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Math Help - Subgroups and Normal Subgroups

  1. #1
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    Subgroups and Normal Subgroups

    Well, my teacher has given us a study guide with no answers. I guess she is busy....anywho...if anyone could help I would really appreciate it.

    Suppose that G is a group and that A is a normal subgroup of G.
    1.) Suppose that A≤J≤G. Show that A is normal to J.
    2.) Suppose that A≤J≤G. show that J/A ≤ G/A.

    Attempt:

    1.) Suppose that A≤J≤G (subgroup notation). We also suppose that A is a normal Subgroup of G. This means that for every x in G, xA = Ax. So really, all I have are definitions and assumptions and not sure where to go to from there. I was thinking along that lines that since A is normal to G it would automatically follow that A is normal to J since J is a subgroup of G and A is a subgroup of J.

    2.) In order to prove that something is a subgroup of another then we have to show 3 things or two things. I also know that

    J/A = { gA | g is in J } and G/A = {gA | g is in G}.....

    a.) first we need to show that J/A is nonempty.
    b.) x in J/A and y in J/A imply xy is also in J/A.
    c.) x in J/A implies that x inverse is also in J/A.
    If I could get any help I would really appreciate it.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by crystalwater View Post
    Well, my teacher has given us a study guide with no answers. I guess she is busy....anywho...if anyone could help I would really appreciate it.

    Suppose that G is a group and that A is a normal subgroup of G.
    1.) Suppose that A≤J≤G. Show that A is normal to J.
    2.) Suppose that A≤J≤G. show that J/A ≤ G/A.

    Attempt:

    1.) Suppose that A≤J≤G (subgroup notation). We also suppose that A is a normal Subgroup of G. This means that for every x in G, xA = Ax. So really, all I have are definitions and assumptions and not sure where to go to from there. I was thinking along that lines that since A is normal to G it would automatically follow that A is normal to J since J is a subgroup of G and A is a subgroup of J.
    You are on the right lines - basically, if you pick any element x \in G then x^{-1}Ax = A. There is no reason you cannot just pick elements from J.

    Quote Originally Posted by crystalwater View Post
    2.) In order to prove that something is a subgroup of another then we have to show 3 things or two things. I also know that

    J/A = { gA | g is in J } and G/A = {gA | g is in G}.....

    a.) first we need to show that J/A is nonempty.
    b.) x in J/A and y in J/A imply xy is also in J/A.
    c.) x in J/A implies that x inverse is also in J/A.
    If I could get any help I would really appreciate it.
    Well, you know that J/A is non-empty and a group as A \unlhd J from part (1), so these 3 things have already been proven.

    What you need to show is that every element from J/A is an element from G/A.
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  3. #3
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    Alrighty. The first one is a little clearer.

    As for the second one, I was working on it and I said something along these lines. Also, if the first is proven correctly, then the second follows straight from that and all i have to do is prove that every element in J/A is an element in G/A?

    Well, i said something like since g is in J and since also g is in G then it follows that g is both in J and G since they are just multiples of the set A. Do I need to think in different ways?

    Again, thank you for the help so far.
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  4. #4
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    Maybe you can do the problem for me mesmo and email it to me ....just kidding. I see both of your points. For the most part, I "think" i understand it, emphasis on think.

    So, instead of picking elements in G, i would be picking elements in J? So if i prove that J is normal to G then A must be normal to J?

    * You know, sometimes I think this kind of math just talks in circles sometimes. And because of that I'm never sure of what to do.*

    Thanks guys.
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by crystalwater View Post
    Maybe you can do the problem for me mesmo and email it to me ....just kidding. I see both of your points. For the most part, I "think" i understand it, emphasis on think.

    So, instead of picking elements in G, i would be picking elements in J? So if i prove that J is normal to G then A must be normal to J?

    * You know, sometimes I think this kind of math just talks in circles sometimes. And because of that I'm never sure of what to do.*

    Thanks guys.
    No, you know next to nothing about J. If we take x \in J then clearly x \in G as J \leq G. Thus, x^{-1}Ax = A as this holds for all x \in G.
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