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Math Help - Third Iso Theorem

  1. #1
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    Third Iso Theorem

    Let G be a group and let H and K be normal subgroups of G with H a subgroup of K. Then K/H\trianglelefteq G/H and (G/H)/(K/H)\cong G/K

    How do I show:
    K/H\trianglelefteq G/H
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    Re: Third Iso Theorem

    most of the work here is showing that K/H is a subgroup of G/H. normality quickly follows.

    (H is clearly normal in K, since it's normal in a larger group G).

    so consider Hk, Hk' in K/H. is Hkk'^-1 is K/H?

    what would it mean for K/H to be normal in G/H?

    we would need (Hg)(K/H)(Hg)^-1 to be in K/H.

    an element of (Hg)(K/H)(Hg)^-1 is the coset product: (Hg)(Hk)(Hg)^-1 = Hgkg^-1.

    but K is normal in G, so....
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    Re: Third Iso Theorem

    define the map \phi: G/H \longrightarrow G/K by \phi(gH)=gK, which is well-defined because H \subseteq K. clearly \phi is a group homomorphism and \ker \phi = K/H. thus, by the first isomorphism theorem, K/H is a normal subgroup of G/H and (G/H)/(K/H) \cong G/K.
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    Re: Third Iso Theorem

    Quote Originally Posted by NonCommAlg View Post
    define the map \phi: G/H \longrightarrow G/K by \phi(gH)=gK, which is well-defined because H \subseteq K. clearly \phi is a group homomorphism and \ker \phi = K/H. thus, by the first isomorphism theorem, K/H is a normal subgroup of G/H and (G/H)/(K/H) \cong G/K.
    is it just me, or does there seem to be a spooky link between this question, and this one:

    http://www.mathhelpforum.com/math-he...es-191667.html ?
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    Re: Third Iso Theorem

    Quote Originally Posted by Deveno View Post
    is it just me, or does there seem to be a spooky link between this question, and this one:

    http://www.mathhelpforum.com/math-he...es-191667.html ?
    not quite! let A=\{v_i + N : \ i \in I\} be a basis for V/N and suppose that \sum_{i \in J} c_i(v_i + M) = 0 for some finite set J \subseteq I and scalars c_i. then \sum_{i \in J} c_iv_i \in M \subseteq N and so \sum_{i \in J} c_i(v_i + N)=0 implying c_i = 0, for all i, because the elements of A are linearly independent. thus the elements of \{v_i + M: \ i \in I \} are linearly independent and so W = span \{v_i + M : \ i \in I\} is a subspace of V/M isomorphic to V/N.
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  6. #6
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    Re: Third Iso Theorem

    well, the way i was looking at it was: suppose we define φ:V/N → V/M by φ(v+N) = v+M. φ is well-defined because M ⊆ N.

    of course, in this case φ(v+N) = 0 (that is, 0+M) implies v is in M, which in turn implies v is in N, so ker(φ) = N, so we have an isomorphism between φ(V/N) and V/N.

    the similarity being, the construction of φ (do we really need a basis? oh, and alex, if you're reading this, in light of what i wrote in the other thread, the fact that i'm asking this question here surely must be somewhat...ironic...)
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    Re: Third Iso Theorem

    Quote Originally Posted by Deveno View Post
    well, the way i was looking at it was: suppose we define φ:V/N → V/M by φ(v+N) = v+M. φ is well-defined because M ⊆ N.

    of course, in this case φ(v+N) = 0 (that is, 0+M) implies v is in M, which in turn implies v is in N, so ker(φ) = N, so we have an isomorphism between φ(V/N) and V/N.

    the similarity being, the construction of φ (do we really need a basis? oh, and alex, if you're reading this, in light of what i wrote in the other thread, the fact that i'm asking this question here surely must be somewhat...ironic...)
    your map,  \phi, is not well-defined.
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    Re: Third Iso Theorem

    Quote Originally Posted by NonCommAlg View Post
    your map,  \phi, is not well-defined.
    go on...
    Last edited by Deveno; November 11th 2011 at 09:21 PM.
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    Re: Third Iso Theorem

    if v \in N \setminus M, then v + N = 0 but \phi(v+N) = v + M \neq 0. for \phi to be well-defined we need to have N \subseteq M not M \subseteq N.
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    Re: Third Iso Theorem

    hmm...so you're saying choosing the v_i + N solves this problem, so we DO need a basis. huh. LIB!MR dux.
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    Re: Third Iso Theorem

    Quote Originally Posted by NonCommAlg View Post
    define the map \phi: G/H \longrightarrow G/K by \phi(gH)=gK, which is well-defined because H \subseteq K. clearly \phi is a group homomorphism and \ker \phi = K/H. thus, by the first isomorphism theorem, K/H is a normal subgroup of G/H and (G/H)/(K/H) \cong G/K.
    Is the whole proof really this compact?
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    Re: Third Iso Theorem

    yup.

    basically, the general strategy for proving G/K is isomorphic to G' is to find a surjective homomorphism G-->G', whose kernel is K, and then apply the FIT.
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    Re: Third Iso Theorem

    Quote Originally Posted by Deveno View Post
    yup.

    basically, the general strategy for proving G/K is isomorphic to G' is to find a surjective homomorphism G-->G', whose kernel is K, and then apply the FIT.
    Why is the kernel K\H?
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  14. #14
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    Re: Third Iso Theorem

    suppose φ(gH) = K, the identity of G/K. since φ(gH) = gK, this means g is in K. so gH is an element of K/H, which are the cosets of H in K.

    so ker(φ) ⊆ K/H.

    on the other hand, suppose we have kH as an element of G/H (so kH is a coset of H in G, where k is an element of K).

    then φ(kH) = kK = K, so every element of G/H of the form kH for k in K, is in ker(φ). thus K/H ⊆ ker(φ), so the two sets are equal.

    (remember, K/H is a subset (even stronger, a subgroup) of G/H consisting of only those elements kH, where k is in K).
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