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Math Help - Cyclic group

  1. #1
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    Cyclic group

    Is (Z3 x Z3,+) cyclic group. Explain
    thanks
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  2. #2
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    Re: Cyclic group

    in Z3 x Z3, 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?
    Thanks from kljoki
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  3. #3
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    Re: 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?
    Last edited by kljoki; August 11th 2012 at 05:07 PM.
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  4. #4
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    Re: 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.
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  5. #5
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    Re: Cyclic group

    I have another questions:

    If i was given (Z3 x Z3,*) how can i check order of elements??
    thanks
    Last edited by kljoki; August 12th 2012 at 04:11 AM.
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  6. #6
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    Re: 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 |Z3xZ3| = |Z3|*|Z3| = 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.

    32 = 9, so the order of 3 is not 2.
    34 = 92 = 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 Z2xZ4. convince yourself of the following two facts:

    a) Z2xZ4 has no element of order greater than 4
    b) Z2xZ4 has at least one element of order 4

    how many elements of order 2 does Z2xZ4 have?
    how many elements of order 4?

    if those questions seem easy, try this: how many different subgroups does Z2xZ4 have? can you find a subgroup of order 4 with no elements of order 4 in it?
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  7. #7
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    Re: Cyclic group

    Quote Originally Posted by Deveno View Post
    a) Z2xZ4 has no element of order greater than 4
    b) Z2xZ4 has at least one element of order 4

    how many elements of order 2 does Z2xZ4 have?
    how many elements of order 4?
    so we have (Z2xZ4,*) 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 = (an(mod 2), bn(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
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  8. #8
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    Re: 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 ZmxZn, 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 Z2xZ4.

    unfortunately, Z2xZ4 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)
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  9. #9
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    Re: 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 (an(mod 3),bn(mod 5)) = (1,1) when the element have (0,b)??
    what should i do here??
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  10. #10
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    Re: Cyclic group

    (Z3xZ5,*) is NOT a group! (0,0) has no inverse. the best we can hope for is ((Z3-{0}) x (Z5 -{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 Z2xZ4, 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, Zn 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, Zn-{0} is a group under multiplication. this only happens when n is prime.

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

    but, back to the original question:

    consider Z2xZ4, 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?
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  11. #11
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    Re: Cyclic group

    first thanks for everything now
    Quote Originally Posted by Deveno View Post
    consider Z2xZ4, 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?
    3 elements of order 2 (0,2),(1,0) and (1,2)

    next
    Quote Originally Posted by Deveno View Post
    how many of order 4?
    4 of order 4 (0,1),(0,3),(1,1) and (1,3)

    Quote Originally Posted by Deveno View Post
    how many subgroups does this group have?
    so Z2xZ4 nave order 8,any subgroup P of Z2xZ4 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 0xZ4 {(0,a) : a in Z4}
    1xZ4 {(1,a) : a in Z4}

    |P| = 8 Z2xZ4

    Quote Originally Posted by Deveno View Post
    does there exist a subgroup of order 4, with no elements of order 4?
    don't know
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  12. #12
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    Re: Cyclic group

    Quote Originally Posted by kljoki View Post
    first thanks for everything now


    3 elements of order 2 (0,2),(1,0) and (1,2)
    this is correct.

    next


    4 of order 4 (0,1),(0,3),(1,1) and (1,3)
    this is also correct.


    so Z2xZ4 nave order 8,any subgroup P of Z2xZ4 must have order that divide 8..... that subgroups are 1,2,4,8
    |P| = 1 {(0,0)}
    so far, so good.
    not sure about this one
    |P| = 2 {(0,0),(1,1)}
    {(0,1),(1,0)}
    {(0,1),(1,1)}
    {(0,2),(1,3)}
    no. EVERY subgroup of ANY group must ALWAYS contain the identity. so a group of order 2 looks like: {e,x}. since |x| has to divide 2, and since x isn't the identity, |x| isn't 1, this means the non-identity element of a group of order 2 has to be an element of order 2. conversely, if we have a group G, and an element x in G of order 2, then {e,x} is a subgroup of G (i urge you to verify this directly). so the subgroups of order 2 in a group G are in 1-1 correspondence with the elements of order 2:

    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)}

    |P| = 4 0xZ4 {(0,a) : a in Z4}
    1xZ4 {(1,a) : a in Z4}
    well, you got one subgroup of order 4 correct:

    {(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....

    |P| = 8 Z2xZ4


    don't know
    yes, there is only one subgroup of order 8.

    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).
    Last edited by Deveno; August 16th 2012 at 11:57 AM.
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  13. #13
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    Re: Cyclic group

    Quote Originally Posted by Deveno View Post

    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....
    the missing element is (0,2)

    Quote Originally Posted by Deveno View Post
    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).
    its {(0,0),(0,2),(1,0),(1,2)}
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  14. #14
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    Re: Cyclic group

    yes, and now you know everything about Z2xZ4 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.
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