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Math Help - some problems about groups

  1. #1
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    some problems about groups

    1. prove no two of C_8, C_4 X C_2, C_2 X C_2 X C_2 are isomorphic.C_n is cyclic group of order n.

    Give, with justification, a group of order 8 that is not isomorphic to any of those groups.

    2. Prove that if m and n are coprime then C_m X C_n is cyclic.
    if m and n are not coprime. can it be cyclic?

    3. H is a normal subgroup of a finite group G. Which of the followings are true?

    (i) if G is cyclic then H and G/H are cyclic.
    (ii) If H and G/H are cyclic then G is cyclic.
    (iii) If G is abelian then H and G/H are abelian
    (iv) If H and G/H are abelian then G is abelian.

    4. find all homomorphisms between C_11 to C_14

    5. H is a subgroup of G and H not equal to G. Show that there is an element of G which does not belong to any subgroup of the form gHg^(-1) for g belongs to G
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  2. #2
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    Quote Originally Posted by szpengchao View Post
    1. prove no two of C_8, C_4 X C_2, C_2 X C_2 X C_2 are isomorphic.C_n is cyclic group of order n.
    C_8 is cyclic while C_4 \times C_2 is not and C_2\times C_2 \times C_2 is not. Now C_4\times C_2 has an element of 4 while C_2\times C_2\times C_2 does not. So all three are non-isomorphic.

    Give, with justification, a group of order 8 that is not isomorphic to any of those groups.
    Use Q_8. It is non-abelian.
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  3. #3
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    Quote Originally Posted by szpengchao View Post
    2. Prove that if m and n are coprime then C_m X C_n is cyclic.
    if m and n are not coprime. can it be cyclic?
    No, it cannot be cyclic. Let d=\gcd(m,n). We claim if (x,y)\in (C_m,C_n) then (x,y)^{mn/d} = (1,1). Note, x^{mn/d} = \left( x^m \right)^{n/d} = 1^{n/d} = 1. Similarly, y^{mn/d} = \left( y^n \right)^{m/d} = 1^{m/d} = 1. Now, finally since n,m are not relatively prime it means d>1 and so mn/d<mn, hence, no element has order mn.
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  4. #4
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    Quote Originally Posted by szpengchao View Post
    3. H is a normal subgroup of a finite group G. Which of the followings are true?
    (i) if G is cyclic then H and G/H are cyclic.
    H is cyclic because every subgroup of a cyclic group is cyclic. Furthermore, G/H is cyclic because there exists a\in G such that \left< a \right> = G this means aH generates G/H.

    (ii) If H and G/H are cyclic then G is cyclic.
    False. Let G=S_3 and H its 3-element subgroup.

    (iii) If G is abelian then H and G/H are abelian
    Yes. Define the natural map \pi: G\mapsto G/H then G/H is a homomorphic image of an abelian group so it must be abelian. Or you can say (xH)(yH) = (yH)(xH) since xy=yx.

    (iv) If H and G/H are abelian then G is abelian.
    False. Use the same example.
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  5. #5
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    Quote Originally Posted by szpengchao View Post
    5. H is a subgroup of G and H not equal to G. Show that there is an element of G which does not belong to any subgroup of the form gHg^(-1) for g belongs to G
    Maybe I am missing something really easy here. But I can only prove it for finite groups , I hope that is what is being ask for. Say H_1,H_2,...,H_n are subgroups such that H=\bigcup_{k=1}^n H_k is a subgroup. Consider h=h_1h_2...h_n\in H where h_i\in H_i because it is closed. So then h \in H_1 without lose of generality. So h_1^{-1}h = h_2...h_n\in H_1 and so on until h_n \in H_1, permuting these products it means h_1,...,h_n\in H_1. Thus, it means H=H_1. Now assume G is finite. If every x\in G is contained in the conjugate subgroup gHg^{-1} for some g\in G then it means \bigcup_{g\in G}gHg^{-1}=G this union is finite for we assumed G was finite. But by above this union is simply g_0Hg_0^{-1} for some g_0\in G. But that means |H| = |g_0Hg_0^{-1}| = |G| and since H\subseteq G means H=G. We have shown that if H\subseteq G has this property (for a finite group) then H=G by above argument.
    -------
    Now it would be nice if this union can generalize, i.e. if G is any group then \bigcup_{i\in I}H_i = H for \{ H_i|i\in I\} implies H_k = H for some k\in I. But sadly this does not work. Consider G = \mathbb{Z}_2^{\infty} to be an \aleph_0-tuple (a,b,c,...). Now let H_1 = \{ (0,0,...),(1,0,...)\}, H_2 = \{ (0,0,...), (0,1,...)\}, ... . Then they unionize to G. But no one groups contains them all.
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