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Thread: Abstract Algebra Help!!!

  1. #1
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    Abstract Algebra Help!!!

    Hey im preparing for my second exam and I need major help with these problems. I would greatly appreciate any help, hints or answers, that could help me understand how to do these problems. So that I can do well thanks!!!

    G denotes a group

    1. Let |G|=p*g*r with different primes p, g, r, and let H,K<=G with |H|=p*g, |K|=g*r. Show:|H (intersection) K|=g.

    2. Let H be the normal subgroup G, a in G and |H|=10. The element aH of the group G/H has order 3; what are the possibilities for |a|? Explain.

    3. Let G be a group and G' be the subgroup generated by S={x^-1*y^-1*x*y| x,y in G}. Then:

    a. Prove that G' is normal in G.
    b. Prove that G/G' is abelian.
    c. If G/N is abelian, prove that G'<=N.
    d. Prove that if H<=G and G'<=H then H is a subgroup of G.

    4. Suppose that G is an abelian group of order 120 and that G has exactly 3 elements of order 2. Determine the isomorphism class of G. Explain.
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  2. #2
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    Quote Originally Posted by GreenandGold View Post
    1. Let |G|=p*g*r with different primes p, g, r, and let H,K<=G with |H|=p*g, |K|=g*r. Show:|H (intersection) K|=g.
    If $\displaystyle H,K$ are subgroups of a finite group $\displaystyle G$ then $\displaystyle |HK|\cdot |H\cap K| = |H||K|$.

    Now since $\displaystyle H\cap K$ is a subgroup of $\displaystyle H$ and $\displaystyle K$ it means its order divides $\displaystyle |H| = pg$ and $\displaystyle gr$. Therefore, $\displaystyle |H\cap K| = 1\text{ or }g$. We cannot have $\displaystyle |H\cap K| = 1$ for that would imply $\displaystyle pgr=|G| \geq |HK| = |H||K| = pg^2r$. A contradiction.

    Therefore, the only possible case is that $\displaystyle |H\cap K| = g$.

    2. Let H be the normal subgroup G, a in G and |H|=10. The element aH of the group G/H has order 3; what are the possibilities for |a|? Explain.
    If $\displaystyle |aH| = 3$ in the group $\displaystyle G/H$ it means $\displaystyle (aH)^3 = H$ and therefore $\displaystyle a^3 \in H$. But $\displaystyle |H| = 10$. Therefore, $\displaystyle (a^3)^{10} = a^{30} = 1$. Since $\displaystyle a\not = 1$ it means the possible values for $\displaystyle |a|$ are $\displaystyle 2,3,5,6,10,15,30$.
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    Quote Originally Posted by GreenandGold View Post
    3. Let G be a group and G' be the subgroup generated by S={x^-1*y^-1*x*y| x,y in G}. Then:

    a. Prove that G' is normal in G.
    b. Prove that G/G' is abelian.
    c. If G/N is abelian, prove that G'<=N.
    d. Prove that if H<=G and G'<=H then H is a subgroup of G.
    For $\displaystyle x,y\in G$ define $\displaystyle [x,y] = xyx^{-1}y^{-1}$ then show that $\displaystyle g[x,y]g^{-1} = [gxg^{-1},gyg^{-1}]$.
    Therefore, we see that $\displaystyle G'$ is a normal subgroup.

    Since this group is normal we can form the factor group $\displaystyle G/G'$. Let $\displaystyle aG',bG' \in G/G'$. To show $\displaystyle (aG')(bG') = (bG')(aG')$ it is necessary and sufficient to prove that $\displaystyle (ab)(ba)^{-1} \in G' \implies aba^{-1}b^{-1} \in G'$. This is true by definition of $\displaystyle G'$.

    If $\displaystyle G/N$ is abelian it means $\displaystyle (aN)(bN) = (bN)(aN)$ and so $\displaystyle aba^{-1}b^{-1} \in N$. This is true for all $\displaystyle a,b\in G$. Therefore, $\displaystyle G'\subseteq N$.
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  4. #4
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    Quote Originally Posted by GreenandGold View Post
    4. Suppose that G is an abelian group of order 120 and that G has exactly 3 elements of order 2. Determine the isomorphism class of G. Explain.
    Let $\displaystyle P$ be the Sylow $\displaystyle 2$-subgroup of order $\displaystyle 8$.

    Consider, $\displaystyle G/P$. This group has order $\displaystyle 15$ and therefore $\displaystyle G/P \simeq \mathbb{Z}_{15}$ because that is the only group up to isomorphism.

    Since $\displaystyle |P| = 8$ and it is abelian implies that $\displaystyle P$ must be isomorphic to one of the following groups: $\displaystyle \mathbb{Z}_8, \mathbb{Z}_2^3, \mathbb{Z}_4\times \mathbb{Z}_2$.
    Elements of order $\displaystyle 2$ must be contained in $\displaystyle P$. Therefore, by hypothesis, we are told that $\displaystyle P$ has exactly $\displaystyle 3$ elements of order $\displaystyle 2$.
    The first two groups (on the list) do not satisfy this requirement.
    However, $\displaystyle \mathbb{Z}_4\times \mathbb{Z}_2$ has that requirement with elements $\displaystyle (0,1),(2,1),(2,0)$.

    Therefore, $\displaystyle G/P \simeq G/(\mathbb{Z}_4 \times \mathbb{Z}_2) \simeq \mathbb{Z}_3 \times \mathbb{Z}_5$.

    Now we "cross multiply" and we get, $\displaystyle G\simeq \mathbb{Z}_2\times \mathbb{Z}_3 \times \mathbb{Z}_4 \times \mathbb{Z}_5$.
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