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Math Help - Ugh! Modules....

  1. #1
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    Ugh! Modules....

    I'll write " Z" to denote the set of whole numbers

    Let n > 1 be a whole number and let a be a fixed whole number. Prove that

    H = {x ∈ Z, such that ax ≡ 0 (mod n)}

    is a subgroup of Z in the sum.

    In the sum is the part where I'm confused... I though you could only multiply with modules...?
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  2. #2
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    Quote Originally Posted by Tall Jessica View Post
    I'll write " Z" to denote the set of whole numbers

    Let n > 1 be a whole number and let a be a fixed whole number. Prove that

    H = {x ∈ Z, such that ax ≡ 0 (mod n)}

    is a subgroup of Z in the sum.

    In the sum is the part where I'm confused... I though you could only multiply with modules...?
    Hi Tall Jessica,

    Yes, you can add them.

    Let x,y \in H, check if xy^{-1} \in H and x^{-1} \in H is true.
    (Since it is under the operation of "+", y^{-1} is the same as -y, so xy^{-1} =x-y)
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  3. #3
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    Wink

    Quote Originally Posted by chabmgph View Post
    Hi Tall Jessica,

    Yes, you can add them.

    Let x,y \in H, check if xy^{-1} \in H and x^{-1} \in H is true.
    (Since it is under the operation of "+", y^{-1} is the same as -y, so xy^{-1} =x-y)

    Okay, so...

    I multiply both sides by y^{-1}, which gives me:

    (ax)y^{-1} ≡ 0 (mod n)

    (ay^{-1})(xy^{-1}) ≡ 0 (mod n)

    (a-y)(x-y) ≡ 0 (mod n)

    ...is this okay? I'm soooo lost... lol!
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  4. #4
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    Quote Originally Posted by Tall Jessica View Post
    Okay, so...

    I multiply both sides by y^{-1}, which gives me:

    (ax)y^{-1} ≡ 0 (mod n)

    (ay^{-1})(xy^{-1}) ≡ 0 (mod n)

    (a-y)(x-y) ≡ 0 (mod n)

    ...is this okay? I'm soooo lost... lol!
    Well, you can't assume that  (xy^{-1}) \equiv 0 \mod n, can you? That is what you are trying to prove. lol. (I know, I have those moments too.)

    You may want to try this:
    Since x,y \in H, we have
    ax \equiv 0 \mod n
    ay \equiv 0 \mod n
    Then, ax - ay \equiv 0 \mod n
    ...
    Does that help?
    Last edited by chabmgph; December 16th 2008 at 08:34 AM.
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  5. #5
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    Quote Originally Posted by chabmgph View Post
    Well, you can't assume that xy^{-1} \equiv 0 \mod n, can you? That is what you are trying to prove. lol. (I know, I have those moments too.)

    You may want to try this:
    Since x,y \in H, we have
    ax \equiv 0 \mod n
    ay \equiv 0 \mod n
    Then, ax - ay \equiv 0 \mod n
    ...
    Does that help?
    Oy, vey! It's NOT my day! lol! I HATE modules! I'll be honest, I can hardly work them.

    Let's see...

    ax - ay \equiv 0 \mod n

    I can take out a common factor a

    a (x - y) \equiv 0 \mod n

    Which would give me

    a xy^{-1} \equiv 0 \mod n

    ...yes?
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  6. #6
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    Quote Originally Posted by Tall Jessica View Post
    Oy, vey! It's NOT my day! lol! I HATE modules! I'll be honest, I can hardly work them.

    Let's see...

    ax - ay \equiv 0 \mod n

    I can take out a common factor a

    a (x - y) \equiv 0 \mod n

    Which would give me

    a xy^{-1} \equiv 0 \mod n

    ...yes?

    So that means xy^{-1} \in H, Now don't forget to check if x^{-1} \in H is true.
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  7. #7
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    I knew it :-)

    I'm awsome! lol!!


    Thank you :-)
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  8. #8
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    Quote Originally Posted by chabmgph View Post

    So that means xy^{-1} \in H, Now don't forget to check if x^{-1} \in H is true.

    Wait. If x \in H is true, doesn't it mean that x^{-1} \in H is also true?
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  9. #9
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    Quote Originally Posted by Tall Jessica View Post
    Wait. If x \in H is true, doesn't it mean that x^{-1} \in H is also true?
    Yup.
    Since ax is divisible by n, -ax = a(-x) is also divisible by n.
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  10. #10
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    Quote Originally Posted by chabmgph View Post
    Yup.
    You're just givin' me the runaround!

    You meanie!

    (Thanx) :-)
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