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Math Help - Abelian subgroups proof

  1. #1
    Senior Member Educated's Avatar
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    Abelian subgroups proof

    If A is an abelian group with A \unlhd G and B is any subgroup of G, prove that A \cap B \unlhd AB
    So I'm guessing AB is just defined as AB : = \{ ab \, \, | \, \, a \in A, b \in B \}

    I started off with:

    Let x \in A \cap B, a \in A, b \in B
    We must show that (ab) x (ab)^{-1} \in A \cap B

    (ab) x (ab)^{-1} = ab x b^{-1} a^{-1}

    And now here I'm stuck...
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  2. #2
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    Re: Abelian subgroups proof

    Hi,
    My earlier post on exponential notation for conjugation is used here. I'm a firm advocate of this notation; it makes lots of proofs much easier.

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  3. #3
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    Re: Abelian subgroups proof

    Thanks, I finally got it after reading it 3 or 4 times
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  4. #4
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    Re: Abelian subgroups proof

    Actually reading that again, I still don't get how it's a subgroup...

    So I realised there's an easier way to prove that it normal, but what about proving that it is an actual subgroup of AB, ie. A \cap B \leq AB

    Also, here's my reasoning for a normal subgroup:
    Since A and B are subgroups, A \cap B is also a subgroup. In particular since A is abelian and A \cap B \leq A then A \cap B is also abelian, hence it is also a normal subgroup of G, ie. A \cap B \unlhd G

    This means gxg^{-1} \in A \cap B \, \, \forall \, \, x \in A \cap B, \, \, \forall \, \, g \in G
    Since AB \subseteq G then
    gxg^{-1} \in A \cap B \, \, \forall \, \, x \in A \cap B, \, \, \forall \, \, g \in AB \subseteq G

    Would that work?
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  5. #5
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    Re: Abelian subgroups proof

    Hi again,
    Thanks from Educated
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  6. #6
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    Re: Abelian subgroups proof

    Suppose we have:

    $abxb^{-1}a^{-1}$ with $a \in A,b \in B, x\in A \cap B$.

    Since $AB$ is a group, $AB = BA$, so we can write this as:

    $b'a'xa'^{-1}b'^{-1}$.

    Now $A$ is abelian, and thus $a'xa'^{-1} = x$, so $b'a'xa'^{-1}b'^{-1} = b'xb'^{-1} \in B$

    Also, $A \unlhd G$, so since $b' \in B \leq G$, $b'xb'^{-1} \in A$, so $abxb^{-1}a^{-1} = b'a'xa'^{-1}b'^{-1} \in A\cap B$

    Proof that $AB$ is a subgroup iff $AB = BA$:

    Suppose $AB$ is a group. Then, by closure, $abab = a'b'$ for any $a \in A,b \in B$ and some $a' \in A, b' \in B$.

    Thus $ba = (a^{-1}a')(b'b^{-1}) \in AB$, so $BA \subseteq AB$.

    Also, since $AB$ is a group, for $ab \in AB$ we have $(ab)^{-1} \in AB$, so $ab = (a'b')^{-1}$ for some $a' \in A, b' \in B$, thus $ab = (a'b')^{-1} = b'^{-1}a'^{-1} \in BA$, so $AB \subseteq BA$.

    Now suppose $AB = BA$. We need to show closure and existence of inverses (that $e \in AB$ is trivial).

    So take any product $aba'b'$. Since $BA = AB$, we can write $ba' = a''b''$, so $aba'b' = (aa'')(b''b') \in AB$, which shows closure.

    Now if $ab \in AB$, then $(ab)^{-1} = b^{-1}a^{-1} \in BA = AB$, which shows $AB$ contains all inverses, and is thus a group.
    Last edited by Deveno; April 30th 2014 at 12:36 PM.
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  7. #7
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    Re: Abelian subgroups proof

    Hi,
    Here's some more discussion that I hope you find helpful:

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