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Thread: Subgroup

  1. #1
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    Subgroup

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

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

    My attempt for for part (a):

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

    Proof:

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

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

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

    Here run into difficulties: Without knowing how to show $\displaystyle b^{-1}\in H$ , 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 $\displaystyle (G, \cdot)$ is an infinite group and $\displaystyle H \subseteq G$ is closed under $\displaystyle \cdot$ then $\displaystyle (H, \cdot)$ is not neccesarily a subgroup of $\displaystyle G$.

    Now, regarding part a:
    Let $\displaystyle h \in H$. $\displaystyle H$ is closed under * and so $\displaystyle h^n \in H \ \forall n \in \mathbb{N}$. $\displaystyle H$ is finite , thus $\displaystyle o(h)$ 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 $\displaystyle H$ of $\displaystyle G$ is a subgroup by "One Step Subgroup Test;" i.e, if $\displaystyle a*b \in H$, then $\displaystyle a*b^{-1}\in H$. This sure makes it a lot easier.


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

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

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

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

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

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

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

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

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

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

    $\displaystyle H\not=\varnothing, H\subset G$ where $\displaystyle G$ is a group, then $\displaystyle H$ is a subgroup of $\displaystyle G$ if and only if $\displaystyle ab\in H$ and $\displaystyle a^{-1} \in H$.
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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 $\displaystyle (G, \cdot)$ is an infinite group and $\displaystyle H \subseteq G$ is closed under $\displaystyle \cdot$ then $\displaystyle (H, \cdot)$ is not neccesarily a subgroup of $\displaystyle G$.

    Now, regarding part a:
    Let $\displaystyle h \in H$. $\displaystyle H$ is closed under * and so $\displaystyle h^n \in H \ \forall n \in \mathbb{N}$. $\displaystyle H$ is finite , thus $\displaystyle o(h)$ 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 $\displaystyle h^n \in H \forall n \in \mathbb{Z}$.
    Don't we need the negative numbers for inverses so that $\displaystyle h^{-z}\cdot h^{+z}=e$?
    Don't we need $\displaystyle {0}\in \mathbb{Z}$ so that $\displaystyle h^0=1$ for identity?

    Finally, what's $\displaystyle o(h)$?
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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 $\displaystyle H$ being a group. It only said $\displaystyle H \not= \varnothing$ and $\displaystyle H \subset G$. Being a subset of $\displaystyle G$ 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:

    $\displaystyle H\not=\varnothing, H\subset G$ where $\displaystyle G$ is a group, then $\displaystyle H$ is a subgroup of $\displaystyle G$ if and only if $\displaystyle ab\in H$ and $\displaystyle a^{-1} \in H$.
    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 $\displaystyle h^n \in H \forall n \in \mathbb{Z}$.
    Don't we need the negative numbers for inverses so that $\displaystyle h^{-z}\cdot h^{+z}=e$?
    Don't we need $\displaystyle {0}\in \mathbb{Z}$ so that $\displaystyle h^0=1$ for identity?

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

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

    After you have that $\displaystyle h^{-1} \in H \ \forall h \in H$ and that $\displaystyle H$ is closed under multiplication, identity is easy..
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  9. #9
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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 $\displaystyle H$ 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 $\displaystyle h^n \in H, \forall n \in \mathbb{Z}$.

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

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

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

    After you have that $\displaystyle h^{-1} \in H \ \forall h \in H$ and that $\displaystyle H$ 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 $\displaystyle (\forall h \in H)(\forall n \in \mathbb{N})h^n \in H$ (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 $\displaystyle h \in H$ and let $\displaystyle o(h)$ be the order of $\displaystyle h$. Then $\displaystyle h^{o(h)} = e$ but also $\displaystyle h^{o(h)} = h^{o(h)-1+1} = h^{o(h)-1}\cdot h = h \cdot h^{o(h)-1}$

    That is, for any $\displaystyle h \in H, ~ e = h\cdot h^{o(h)-1}=h^{o(h)-1}\cdot h$ and therefore the inverse of $\displaystyle h$ is $\displaystyle h^{o(h)-1}$. 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 $\displaystyle (\mathbb{Z},+)$ that would be closed under addition but not have the identity element?
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