Is (Z_{3 }x Z_{3},+) cyclic group. Explain :)

thanks

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- Jun 17th 2012, 03:31 AMkljokiCyclic group
Is (Z

_{3 }x Z_{3},+) cyclic group. Explain :)

thanks - Jun 17th 2012, 12:46 PMDevenoRe: Cyclic group
in Z

_{3}x Z_{3}, we define (a,b) + (a',b') as (a+a' (mod 3), b+b' (mod 3)).

this group has 9 elements, so it is cyclic if and only if it has an element of order 9.

can you prove no element has order greater than 3? - Aug 11th 2012, 05:03 PMkljokiRe: Cyclic group
i cant prove :/

but i think i can show :)

so the elements are {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),( 2,2)}

to check order i should multiply each element with number that gives me (0,0)(mod 3)

(0,0) has order 1

(0,1)*3=(0,3) = (0,0)(mod 3) order 3

(0,2)*3=(0,6) = (0,0)(mod 3) order 3

(1,0)*3=(3,0) = (0,0)(mod 3) order 3

(1,1)*3=(3,3) = (0,0)(mod 3) order 3

(1,2)*3=(3,6) = (0,0)(mod 3) order 3

(2,0)*3=(6,0) = (0,0)(mod 3) order 3

(2,1)*3=(6,3) = (0,0)(mod 3) order 3

(2,2)*3=(6,6) = (0,0)(mod 3) order 3

so there isn't element with order bigger then 3

am i correct? - Aug 11th 2012, 06:09 PMDevenoRe: Cyclic group
that is certainly one way to prove it (exhaustively).

i would merely note that:

(a,b) + (a,b) + (a,b) = (3a (mod 3), 3b (mod 3)) = (0,0), so that the maximum possible order is 3. - Aug 12th 2012, 04:05 AMkljokiRe: Cyclic group
I have another questions:

If i was given (Z_{3}x Z_{3},*) how can i check order of elements??

thanks :) - Aug 12th 2012, 09:08 AMDevenoRe: Cyclic group
well, clearly the only element of order 1 is the identity, (0,0) (this is true of any group).

we know that the order of an element has to divide the order of the group, in this case since |Z_{3}xZ_{3}| = |Z_{3}|*|Z_{3}| = 9, an element can only have order 1,3, or 9. this means any non-identity element can only have order 3 or 9.

but as we saw above, the maximum order is 3, so *every* non-identity element has order 3. this is the power of theorems such as lagrange, they allow us to deduce such things without actually calculating the order of specific elements.

in general, determining an element's order is not always an easy thing to do. for example, take the group U(20) = {1,3,7,9,11,13,17,19} (under multiplication mod 20). this has order 8, so the only orders we need to check are 1,2,4 and 8. let's say we want to know the order of 3.

3^{2}= 9, so the order of 3 is not 2.

3^{4}= 9^{2}= 81 = 1 (mod 20), so the order of 3 is 4.

it turns out that U(20) = U(4)xU(5), which is isomorphic to Z_{2}xZ_{4}. convince yourself of the following two facts:

a) Z_{2}xZ_{4}has no element of order greater than 4

b) Z_{2}xZ_{4}has at least one element of order 4

how many elements of order 2 does Z_{2}xZ_{4}have?

how many elements of order 4?

if those questions seem easy, try this: how many different subgroups does Z_{2}xZ_{4}have? can you find a subgroup of order 4 with no elements of order 4 in it? - Aug 14th 2012, 05:43 AMkljokiRe: Cyclic group
so we have (Z

_{2}xZ_{4},*) with 8 elements {(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)}. Now every element (a,b)^{n}= (a^{n}(mod 2), b^{n}(mod 4) = (1,1) for n=1,2,4,8

(0,0) has order 1(identity)

now (0,1)^{n}always the answer will be (0,1) so the order is ???

and element (1,2)

(1,2)^{2}= (1(mod2),4(mod4)) = (1,0) order??

same with

(1,2)^{4}and (1,2)^{8}

so i'm stuck :( - Aug 14th 2012, 02:59 PMDevenoRe: Cyclic group
oops! in abelian groups like the one i gave, the operation is usually written

*additively*(and the identity element is a 0, not a 1).

so instead of (a,b)^{n}= (a,b)*(a,b)*....*(a,b) (n times) one writes:

n(a,b) = (a,b) + (a,b) +...+ (a,b) n times.

in Z_{m}xZ_{n}, the first coordinate operation is addition mod m, and the second coordinate operation is addition mod n:

(a,b) + (a',b') = ((a+a') (mod m), (b+b') (mod b)).

for example: (1,0) + (1,3) = (0,3) in Z_{2}xZ_{4}.

unfortunately, Z_{2}xZ_{4}is not a group under "point-wise multiplication". this is because 4 is not a prime number, so:

(0,2) has no inverse: (the identity would have to be (1,1))

(0,2)*(0,0) = (0,0)

(0,2)*(0,1) = (0,2)

(0,2)*(0,2) = (0,0) <---this is bad

(0,2)*(0,3) = (0,2)

(0,2)*(1,0) = (0,0) <---also bad

(0,2)*(1,1) = (0,2)

(0,2)*(1,2) = (0,0) <---and still more bad stuff

(0,2)*(1,3) = (0,2)

as you can see there isn't ANY (a,b) with (0,2)*(a,b) = (1,1) - Aug 16th 2012, 03:46 AMkljokiRe: Cyclic group
ok another thing (Z3 x Z5,*), 3 and 5 are prime numbers so we can have multiplication

(0,0),(0,1),(0,2),(0,3),(0,4) will have order 1 or what??

how can (a^{n}(mod 3),b^{n}(mod 5)) = (1,1) when the element have (0,b)?? :D

what should i do here?? - Aug 16th 2012, 07:42 AMDevenoRe: Cyclic group
(Z

_{3}xZ^{5},*) is NOT a group! (0,0) has no inverse. the best we can hope for is ((Z_{3}-{0}) x (Z_{5}-{0}),*), which *is* a group.

its elements are as follows:

(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4) (we have 8 of them)

and we multiply like so:

(a,b)*(a',b') = ((aa' mod 3), (bb' mod 5)).

so, for example: (2,1)*(2,4) = (4 (mod 3), 4 (mod 5)) = (1,4).

as it turns out, this group is isomorphic to Z_{2}xZ_{4}, under addition, but the isomorphism is NOT this mapping:

(a,b) → ((a-1),(b-1)) (going from the * group to the + group).

suppose that mapping WAS a homomorphism:

we would have to have:

(a,b)*(a',b') → ((a-1),(b-1)) + ((a'-1),(b'-1)) = (a+a'-2,b+b'-2).

let a = a' = 1, b = b' = 2.

then (1,2)*(1,2) = (1,4) which maps to (0,3).

but (1,2) maps to (0,1), and (0,1) + (0,1) = (0,2) ≠ (0,3).

*************************

in general, Z_{n}is not a group under multiplication (mod n), for n > 1. why? because 0 never has an inverse. 0*k = 0. always. so 0*k never equals 1. sometimes, Z_{n}-{0} is a group under multiplication. this only happens when n is prime.

*************************

but, back to the original question:

consider Z_{2}xZ_{4}, where the operation is "+" defined like so:

(a,b) + (a',b') = ((a+a' mod 2),(b+b' mod 4))

how many elements of order 2 are there? how many of order 4? how many subgroups does this group have? does there exist a subgroup of order 4, with no elements of order 4? - Aug 16th 2012, 10:51 AMkljokiRe: Cyclic group
first thanks for everything now

3 elements of order 2 (0,2),(1,0) and (1,2)

next

4 of order 4 (0,1),(0,3),(1,1) and (1,3)

so Z_{2}xZ_{4}nave order 8,any subgroup P of Z_{2}xZ_{4}must have order that divide 8..... that subgroups are 1,2,4,8

|P| = 1 {(0,0)}

not sure about this one

|P| = 2 {(0,0),(1,1)}

{(0,1),(1,0)}

{(0,1),(1,1)}

{(0,2),(1,3)}

|P| = 4 0xZ_{4}{(0,a) : a in Z_{4}}

1xZ_{4}{(1,a) : a in Z_{4}}

|P| = 8 Z_{2}xZ_{4}

don't know :) - Aug 16th 2012, 11:55 AMDevenoRe: Cyclic group
this is correct.

Quote:

next

4 of order 4 (0,1),(0,3),(1,1) and (1,3)

Quote:

so Z_{2}xZ_{4}nave order 8,any subgroup P of Z_{2}xZ_{4}must have order that divide 8..... that subgroups are 1,2,4,8

|P| = 1 {(0,0)}

Quote:

not sure about this one

|P| = 2 {(0,0),(1,1)}

{(0,1),(1,0)}

{(0,1),(1,1)}

{(0,2),(1,3)}

x ↔ {e,x}

you found 3 elements of order 2. therefore, you should be able to come up easily with the corresponding subgroups of order 2:

{(0,0), (0,2)}, {(0,0), (1,0)} and {(0,0), (1,2)}

Quote:

|P| = 4 0xZ_{4}{(0,a) : a in Z_{4}}

1xZ_{4}{(1,a) : a in Z_{4}}

{(0,0),(0,1),(0,2),(0,3)} is indeed a subgroup of order 4. however:

{(1,0),(1,1),(1,2),(1,3)} is *not*, it isn't even closed under addition: (1,1) + (1,1) = (0,2), which isn't in this set. also, this set does *not* contain the identity.

here is one way:

suppose a subgroup of order 4 contains an element of order 4, call it (a,b). then this subgroup also contains:

(a,b) + (a,b)

(a,b) + (a,b) + (a,b)

(a,b) + (a,b) + (a,b) + (a,b) = (0,0) (since (a,b) is of order 4).

it is easy to see that (a,b) + (a,b) has order 2. what is the order of 3(a,b) = (a,b) + (a,b) + (a,b)?

well:

2(3(a,b)) = 6(a,b) = 4(a,b) + 2(a,b) = (0,0) + 2(a,b) = 2(a,b), and 2(a,b) = (a,b) + (a,b) is not the identity.

3(3(a,b)) = 9(a,b) = 4(a,b) + 4(a,b) + (a,b) = (a,b), which is not the identity.

4(3(a,b)) = 12(a,b) = 4(a,b) + 4(a,b) + 4(a,b) = (0,0) + (0,0) + (0,0) = (0,0), so 3(a,b) also has order 4.

this means that if we have a subgroup of order 4, we get 2 elements of order 4, either of which generates the subgroup. you found 4 elements of order 4, so that should give you the suspicion that there are at least 2 different subgroups of order 4:

{(0,0),(0,1),(0,2),(,0,3)} <---you found this one. note that this takes care of the two elements of order 4 (0,1) and (0,3).

the other subgroup should contain "the other two elements of order 4", namely, (1,3) and (1,1). and of course, a subgroup has to contain the identity. so that gives us 3 elements:

{(0,0),(1,3),(1,1),...?}

what could the missing element be? well, by closure, both (1,1) + (1,1) and (1,3) + (1,3) have to be in our subgroup. let's hope they're the same....

Quote:

|P| = 8 Z_{2}xZ_{4}

don't know :)

it might appear we are done. however, there is one more subgroup we haven't found yet (hint: it is of order 4, but has no elements of order 4). - Aug 16th 2012, 01:37 PMkljokiRe: Cyclic group
- Aug 16th 2012, 04:17 PMDevenoRe: Cyclic group
yes, and now you know everything about Z

_{2}xZ_{4}there is to know:

it is of order 8.

it is abelian.

it is not cyclic (no elements of order 8).

it has 3 cyclic subgroups of order 2, 2 cyclic subgroups of order 4, and 1 non-cyclic subgroup of order 4.