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Thread: abelian normal group

  1. #1
    Senior Member abhishekkgp's Avatar
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    abelian normal group

    Let $\displaystyle G$ be a group of order $\displaystyle 3825$. Prove that if $\displaystyle H$ is a normal subgroup of order $\displaystyle 17$ in $\displaystyle G$ then $\displaystyle H \leq Z(G)$, where $\displaystyle Z(G)$ represents the center of $\displaystyle G$.
    (Please do not use sylow's or cauchy's theorems.)
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    Quote Originally Posted by abhishekkgp View Post
    Let $\displaystyle G$ be a group of order $\displaystyle 3825$. Prove that if $\displaystyle H$ is a normal subgroup of order $\displaystyle 17$ in $\displaystyle G$ then $\displaystyle H \leq Z(G)$, where $\displaystyle Z(G)$ represents the center of $\displaystyle G$.
    (Please do not use sylow's or cauchy's theorems.)


    If you don't want Sylow and Cauchy theorems, then I suppose you want even less semidirect

    products and stuff, thus the question is: what do you want? What stuff is allowed or wanted by you?

    Tonio
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    Senior Member roninpro's Avatar
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    Is there a nice way to do this if the Sylow theorems are allowed? I can't see anything short of a group classification argument.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by roninpro View Post
    Is there a nice way to do this if the Sylow theorems are allowed? I can't see anything short of a group classification argument.
    Indeed, I feel as though the simplest way to do this is to note that it is of the form $\displaystyle p^2q$ with $\displaystyle p,q$ prime and $\displaystyle q\notequiv 1\text{ mod }p$ and thus abelian...that said the only ways I know how to prove this are Sylow's theorems or rep. theory.
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  5. #5
    Senior Member roninpro's Avatar
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    Just noting the arithmetic: $\displaystyle 3825=3^2\cdot 5^2\cdot 17$.
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    Quote Originally Posted by abhishekkgp View Post
    Let $\displaystyle G$ be a group of order $\displaystyle 3825$. Prove that if $\displaystyle H$ is a normal subgroup of order $\displaystyle 17$ in $\displaystyle G$ then $\displaystyle H \leq Z(G)$, where $\displaystyle Z(G)$ represents the center of $\displaystyle G$.
    (Please do not use sylow's or cauchy's theorems.)
    (*) If H is a subgroup of G, then the factor group $\displaystyle N_G(H)/C_G(H)$ is isomorphic to a subgroup of Aut H (Hungerford, "Algebra", p92).

    By assumption, $\displaystyle N_G(H)=G$. Since H is a group of order 17, H is isomorphic to the cyclic group of order 17, i.e., $\displaystyle C_{17}$. Thus, |Aut(H)|=16.

    Now, $\displaystyle |G/C_G(H)|$ divides both 16 and 3825 by Lagrange's theorem and (*), it follows that $\displaystyle |G/C_G(H)|=1$. Can you conclude from here?
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