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    I was hoping someone could check my responses to the above question as I'm pretty much self-studying the subject and hence have nobody to correct me nor a markscheme to look through.

    (a)
    1. The group should also exhibit closure.
    2. There should exist an identity element in the group.
    3. Every element in the group should possess an inverse.


    (b) (i)

    Commutativity is not satisfied, consider the counterexample: p \odot q = r whereas q \odot p = t hence implying that p \odot q \neq q \odot p and hence we can state that the group is not commutative. We also note that the Cayley table is not symetric about the leading diagonal, and hence not commutative.

    Associativity is also not satisfied, consider the counterexample: (p \odot q) \odot t = p whereas p \odot (q \odot t) = t, implying that (p \odot q) \odot t \neq p \odot (q \odot t), allowing us to state the group is not associative.

    Closure, however is satisfied, since the Cayley table shows us that for all a, b \in T we have that (a \odot b) \in T.

    There exists the identity element s in the group.

    Every element has an inverse, in fact, every element is a self-inverse, this can be seen by the presence of the identity element, s in every row/column of the Cayley table.

    (ii) I don't actually really understand Lagrange's theorem. Can anybody explain it to me, none of the resources I've sourced out seem to help me understand.

    (iii)(p \odot x) \odot x = x \odot p \Rightarrow x=q,r,t

    Thank you!
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    (Original post by Zacken)

    I was hoping someone could check my responses to the above question as I'm pretty much self-studying the subject and hence have nobody to correct me nor a markscheme to look through.

    (a)
    1. The group should also exhibit closure.
    2. There should exist an identity element in the group.
    3. Every element in the group should possess an inverse.


    (b) (i)

    Commutativity is not satisfied, consider the counterexample: p \odot q = r whereas q \odot p = t hence implying that p \odot q \neq q \odot p and hence we can state that the group is not commutative. We also note that the Cayley table is not symetric about the leading diagonal, and hence not commutative.

    Associativity is also not satisfied, consider the counterexample: (p \odot q) \odot t = p whereas p \odot (q \odot t) = t, implying that (p \odot q) \odot t \neq p \odot (q \odot t), allowing us to state the group is not associative.

    Closure, however is satisfied, since the Cayley table shows us that for all a, b \in T we have that (a \odot b) \in T.

    There exists the identity element s in the group.

    Every element has an inverse, in fact, every element is a self-inverse, this can be seen by the presence of the identity element, s in every row/column of the Cayley table.

    (ii) I don't actually really understand Lagrange's theorem. Can anybody explain it to me, none of the resources I've sourced out seem to help me understand.

    (iii)(p \odot x) \odot x = x \odot p \Rightarrow x=q,r,t

    Thank you!
    a) Strictly speaking you should specify that the identity is unique, but I don't know how finnicky they're being.

    b)i) Looks good.
    ii) Lagrange's theorem states that for a finite group G, the order of any subgroup H must divide the order of G. The proof involves defining cosets and the like, depending on how advanced your study is you may not need to know the proof.
    Applying this to your example should be relatively straightforward.

    iii) I haven't checked your working but for a small group like this, picking out values should be straightforward enough without doing anything too fancy.
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    (Original post by joostan)
    a) Strictly speaking you should specify that the identity is unique, but I don't know how finnicky they're being.

    b)i) Looks good.
    ii) Lagrange's theorem states that for a finite group G, the order of any subgroup H must divide the order of G. The proof involves defining cosets and the like, depending on how advanced your study is you may not need to know the proof.
    Applying this to your example should be relatively straightforward.

    iii) I haven't checked your working but for a small group like this, picking out values should be straightforward enough without doing anything too fancy.
    Identity is unique, good to note, thanks!

    Okaaaay, but then wouldn't the order of T here be 5, and the order of any of the subgroups be 2? And 2 does not divide 5? :confused:

    Are the subgroups meant to be: \{p,s\}, \{q,s\}, \{r,s\}, \{t,s\}

    Yups, we're expected to be able to define cosets and the like.

    iii) Great, we're only ever going to get small groups. But what method would you use if you had a group that was too big to pick the solutions out like that?

    Thanks for the help!
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    (Original post by Zacken)
    Identity is unique, good to note, thanks!

    Okaaaay, but then wouldn't the order of T here be 5, and the order of any of the subgroups be 2? And 2 does not divide 5? :confused:

    Are the subgroups meant to be: \{p,s\}, \{q,s\}, \{r,s\}, \{t,s\}

    Yups, we're expected to be able to define cosets and the like.

    iii) Great, we're only ever going to get small groups. But what method would you use if you had a group that was too big to pick the solutions out like that?

    Thanks for the help!
    Since 2 does not divide 5 you can deduce that there are no such subgroups of order 2.

    As for iii) the only thing that's jumping out at me is that the equation.
    px^2=xp can be rearranged marginally to x^2=p^{-1}xp and thus we require x^2 \in ccl(x)
    If there is any slick way of doing it for large groups then I don't know it.

    EDIT: I did some checking, I didn't notice that T is not a group, and so there can be a group of order 2 in the set, so you are correct with your statement of the 4 subsets being groups of order 2.
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    (Original post by joostan)
    Since 2 does not divide 5 you can deduce that there are no such subgroups of order 2.

    As for iii) the only thing that's jumping out at me is that the equation.
    px^2=xp can be rearranged marginally to x^2=p^{-1}xp and thus we require x^2 \in ccl(x)
    If there is any slick way of doing it for large groups then I don't know it.

    EDIT: I did some checking, I didn't notice that T is not a group, and so there can be a group of order 2 in the set.
    Would the equation not imply that you are using the property of associativity, that the particular group here doesn't have?

    Wait, what?! Aren't these subgroups?: \{p,s\}, \{q,s\}, \{r,s\}, \{t,s\} :eek:
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    (Original post by joostan)
    Since 2 does not divide 5 you can deduce that there are no such subgroups of order 2.

    As for iii) the only thing that's jumping out at me is that the equation.
    px^2=xp can be rearranged marginally to x^2=p^{-1}xp and thus we require x^2 \in ccl(x)
    If there is any slick way of doing it for large groups then I don't know it.

    EDIT: I did some checking, I didn't notice that T is not a group, and so there can be a group of order 2 in the set, so you are correct with your statement of the 4 subsets being groups of order 2.
    Oops, sorry, replied too fast!

    What would your full blown answer to (ii) be then?
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    (Original post by Zacken)
    Oops, sorry, replied too fast!

    What would your full blown answer to (ii) be then?
    I guess I'd say that it's another way of deducing that the set T is not a group.
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    (Original post by joostan)
    I guess I'd say that it's another way of deducing that the set T is not a group.
    Ah, I see, makes sense. Thank you!
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    (Original post by joostan)
    a) Strictly speaking you should specify that the identity is unique, but I don't know how finnicky they're being.
    No. Uniqueness can be proved it is not an axiom.
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    (Original post by joostan)
    a) Strictly speaking you should specify that the identity is unique, but I don't know how finnicky they're being.
    Uniqueness of identity is not a group axiom, it can be derived from them though.
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    (Original post by Zacken)
    Would the equation not imply that you are using the property of associativity, that the particular group here doesn't have?

    Wait, what?! Aren't these subgroups?: \{p,s\}, \{q,s\}, \{r,s\}, \{t,s\} :eek:
    T is not a group, so the theorem does not actually hold.
 
 
 
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