arbitrary, then our ring has to be commutative).
perhaps i overlooked something, but i did not see a demonstration that multiplication is indeed commutative in your posts. this isn't terribly difficult, and it essentially reduces to the fact that multiplication is commutative in Z. but it is the crucial step in showing 4), every other part follows from distributivity.
finally, concerning question 5: Z[i] itself is neither finite, nor cyclic. however; {-1,1} is a multiplicative subgroup of the multiplicative semi-group (Z[i],*).
and as is readily verified, it is a cyclic group of order 2, generated by -1. {i,-1,-i,1} is a cyclic group of order 4 (generated by either i or -i). i will agree that
the wording of question 5 is vague-some indication of what the group operation is, should be indicated. also ({0}, +) is a cyclic subgroup of order 1, of the
additive group. so the answer to 5) is "yes".
self-study is the kind of thing these forums ARE designed to help with. and no one can fault you for wanting to learn the material better, so as to improve your chances on future exams. but if you read the forum rules, you will see that academic integrity is something taken very seriously on these forums. so if some answers of mine, or of anyone else on these forums seem unclear, it is not because we assume you are a "math guru" but because the intention is to help, not "do for".
and you have to understand, often people cut and paste their entire homework questions here (10 questions at a time). your first post merely said: "this was on my exam" making it unclear if these were study questions, homework, exam questions on a future exam, or questions on a previous exam. in the first case and fourth cases, forum rules allow us to help considerably, in the second and third cases, we ought to say as little as possible, the work should be your own.
Thanks for the replay, and now let me explain:
The task is from past exam, and i took the paper with me cause the professor said that we can take the papers, so i try to solve those tasks and do other tasks from a book too, neither is homework nor future exam (how could i possibly take future exam's task from the professor, except the professor is in tight with me but that's not the case nor i want to take tasks an prepare my self just for the exam, as i said i want to get into this more not just the exam stuff..)..
Now enough about the forum rules and all the other stuff, i read the rules and i'm not here to break them, but contribute to mhf if i can someday, and help others if i can..
About the subgroups, are trivial groups counted in the number of subgroups, i mean, if there's no other subgroup but the trivial, cause of course trivial i.e. {0} is subgroup to every group considered operation +, or the group it self, then i must consider these two trivial subgroups, in this case for Z[i], + only cyclic and finite is {0} as you said, i knew this but i wasn't quite sure if those count cause in the book i read i can't find any example similar to this, and for the multiplicative
i quote: "{-1,1} is a multiplicative subgroup of the multiplicative semi-group (Z[i],*)." , how come you find subgroup for a semi-group ehm , you meant Z[i],* is group i'm sure, cause it must contain the inverse elements for every element in Z[i] according to *, and {i, -i, 1, -1} could be subgroup, and subgroup of Z[i], *, but Z[i], * is not group, cause you can't find inverse to 0+i0, and here another note i got: no subgroup is possible for structure that is neither group it self...
Thanks again for the useful post!
Deveno: OH! I'm so embarrassed: I got the names wrong. I apologize profusely for my mistake. Let me try this again:
goroner: That message was intended for you, as you realized. For the record, your most recent posts have been much nicer, and I, at least, thank you for the explanations
it is quite possible to find a subgroup of a semi-group. the normal example is the group of units of a ring.
for example, in Z6*, {1,5} form a subgroup under multiplication modulo 6.
perhaps saying: "subgroup of a semi-group" seems non-sensical. what i mean by such is: sub-semi-group of a semi-group, that is also a group.
in fact, that is what we have here: the units of Z[i] are {1,-1,i,-i}. i am not saying (Z[i],*) is a group, for example, 2 = 2+0i has no multiplicative inverse.
Okay no more questions from me i got what i needed now thread can be closed if mods. like to and big thanks to all that replayed and helped me out, i got a lot of infractions top there, it's fine with me you might thought i was being rude, but i never intended to sound like i'm rude, i didn't point anyone and i'm sure i didn't break the rules which i read about 3 times i think, i just asked for you guys to be more descriptive and if this question is rude fine i wont ask it anymore, might be my lack of picking the right words but that's me..
I'm not quick minded and i need better explanation then someone else might need, so i just try to learn.
Thanks, see you soon!
Since Deveno hasn't made it perfectly clear, I will add that "subgroup of a semigroup" is a genererally accepted term in semigroup theory. (And it is one of those irritating little problems in terminology. I don't know how many times I mistook "subgroup" for "subsemigroup" or the other way around. Deveno's hyphens seem very reasonable. I think I will try to stick to them. Thanks, Deveno.)