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Math Help - Abstract Algebra

  1. #1
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    Abstract Algebra

    I have no clue how to attack these question one. All help or hints is appreciated.

    G always denotes a group

    1. Let G be finite and G not equal to {e}. Show that G has an element of prime order.

    2. Prove that isomorphic groups have isomorphic automorphism groups.

    3. Let a, b be in G . If |a| and |b| are relatively prime (i.e. gcd (|a|, |b|)=1), then <a> intersection <b> ={e}. Prove the last statement.

    Thanks!
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  2. #2
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    Quote Originally Posted by GreenandGold View Post
    1. Let G be finite and G not equal to {e}. Show that G has an element of prime order.
    Let a\in G - \{ e \} and construct \left< a \right>. Let |\left< a \right> | = n>1. The integer n has a prime divisor p. The element a^{n/p} has order p.

    2. Prove that isomorphic groups have isomorphic automorphism groups.
    Let \phi : G_1 \to G_2 be a group isomorphism.
    Define \hat \phi : \text{Aut}(G_1)\to \text{Aut}(G_2) by \hat \phi (\theta)(a) = \phi(\theta(a)).
    Show this is an isomorphism.

    3. Let a, b be in G . If |a| and |b| are relatively prime (i.e. gcd (|a|, |b|)=1), then <a> intersection <b> ={e}. Prove the last statement.
    Notice that \left< a\right> \cap \left< b \right> is a subgroup of \left< a \right> and \left< b\right>. By Lagrange's theorem it means | \left< a\right> \cap \left< b\right>| divides |\left< a\right>| = |a| and \left< b\right>| = |b|. Therefore, |\left< a\right> \cap \left< b \right>| = 1 \implies \left< a \right> \cap \left< b \right> = \{ e \}.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Let a\in G - \{ e \} and construct \left< a \right>. Let |\left< a \right> | = n>1. The integer n has a prime divisor p. The element a^{n/p} has order p.


    Let \phi : G_1 \to G_2 be a group isomorphism.
    Define \hat \phi : \text{Aut}(G_1)\to \text{Aut}(G_2) by \hat \phi (\theta)(a) = \phi(\theta(a)).
    Show this is an isomorphism.


    Notice that \left< a\right> \cap \left< b \right> is a subgroup of \left< a \right> and \left< b\right>. By Lagrange's theorem it means | \left< a\right> \cap \left< b\right>| divides |\left< a\right>| = |a| and \left< b\right>| = |b|. Therefore, |\left< a\right> \cap \left< b \right>| = 1 \implies \left< a \right> \cap \left< b \right> = \{ e \}.

    So need to show that \hat \phi (\theta)(a) = \phi(\theta(a)) is a structure preserving function?
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  4. #4
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    Quote Originally Posted by GreenandGold View Post
    So need to show that \hat \phi (\theta)(a) = \phi(\theta(a)) is a structure preserving function?
    Yes
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