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Math Help - Direct product

  1. #1
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    Commutator subgroup

    Let G_{1} \times G_{2}=Z_{12} \times S_{24}.
    1. The commutator subgroup of Z_{12} \times S_{24} I got is e \times A_{24}.
    2. The quotient group of \frac{Z_{12} \times S_{24}}{e \times A_{24}} = Z_{12} \times Z_{2}.

    Is above 1&2 correct?
    Since the maximum order of elements in Z_{12} \times Z_{2} is 12, the above 2 seems wrong. ((12 * 24!) / (0.5 *24!) = 24 ).
    If 2 is wrong, what is the quotient group of 2?
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  2. #2
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    Quote Originally Posted by aliceinwonderland View Post
    Let G_{1} \times G_{2}=Z_{12} \times S_{24}.
    1. The commutator subgroup of Z_{12} \times S_{24} I got is e \times A_{24}.
    2. The quotient group of \frac{Z_{12} \times S_{24}}{e \times A_{24}} = Z_{12} \times Z_{2}.

    Is above 1&2 correct?
    Since the maximum order of elements in Z_{12} \times Z_{2} is 12, the above 2 seems wrong. ((12 * 24!) / (0.5 *24!) = 24 ).
    If 2 is wrong, what is the quotient group of 2?
    We need to know that (G_1\times G_2)' = G_1' \times G_2'. To prove this just consider a commutator, c=(a_1,a_2)(b_1,b_2)(a_1^{-1},a_2^{-1})(b_1^{-1},b_2^{-1}) where a_1,b_1\in G_1 and a_2,b_2\in G_2. We can rewrite c = (a_1b_1a_1^{-1}b_1^{-1}, a_2b_2a_2^{-1}b_2^{-1}). But a_1b_1a_1^{-1}b_1^{-1}\in G_1' and a_2b_2a_2^{-1}b_2^{-1} \in G_2'. Therefore, the subgroup generated by the commutators i.e. (G_1\times G_2)' is equal to G_1'\times G_2'.

    Now the commutator subgroup of \mathbb{Z}_{12} is \{ e \} (not e !) While it is an established result that the commutator of S_{24} is A_{24}.
    Therefore, I agree with your result.
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  3. #3
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    2. is definitely correct. And note that the order of (Z/12)X(Z/2) is 24 not 12. The highest order of a single element is 12 but there are 24 distinct elements in the whole group.
    Last edited by bulls6x; January 8th 2009 at 04:15 PM.
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