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  1. #1
    Super Member craig's Avatar
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    Groups & Subgroups help

    Got a tricky question here that I'm not too sure how to approach:

    F is a group of order 12 under multiplication and with e the neutral element.

    i) H = \{ f \in F : f^{-1} = f \} is the set of self inverse elements of F.

    I need to show that H is a subgroup of F if F is abelian.

    ii) We now need to assume that F is cyclic, with generator \alpha and find the order and all the elements of H

    iii) Give an example of a group F of order 12 for which H = \{ f \in F : f^{-1} = f \} isn't a subgroup. (For this one obviously you need to use the sub-group criteria, just not sure how to apply them).

    Greatly appreciate any help/pointers with this.

    Thanks in advance

    Craig
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    i) H = \{ f \in F : f^{-1} = f \} is the set of self inverse elements of F.

    I need to show that H is a subgroup of F if F is abelian.

    SG1 = Since neutral element is its self inverse, H is not empty

    SG2 = a*b member of H, for all a,b member of H
    (a*b)^{-1}= (b^{-1})*(a^{-1})= b*a
    since f is abelian  b*a=a*b
    (a*b)^{-1}= a*b
    Hence it satisfies conditon of H
    SG2 holds
    SG3 = every element is its self inverse


     a^{-1}=a,
    Thus a^{-1} member of H
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  3. #3
    Super Member craig's Avatar
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    Thanks for the reply, it actually sounds quite simple when you put it like that

    For the part ii). I need to try and find a way to list the elements of F in terms of \alpha, how would I go about this, something like x (mod \alpha) or would it involve powers of \alpha, considering the group is under multiplication?

    And for the third section, obviously need to find a group that's not a subgroup. Seeing as SG1 and 3 hold, need to find a non-abelian group with order 12. Would the Dihedral Group of order 12, D_{12} be such a group. If so how would I list the elements, in terms of rot_{\beta} where \beta are the 12 degrees of rotation?

    Thanks again for the help, really appreciate it.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by craig View Post
    Got a tricky question here that I'm not too sure how to approach:

    F is a group of order 12 under multiplication and with e the neutral element.

    i) H = \{ f \in F : f^{-1} = f \} is the set of self inverse elements of F.

    I need to show that H is a subgroup of F if F is abelian.

    ii) We now need to assume that F is cyclic, with generator \alpha and find the order and all the elements of H

    iii) Give an example of a group F of order 12 for which H = \{ f \in F : f^{-1} = f \} isn't a subgroup. (For this one obviously you need to use the sub-group criteria, just not sure how to apply them).

    Greatly appreciate any help/pointers with this.

    Thanks in advance

    Craig
    For ii) look at the cyclic group of order 12. This, well, is the only cyclic group of order 12...!

    So, for which n does (\alpha^n)^2 = 1 hold?

    For ii) you clearly have that any subgroup is closed under inverses. So, you need to look at either your favourite or second-favourite non-abelian group of order 12 (it definately doesn't hold for your third favourite non-abelian group of order 12, T) and find two elements in it of order 2 which, when multiplied together, give you an element not of order 2.

    I would suggest looking in D_{12}, the dihedral group of order 12 (it contains a copy of S_3 in it, and S_3 is generated by its 2-cycles. For example, (12)(23)=(132)).
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  5. #5
    Super Member craig's Avatar
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    Thanks for the reply.

    Quote Originally Posted by Swlabr View Post
    For ii) look at the cyclic group of order 12. This, well, is the only cyclic group of order 12...!

    So, for which n does (\alpha^n)^2 = 1 hold?
    With it being cyclic, you have the following.

    \{ e \alpha, \alpha^2, \alpha^3 ... \alpha^{11} \}

    \{ e, \alpha^2, \alpha^4 ... \alpha^{10} \}

    ... etc

    \{ e = \alpha^{12} \}

    So the only element of H would be \alpha^{12} = \langle e \rangle, which has order 1?

    For part 3, I considered D_{12}, the symmetries of a regular hexagon.

    For this would I just list the 6 symmetries and the 6 reflections in D_{12}, and to show that it is not a subgroup explain how it's not commutative?

    Thanks again for the pointers.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by craig View Post
    Thanks for the reply.



    With it being cyclic, you have the following.

    \{ e \alpha, \alpha^2, \alpha^3 ... \alpha^{11} \}

    \{ e, \alpha^2, \alpha^4 ... \alpha^{10} \}

    ... etc

    \{ e = \alpha^{12} \}

    So the only element of H would be \alpha^{12} = \langle e \rangle, which has order 1?

    For part 3, I considered D_{12}, the symmetries of a regular hexagon.

    For this would I just list the 6 symmetries and the 6 reflections in D_{12}, and to show that it is not a subgroup explain how it's not commutative?

    Thanks again for the pointers.
    In \mathbb{Z}_{12} there is one element of order 2. In fact, \mathbb{Z}_{n} has a subgroup of order m if and only if m \mid n.

    Write down all the elements of \mathbb{Z}_{12} in a list, and beside them write down their squares. This will reveal the element of order 2.

    For D_{12}, again list all the elements and their squares. This will give you the elements of order 2. Then, take two elements of order 2 and multiply them together. Do this until you find a pair which does not appear in your list of elements of order 2.
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  7. #7
    Super Member craig's Avatar
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    Quote Originally Posted by Swlabr View Post
    In \mathbb{Z}_{12} there is one element of order 2. In fact, \mathbb{Z}_{n} has a subgroup of order m if and only if m \mid n.

    Write down all the elements of \mathbb{Z}_{12} in a list, and beside them write down their squares. This will reveal the element of order 2.
    Ahh thankyou, the only element of order 2 in \mathbb{Z}_{12} would be 6 right, because this cycle is (0,6)?

    Would this mean that 6 is it's own inverse in \mathbb{Z}_{12} then?


    Quote Originally Posted by Swlabr View Post
    For D_{12}, again list all the elements and their squares. This will give you the elements of order 2. Then, take two elements of order 2 and multiply them together. Do this until you find a pair which does not appear in your list of elements of order 2.

    When you say the elements, do you mean the different rotations and reflections in D_{12}?
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by craig View Post
    Ahh thankyou, the only element of order 2 in \mathbb{Z}_{12} would be 6 right, because this cycle is (0,6)?

    Would this mean that 6 is it's own inverse in \mathbb{Z}_{12} then?
    yup

    Quote Originally Posted by craig View Post
    When you say the elements, do you mean the different rotations and reflections in D_{12}?
    Yes - the elements of the group are the things generated by the rotations and reflections.
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  9. #9
    Super Member craig's Avatar
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    Thanks again!
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