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Math Help - proof about the order of subgroup

  1. #1
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    proof about the order of subgroup

    Here is a simple question and I have the answer for it but I have trouble understanding the answer.


    Question.

    Suppose that G is a group with |G|=27. Prove that G contains a subgroup H satisfying |H|=3.


    Answer.

    Clear so far: applying Lagrange theorem, possible orders of subgroups are 1, 3, 9, 27. Possible orders of elements are the same (1, 3, 9, 27). There can only be one element of order 1, namely the identity element, so all other elements msut have order 3, 9 or 27.


    Then it gets muddy.

    "Let x be any non-identity element of the group. If x has order 3, we can take H=<x>." - then |H|=3?? because H={x^1, x^2, x^3}?

    "If x has order 9, we can take H=<x^3> and this will satisfy |H|=3 since x^3 will have order 3." - is that because H={x^1, x^2, x^3} again? so that we can multiply 9 by 1, 2 or 3 to get to a number less than 27.

    "If x has order 27, we may take H=<x^9>" - same logic as in the previous statement, I guess.

    Thanks for your comments!
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  2. #2
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    The thing is that an element of order 3 will always generate a subgroup of order 3.

    If x has order 3, then x^3 = e, where e is the identity. The group generated by x consists of all powers of x, and is thus {e,x,x^2}. Notice that you forgot the identity in your set, while you included x^3 (which is equal to the identity, so strictly speaking you didn't forget it after all ).

    If x has order 9, then x^3 has order 3. Indeed, we know that x^9=e. Since (x^3)^3 = x^9=e, we see that x^3 has order 3, whereas the subgroup generated by x^3 will be of order 3. This subgroup is {e,x^3,x^6}.

    If x has order 27, then x^9 has order 3. The subgroup generated by x^9 is thus {e,x^9,x^18}.
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  3. #3
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    Thank you very much HappyJoe! It is all clear now.

    You are right, I forgot that e had to be in eery subgroup! Even though it was there by virtue of being a power of x )))

    So, just to sum up for myself, what we did here is: we took each element of order that divides the group order, and specified a subgroup generated by each such element. Once we list all elements of this sugroup, we see that the order of such subgroup is 3.
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  4. #4
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    not quite. any element of a group of finite order, say x of order n, generates a subgroup of order n, <x> = {e,x,x^2,...,x^(n-1)}.

    so if x is of order 9, <x> is of order 9. but the special thing about the subgroup <x>, is that it is cyclic.

    and it is a special property of cyclic groups, that if G is cyclic, and d divides |G|, that G has one (and only one, which is besides the point,here)

    subgroup of order d. if G = <x>, then the subgroup of order d is <x^(|G|/d)>.

    it is somewhat easier to immediately comprehend this fact if you see that any (finite) cyclic group G is isomorphic to (Zn,+), for n = |G|.

    for example, Z9 = {0,1,2,3,4,5,6,7,8}. 3 divides 9, and the only subgroup of order 3 in Z9 is {0,3,6}.

    notice the similarity between x of order 9 so that <x> = {e,x,x^2,x^3,x^4,x^5,x^6,x^7,x^8} and <x^3> = {e,x^3,x^6}.

    so we didn't see that the order of the subgroup generated by any element is 3, but rather there is a subgroup of the subgroup generated by the element x,

    that is of order 3. if x is of order 3, this is <x>, if x is of order 9, this is <x^3> = <x^(9/3)>, if x is of order 27, this is <x^9> = <x^(27/3)>.

    so the thing to see is that G has some element of order 3. if x is of order 3, we take H = <x>.

    if x is of order 9, x^3 is of order 3, and we take H = <x^3>.

    if x is of order 27, x^9 is of order 3, and we take H = <x^9>. and any non-identity element of G must have one of these 3 orders.
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  5. #5
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    OK. Let me take another value, not 27, to see if I got the point.

    Suppose that G is a group with |G|=6. Prove that G contains a subgroup H satisfying |H|=3.

    To prove this, I need to find at least one such subgroup.
    Element of g with power of 2 generates a subgroup H={g^2, g^4, g^6=e}. It has three elements, thus |H|=3.
    (I took the g^2 because 6/3=2).

    Is this it?

    Thanks.
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  6. #6
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    that proof only works if G is cyclic. not all groups of order 6 are cyclic (for example, S3 is order 6, and it isn't even abelian).

    if G HAS an element of order 6, then it IS cyclic, in which case if g is that element, g^2 has order 3, as in your reply.

    however, you need to rule out the possibility that every element of G besides the identity is of order 2.

    in actual fact, every group of order 6 does indeed have an element of order 3.

    suppose that a,b were two elements of a group G of order 6, and that both of them have order 2.

    if ab has order 2 as well, then abab = (ab)^2 = e. but if abab = e, then ababb = eb, so

    aba = b, and then aaba = ab, so ba = ab. but then {e,a,b,ab} is a subgroup of G of order 4, and 4 does not divide 6.

    so it must be, that if we have a group of order 6, with 2 elements of order 2, that ab has either order 3, or order 6.

    if ab has order 3, it is the element we're looking for, if not, then (ab)^2 is the element we're looking for.

    this rules out the possibility that a group of order 6 can be the identity and 5 elements of order 2.
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