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Math Help - Modern Algebra, Groups, Integral domains, fields...

  1. #1
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    Modern Algebra, Groups, Integral domains, fields...

    I have an assgnment due on Thursday in Mordern Algebra and I must say that I am very lost I got (I think) 3 problem out of the 6 and I was wondering if someone could help me out and check the work I have done already.
    Ok here it is, any suggestion for 1, 2 and 3????

    1. Suppose that  \phi : I50 \to I50 is a group isomorphism  \phi(7) = 13 .
    Determine  \phi(x)

    I was trying to find generator of I50 all I found is that 3^{15}= 7 and that 3^{17}=13 but <3> doesn’t map all of I50 so I am not sure what to do.


    2. Suppose that G is a finite Abelian Group and G has no element of order 2. Show that the mapping g \to g^2 is an automorphism of G.

    3. Le G be a group of order  p^n where p is prime. Prove that the center G cannot have order  p^{(n-1)}.

    4. Let G be a group. If  H={g^5|g \in G} is a subgroup of G, prove that it is a normal subgroup of G.

    My answer: Given  H={g^5|g \in G} is a subgroup of G we must show that  xHx^{-1} \subset H, \forall x \in G
    Let  x \in G and  y^5 \in H for some  z \in G.
    Since  y^5 \in H then by definition  y \in H
    Let  z = xyx^{-1}, with  xyx^{-1} \in G according to the definition of a conjugate.
    Then let  z^5=(xyx^{-1})^5 = (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1})
     = xy(x^{-1}x)y(x^{-1}x)y(x^{-1}x)y(x^{-1}x)yx^{-1}=xy^5x^{-1}
    By definition  z^5 \in H, thus xy^5x^{-1} \in H.
    Therefore H is a normal subgroup.


    5.Show that there is an integral domain with exactly 4 elements.

    My answer: Let F4 = { \begin{bmatrix}x & y\\x & (x+y)\end{bmatrix} , x,y \in F2}
     F4 = { \begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}1&1\\1&0\end{bmatrix}, \begin{bmatrix}0&1\\1&1\end{bmatrix}}

    F4 is a field of 4 elements because it is a commutative ring (close under multiplication and addition), it has a unity  \begin{bmatrix}1&0\\0&1\end{bmatrix} and each non zero elements has a multiplicative inverse. Since F4 is a field then it is an integral domain with 4 elements.

    6.Let R be an integral domain and  a,b \in R. Suppose  a^3 = b^3 and  a^7=b^7. Prove that  a= b

    My answer:  a^7=b^7
    a.a^6=b.b^6
    a.(a^3)^2= b.(b^3)^2
    a=b since  a^3=b^3.
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  2. #2
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    Quote Originally Posted by ynn6871 View Post
    2. Suppose that G is a finite Abelian Group and G has no element of order 2. Show that the mapping g \to g^2 is an automorphism of G.
    Define \phi : G\to G by \phi (g) = g^2.
    Since G is finite it is sufficient to prove that \phi is injective for \phi to be a bijection.
    If \phi(g_1) = \phi(g_2) \implies g_1^2 = g_2^2 \implies (g_1g_2^{-1})^2 \implies g_1g_2^{-1} = e\implies g_1=g_2
    And \phi(g_1g_2) = (g_1g_2)^2 = g_1^2g_2^2 = \phi (g_1) \phi (g_2).

    3. Le G be a group of order  p^n where p is prime. Prove that the center G cannot have order  p^{(n-1)}.
    Say Z(G) has order p^{n-1}.
    Pick a\in G - Z(G) and construct C(a) = \{x\in G | xa = ax \}.
    Now prove that C(a) is a subgroup of G.
    Notice that Z(G) is properly contained in C(a).
    This forces |C(a)| = p^n because Z(G) \subset C(a) \subseteq G by Lagrange's theorem.
    But then it must mean that a commutes with all of G.
    Therefore, a\in Z(G). A contradiction!
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  3. #3
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    Hi.

    For 2), verfying that \Phi_{2}:g\mapsto g^{2} is a group morphism isn't hard because of commutativity. What is Ker(\Phi_{2})? The set of elements in G whose square is the identity, i.e. the identity and all elements of order 2. So \Phi_{2} is injective and, since G is finite, bijective.

    For 3), let Z(G) be the center of G. With the hypothesis, as Z(G) is normal in G, G/Z(G) order is p, so is a cyclic group. Hence G is abelian, and Z(G)=G. Contradiction.
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  4. #4
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    For number 2. can you tell me what are the implication of
    G has no element of order 2
    ??? Did you use it to prove it is a bijection???
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  5. #5
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    Quote Originally Posted by ynn6871 View Post
    4. Let G be a group. If  H={g^5|g \in G} is a subgroup of G, prove that it is a normal subgroup of G.

    My answer: Given  H={g^5|g \in G} is a subgroup of G we must show that  xHx^{-1} \subset H, \forall x \in G
    Let  x \in G and  y^5 \in H for some  z \in G.
    Since  y^5 \in H then by definition  y \in H
    Let  z = xyx^{-1}, with  xyx^{-1} \in G according to the definition of a conjugate.
    Then let  z^5=(xyx^{-1})^5 = (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1}) (xyx^{-1})
     = xy(x^{-1}x)y(x^{-1}x)y(x^{-1}x)y(x^{-1}x)yx^{-1}=xy^5x^{-1}
    By definition  z^5 \in H, thus xy^5x^{-1} \in H.
    Therefore H is a normal subgroup.
    I do not like how you wrote up your solutions. Let h\in H and g\in G we wish to show ghg^{-1} \in H. Notice that h=x^5 for some x\in G. Therefore, ghg^{-1} = gx^5g^{-1} = (gxg^{-1})^5 \in H.

    5.Show that there is an integral domain with exactly 4 elements.
    Many ways to answer (all of which are isomorphic), one can be found here.

    ??? Did you use it to prove it is a bijection???
    Yes. Because if g_1^2 = g_2^2 \implies g_1 = g_2 since there is no element of order 2.
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  6. #6
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    I am sorry but I am still having a hard time understanding the no elements of order 2.
    Does it mean that since there exists no elements of order 2 then g^2=g?? I just don't understand the concept behind, I am sorry I am having a hard time with it all!

    Also do you have any suggestion as to how to approach problem number 1 ??
    1. Suppose that is a group isomorphism .
    Determine
    Clic-clac: when you said For 3),
    let be the center of . With the hypothesis, as is normal in , order is , so is a cyclic group. Hence is abelian, and . Contradiction.
    , can I just assume that is normal in G or do I have to prove it somehow, the rest of your proof makes total sense to me (I think) it's just that one part.
    Let me rephrase the solution really quick:

    We assume that Z(G) has order p^{n-1} and that it is normal in G, then G/Z(G) has order p, since  p / p^{n-1} is equal to p. Then since p is prime then G is abelian, and Z(G)=G which is a contradiction. But what happens in the case that Z(G) is not normal, is Z(G) ever not normal???

    Thank you very much for your help guys, I am getting really confused in this class!!!
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  7. #7
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    Quote Originally Posted by ynn6871 View Post
    I am sorry but I am still having a hard time understanding the no elements of order 2.
    Does it mean that since there exists no elements of order 2 then g^2=g?? I just don't understand the concept behind, I am sorry I am having a hard time with it all!
    Define the function \phi : G\to G by \phi (x) = x^2. Since G has no element of order two it means that if x^2 = e then x=e. Therefore, to show that \phi is one-to-one we need to show that if \phi (a) = \phi( b ) then a=b. Now if a^2 = b^2 then a^2b^{-2} = e and so (ab^{-1})^2 = e because the group is abelian. But that means ab^{-1} =e and finally a=b.

    Let me rephrase the solution really quick:

    We assume that Z(G) has order p^{n-1} and that it is normal in G, then G/Z(G) has order p, since  p / p^{n-1} is equal to p. Then since p is prime then G is abelian, and Z(G)=G which is a contradiction. But what happens in the case that Z(G) is not normal, is Z(G) ever not normal???
    Try to prove the following theorem: If H is normal in K so that K/H is cyclic and H is abelian then K has to be abelian. You will realize this is a special case when K=G and H = Z(G). But how do we show that Z(G) is normal. That is easy! Let g\in G then gZ(G) = Z(G)g because Z(G) contains all elements which commute with everyone. We see that the left cosets of Z(G) are equal to the right cosets of Z(G). That is one of the equivalent conditions for a subgroup to be normal. Thus, Z(G) is normal.
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  8. #8
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    Any suggestion on how to go about problem 1???
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  9. #9
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    What is I50 ?
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  10. #10
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    Integers modulo 50
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  11. #11
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    Ok! Then, in I50, we get:

    7^{2}=-1\ , so 7^{4}=1.

    \phi(1)=\phi(7^{3}\times 7)=7^{3}\times \phi(7)=-7\times 13=-41=9

    So \phi:I50\rightarrow I5O:n\mapsto 9n
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  12. #12
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    Quote Originally Posted by clic-clac View Post
    I50, we get:

    7^{2}=-1\ , so 7^{4}=1.

    \phi(1)=\phi(7^{3}\times 7)=7^{3}\times \phi(7)=-7\times 13=-41=9

    So \phi:I50\rightarrow I5O:n\mapsto 9n

    Can you explain your thought process a little bit??? Why did you pick powers of 7??? And why did you start with \phi(1)????
    Will this function be a bijection???? is everything mapped?
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  13. #13
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    Well a group endomorphism f of I50 will be determined by the range of 1, which is a generator, and would be like : n\mapsto f(1)n (that's easy to show with morphism properties).

    Here, we can see that 7^{2}\equiv -1\ (50). So 7^{4}\equiv 1\ (50).

    Then I just calculate \phi(1) (modulo 50 of course).

    As you may see, with that result, we have \phi(7)=7\phi(1)=7\times 9=63 \equiv 13\ (50), that sounds not too bad.

    Finally, pgcd(9,50)=1, so 9 is a generator of I50, hence your morphism is bijective.


    Ps. Since 7 and 13 are also relative primes with 50, \phi(7)=\phi(13) means the range of a generator is a generator, with morphism properties, that implies \phi is an isomophism!
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  14. #14
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    <br />
\phi(1)=\phi(7^{3}\times 7)=7^{3}\times \phi(7)<br />
    Ok I understand where you are coming from a little better, the only other thing I don't understand is how can you just pull out the  7^3 outside of \phi() ??? shouldn't it be something like <br />
\phi(1)=\phi(7^{3}\times 7)=\phi(7)\times\phi(7)\times\phi(7)\times\phi(7)=  11  <br />
which gives you \phi(n)=11nbut then I don't get the desired  \phi(7) = 13 but \phi(7) = 27 which is not what I want.
    Last edited by ynn6871; December 3rd 2008 at 12:03 PM.
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  15. #15
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    Oh that's a tempting error

    The group is (I50,+), so \phi(a+b)=\phi(a)+\phi(b) \\(not \phi(ab)=\phi(a)\phi(b) !)

    Therefore \phi(n\times7)=\phi(7+7+...+7)=\phi(7)+\phi(7)+...  +\phi(7)=n\phi(7)

    In our case, n=7^{3}
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