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Thread: 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 $\displaystyle \phi : I50 \to I50 $ is a group isomorphism $\displaystyle \phi(7) = 13 $.
    Determine $\displaystyle \phi(x)$

    I was trying to find generator of I50 all I found is that $\displaystyle 3^{15}= 7$ and that $\displaystyle 3^{17}=13$ but $\displaystyle <3>$ 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 $\displaystyle g \to g^2$ is an automorphism of G.

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

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

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


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

    My answer: Let $\displaystyle F4 = { \begin{bmatrix}x & y\\x & (x+y)\end{bmatrix} , x,y \in F2}$
    $\displaystyle 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 $\displaystyle \begin{bmatrix}1&0\\0&1\end{bmatrix}$ 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 $\displaystyle a,b \in R$. Suppose $\displaystyle a^3 = b^3$ and $\displaystyle a^7=b^7$. Prove that $\displaystyle a= b$

    My answer: $\displaystyle a^7=b^7$
    $\displaystyle a.a^6=b.b^6$
    $\displaystyle a.(a^3)^2= b.(b^3)^2$
    $\displaystyle a=b$ since $\displaystyle a^3=b^3$.
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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 $\displaystyle g \to g^2$ is an automorphism of G.
    Define $\displaystyle \phi : G\to G$ by $\displaystyle \phi (g) = g^2$.
    Since $\displaystyle G$ is finite it is sufficient to prove that $\displaystyle \phi$ is injective for $\displaystyle \phi$ to be a bijection.
    If $\displaystyle \phi(g_1) = \phi(g_2) \implies g_1^2 = g_2^2 \implies (g_1g_2^{-1})^2 \implies g_1g_2^{-1} = e\implies g_1=g_2$
    And $\displaystyle \phi(g_1g_2) = (g_1g_2)^2 = g_1^2g_2^2 = \phi (g_1) \phi (g_2)$.

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

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

    For 3), let $\displaystyle Z(G)$ be the center of $\displaystyle G$. With the hypothesis, as $\displaystyle Z(G)$ is normal in $\displaystyle G$, $\displaystyle G/Z(G)$ order is $\displaystyle p$, so is a cyclic group. Hence $\displaystyle G$ is abelian, and $\displaystyle Z(G)=G$. 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 $\displaystyle H={g^5|g \in G}$ is a subgroup of G, prove that it is a normal subgroup of G.

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

    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 $\displaystyle g_1^2 = g_2^2 \implies g_1 = g_2$ 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 $\displaystyle g^2=g$?? 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 $\displaystyle p^{n-1}$ and that it is normal in G, then G/Z(G) has order p, since $\displaystyle p / p^{n-1}$ 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 $\displaystyle g^2=g$?? I just don't understand the concept behind, I am sorry I am having a hard time with it all!
    Define the function $\displaystyle \phi : G\to G$ by $\displaystyle \phi (x) = x^2$. Since $\displaystyle G$ has no element of order two it means that if $\displaystyle x^2 = e$ then $\displaystyle x=e$. Therefore, to show that $\displaystyle \phi$ is one-to-one we need to show that if $\displaystyle \phi (a) = \phi( b )$ then $\displaystyle a=b$. Now if $\displaystyle a^2 = b^2$ then $\displaystyle a^2b^{-2} = e$ and so $\displaystyle (ab^{-1})^2 = e$ because the group is abelian. But that means $\displaystyle ab^{-1} =e$ and finally $\displaystyle a=b$.

    Let me rephrase the solution really quick:

    We assume that Z(G) has order $\displaystyle p^{n-1}$ and that it is normal in G, then G/Z(G) has order p, since $\displaystyle p / p^{n-1}$ 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 $\displaystyle H$ is normal in $\displaystyle K$ so that $\displaystyle K/H$ is cyclic and $\displaystyle H$ is abelian then $\displaystyle K$ has to be abelian. You will realize this is a special case when $\displaystyle K=G$ and $\displaystyle H = Z(G)$. But how do we show that $\displaystyle Z(G)$ is normal. That is easy! Let $\displaystyle g\in G$ then $\displaystyle gZ(G) = Z(G)g$ because $\displaystyle Z(G)$ contains all elements which commute with everyone. We see that the left cosets of $\displaystyle Z(G)$ are equal to the right cosets of $\displaystyle Z(G)$. That is one of the equivalent conditions for a subgroup to be normal. Thus, $\displaystyle Z(G)$ 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 $\displaystyle 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 $\displaystyle I50$, we get:

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

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

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

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

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

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

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

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

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

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

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


    Ps. Since $\displaystyle 7$ and $\displaystyle 13$ are also relative primes with $\displaystyle 50$, $\displaystyle \phi(7)=\phi(13)$ means the range of a generator is a generator, with morphism properties, that implies $\displaystyle \phi$ is an isomophism!
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  14. #14
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    $\displaystyle
    \phi(1)=\phi(7^{3}\times 7)=7^{3}\times \phi(7)
    $
    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 $\displaystyle 7^3$ outside of $\displaystyle \phi()$ ??? shouldn't it be something like $\displaystyle
    \phi(1)=\phi(7^{3}\times 7)=\phi(7)\times\phi(7)\times\phi(7)\times\phi(7)= 11
    $which gives you $\displaystyle \phi(n)=11n$but then I don't get the desired $\displaystyle \phi(7) = 13$ but $\displaystyle \phi(7) = 27$ which is not what I want.
    Last edited by ynn6871; Dec 3rd 2008 at 11:03 AM.
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  15. #15
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    Oh that's a tempting error

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

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

    In our case, $\displaystyle n=7^{3}$
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