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Math Help - Subgroup

  1. #1
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    Subgroup

    (a) Let (G,*) be a finite group. Prove that if H is a nonempty subset of G that is closed under *, then H is a subgroup of G.

    (b) Show that the result in (a) is false if (G,*) is an infinite group.

    My attempt for for part (a):

    Given:
    H\not = \varnothing,
    H \subset G,
    H is closed under *,
    and G is a finite group.

    Proof:

    If we can show H satisfies 4 properties; namely, that it is closed under *; it is associative; it has an identity, and each of its element has an inverse, then H is a subgroup of G.

    Since H is closed under *, then if a, b \in H, then a*b \in H. Let a,b,c \in H. Since H \subset G, a,b,c \in G, (a*b)*c = a*(b*c), so G is associative, and H as well.

    For all b \in H, there is an inverse b^{-1} \in G.

    Here run into difficulties: Without knowing how to show b^{-1}\in H , I could go nowhere.

    For part (b), I find it hard to believe that an infinite group cannot have a subgroup.
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  2. #2
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    Part b doesn't mean that infinite groups don't have any subgroups -- it means that the condition, which you have proved sufficient for finite groups, does not work for infinite groups ie. if (G, \cdot) is an infinite group and H \subseteq G is closed under \cdot then (H, \cdot) is not neccesarily a subgroup of G.

    Now, regarding part a:
    Let h \in H. H is closed under * and so h^n \in H \  \forall n \in \mathbb{N}. H is finite , thus o(h) is finite (this is also why this condition is not sufficient for infinite groups) and both conclusions easily follow (inverse and identity).
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  3. #3
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    Dr. Math says, there is a simple way to proof that a subset H of G is a subgroup by "One Step Subgroup Test;" i.e, if a*b \in H, then a*b^{-1}\in H. This sure makes it a lot easier.


    If that's the case, we can say since for a,b \in H ,
    a*b\in H \Rightarrow a*b^{-1}\in H .
    Since for all b\in H, there is b^{-1}\in H, and since a*b\in H, it follows that b*b^{-1}=e \in H, and so
    H is a subgroup of G.

    Since H\not = \varnothing, there exist x\in H for which x\cdot e = e\cdot x = x for all x\in H

    What do you think?
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  4. #4
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    I think you mean "if a,b \in H then a \cdot b^{-1} \in H". This is equivalent to what you're trying to prove, since if you prove this then:

    1) You prove that H is closed under *
    2) You prove that  e \in H (why?)
    3) You prove that h \in H \Rightarrow h^{-1} \in H (why?)

    Also, your last statement is illogical - first you're saying that there exists such an x \in H (ie. you take a specific x) , but then you say that the statement is true for all x \in H!
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    I think you mean "if a,b \in H then a \cdot b^{-1} \in H". This is equivalent to what you're trying to prove, since if you prove this then:

    1) You prove that H is closed under *
    2) You prove that  e \in H (why?)
    3) You prove that h \in H \Rightarrow h^{-1} \in H (why?)
    I did this because nothing is said about H being a group. It only said H \not= \varnothing and H \subset G. Being a subset of G does not necessarily make it a group.

    In this case, I don't need to prove that H is closed because it's given.

    Also, your last statement is illogical - first you're saying that there exists such an x \in H (ie. you take a specific x) , but then you say that the statement is true for all x \in H!
    Yes, \exist x and \forall x are contradiction.

    I wrote this step because I felt that H has an identity only if x\cdot e = e \cdot x = x, \forall x \in H.

    By the way, in another book, the Subgroup Test says:

    H\not=\varnothing, H\subset G where G is a group, then H is a subgroup of G if and only if ab\in H and a^{-1} \in H.
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  6. #6
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    Quote Originally Posted by Defunkt View Post
    Part b doesn't mean that infinite groups don't have any subgroups -- it means that the condition, which you have proved sufficient for finite groups, does not work for infinite groups ie. if (G, \cdot) is an infinite group and H \subseteq G is closed under \cdot then (H, \cdot) is not neccesarily a subgroup of G.

    Now, regarding part a:
    Let h \in H. H is closed under * and so h^n \in H \  \forall n \in \mathbb{N}. H is finite , thus o(h) is finite (this is also why this condition is not sufficient for infinite groups) and both conclusions easily follow (inverse and identity).
    I can see there be inverses and identity if h^n \in H \forall n \in \mathbb{Z}.
    Don't we need the negative numbers for inverses so that h^{-z}\cdot h^{+z}=e?
    Don't we need {0}\in \mathbb{Z} so that h^0=1 for identity?

    Finally, what's o(h)?
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  7. #7
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    Quote Originally Posted by novice View Post
    I did this because nothing is said about H being a group. It only said H \not= \varnothing and H \subset G. Being a subset of G does not necessarily make it a group.
    You did what? Sorry, I didn't follow.

    By the way, in another book, the Subgroup Test says:

    H\not=\varnothing, H\subset G where G is a group, then H is a subgroup of G if and only if ab\in H and a^{-1} \in H.
    That is virtually the same test. The only difference is that here, you need to explicitly check for the inverse.
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  8. #8
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    Quote Originally Posted by novice View Post
    I can see there be inverses and identity if h^n \in H \forall n \in \mathbb{Z}.
    Don't we need the negative numbers for inverses so that h^{-z}\cdot h^{+z}=e?
    Don't we need {0}\in \mathbb{Z} so that h^0=1 for identity?

    Finally, what's o(h)?
    o(h) is the order of h \in H, ie. the smallest integer n \in \mathbb{N} such that h^n = e.

    Since the order of h is finite and h^n \in H \ \forall n \in \mathbb{N}, you can write h^{o(h)} = h^{o(h) - 1 + 1} = h^{o(h)-1} \cdot h = e and so h^{o(h)-1} = h^{-1} and h^{o(h)-1} \in H.

    After you have that h^{-1} \in H \ \forall h \in H and that H is closed under multiplication, identity is easy..
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    Quote Originally Posted by Defunkt View Post
    You did what? Sorry, I didn't follow.
    Sorry, I meant I did those two things, showing the inverse and identity, because we didn't know whether H is a group or not until it is shown that it's associative, has identity, and inverses for all elements.

    I understand the example you gave has all those things; except I have questions about whether or not h^n \in H, \forall n \in \mathbb{Z}.

    I ask because I can't see h^n \inH, \forall n \in \mathbb{N} would have an inverse or identity.

    Don't we need \mathbb{Z}^{-} for the inverses and h^0 for identity?
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  10. #10
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    Quote Originally Posted by Defunkt View Post
    o(h) is the order of h \in H, ie. the smallest integer n \in \mathbb{N} such that h^n = e.

    Since the order of h is finite and h^n \in H \ \forall n \in \mathbb{N}, you can write h^{o(h)} = h^{o(h) - 1 + 1} = h^{o(h)-1} \cdot h = e and so h^{o(h)-1} = h^{-1} and h^{o(h)-1} \in H.

    After you have that h^{-1} \in H \ \forall h \in H and that H is closed under multiplication, identity is easy..
    I see it now. You are talking about cyclic group, which I have not learned yet, but luckily, I have read it on Wikipedia. Truly understood your example now. Thanks for being so patient.
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  11. #11
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    Quote Originally Posted by novice View Post
    I see it now. You are talking about cyclic group, which I have not learned yet, but luckily, I have read it on Wikipedia. Truly understood your example now. Thanks for being so patient.
    You're very welcome, but I wasn't talking about cyclic groups at all!
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  12. #12
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    Quote Originally Posted by novice View Post
    I see it now. You are talking about cyclic group, which I have not learned yet, but luckily, I have read it on Wikipedia. Truly understood your example now. Thanks for being so patient.
    Ah, it seems you got a bit confused. Since H is closed to *, we get that  (\forall h \in H)(\forall n \in \mathbb{N})h^n \in H (try to prove this). This, of course, does not mean that H is cyclic.

    We want to show that H is a subgroup of G. In order to show this, we want to show that H has the identity, is closed under * (this we are already given) and that each element in H has an inverse.

    To show each element has an inverse - since H is a finite group, every element in it has finite order. Let h \in H and let o(h) be the order of h. Then h^{o(h)} = e but also h^{o(h)} = h^{o(h)-1+1} = h^{o(h)-1}\cdot h = h \cdot h^{o(h)-1}

    That is, for any h \in H, ~ e = h\cdot h^{o(h)-1}=h^{o(h)-1}\cdot h and therefore the inverse of h is h^{o(h)-1}. Now it is just left to show that the identity element is in H (I will leave that for you). This concludes (a).

    For (b), can you think about some subset of (\mathbb{Z},+) that would be closed under addition but not have the identity element?
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