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Thread: 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.
    $\displaystyle C_8$ is cyclic while $\displaystyle C_4 \times C_2$ is not and $\displaystyle C_2\times C_2 \times C_2$ is not. Now $\displaystyle C_4\times C_2$ has an element of $\displaystyle 4$ while $\displaystyle C_2\times C_2\times C_2$ 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 $\displaystyle Q_8$. It is non-abelian.
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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 $\displaystyle d=\gcd(m,n)$. We claim if $\displaystyle (x,y)\in (C_m,C_n)$ then $\displaystyle (x,y)^{mn/d} = (1,1)$. Note, $\displaystyle x^{mn/d} = \left( x^m \right)^{n/d} = 1^{n/d} = 1$. Similarly, $\displaystyle y^{mn/d} = \left( y^n \right)^{m/d} = 1^{m/d} = 1$. Now, finally since $\displaystyle n,m$ are not relatively prime it means $\displaystyle d>1$ and so $\displaystyle mn/d<mn$, hence, no element has order $\displaystyle mn$.
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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.
    $\displaystyle H$ is cyclic because every subgroup of a cyclic group is cyclic. Furthermore, $\displaystyle G/H$ is cyclic because there exists $\displaystyle a\in G$ such that $\displaystyle \left< a \right> = G$ this means $\displaystyle aH$ generates $\displaystyle G/H$.

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

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

    (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 $\displaystyle H_1,H_2,...,H_n$ are subgroups such that $\displaystyle H=\bigcup_{k=1}^n H_k $ is a subgroup. Consider $\displaystyle h=h_1h_2...h_n\in H$ where $\displaystyle h_i\in H_i$ because it is closed. So then $\displaystyle h \in H_1$ without lose of generality. So $\displaystyle h_1^{-1}h = h_2...h_n\in H_1$ and so on until $\displaystyle h_n \in H_1$, permuting these products it means $\displaystyle h_1,...,h_n\in H_1$. Thus, it means $\displaystyle H=H_1$. Now assume $\displaystyle G$ is finite. If every $\displaystyle x\in G$ is contained in the conjugate subgroup $\displaystyle gHg^{-1}$ for some $\displaystyle g\in G$ then it means $\displaystyle \bigcup_{g\in G}gHg^{-1}=G$ this union is finite for we assumed $\displaystyle G$ was finite. But by above this union is simply $\displaystyle g_0Hg_0^{-1}$ for some $\displaystyle g_0\in G$. But that means $\displaystyle |H| = |g_0Hg_0^{-1}| = |G|$ and since $\displaystyle H\subseteq G$ means $\displaystyle H=G$. We have shown that if $\displaystyle H\subseteq G$ has this property (for a finite group) then $\displaystyle H=G$ by above argument.
    -------
    Now it would be nice if this union can generalize, i.e. if $\displaystyle G$ is any group then $\displaystyle \bigcup_{i\in I}H_i = H$ for $\displaystyle \{ H_i|i\in I\}$ implies $\displaystyle H_k = H$ for some $\displaystyle k\in I$. But sadly this does not work. Consider $\displaystyle G = \mathbb{Z}_2^{\infty}$ to be an $\displaystyle \aleph_0$-tuple $\displaystyle (a,b,c,...)$. Now let $\displaystyle H_1 = \{ (0,0,...),(1,0,...)\}, H_2 = \{ (0,0,...), (0,1,...)\}, ... $. Then they unionize to $\displaystyle G$. But no one groups contains them all.
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