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Thread: Sylow subgroup of group of order 56

  1. #1
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    Sylow subgroup of group of order 56

    Let $\displaystyle G$ be a group of order 56, $\displaystyle T \in \text{Syl}_2(G), S \in \text{Syl}_7(G)$ be given. Then $\displaystyle |T| = 8, |S| = 7$. Assume $\displaystyle S$ is not normal, so then $\displaystyle T \lhd G$. Show that $\displaystyle T \cong C_2^3$.

    Here's what I have so far. My idea is to show that each element of $\displaystyle T$ has order dividing 2. Let $\displaystyle z \in S$ be given such that $\displaystyle \langle z \rangle = S$. Then $\displaystyle G / T = \{T z^i : i=0,\dots,6\}$ and $\displaystyle \forall x \in T, (Tz)^x = Tz$. I've also got that $\displaystyle G = T \rtimes S$, although I fail to seen how any of this is helping.

    Any advice?
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  2. #2
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    Quote Originally Posted by Giraffro View Post
    Let $\displaystyle G$ be a group of order 56, $\displaystyle T \in \text{Syl}_2(G), S \in \text{Syl}_7(G)$ be given. Then $\displaystyle |T| = 8, |S| = 7$. Assume $\displaystyle S$ is not normal, so then $\displaystyle T \lhd G$. Show that $\displaystyle T \cong C_2^3$.

    Here's what I have so far. My idea is to show that each element of $\displaystyle T$ has order dividing 2. Let $\displaystyle z \in S$ be given such that $\displaystyle \langle z \rangle = S$. Then $\displaystyle G / T = \{T z^i : i=0,\dots,6\}$ and $\displaystyle \forall x \in T, (Tz)^x = Tz$. I've also got that $\displaystyle G = T \rtimes S$, although I fail to seen how any of this is helping.

    Any advice?

    So you have that $\displaystyle T\rtimes S\Longrightarrow$ there exists a non-trivial homomorphism $\displaystyle S\rightarrow Aut(T)$ . Well, what are the possible orders of automorphisms groups of groups of order 8?

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    So you have that $\displaystyle T\rtimes S\Longrightarrow$ there exists a non-trivial homomorphism $\displaystyle S\rightarrow Aut(T)$ . Well, what are the possible orders of automorphisms groups of groups of order 8?

    Tonio
    So $\displaystyle \exists \phi : S \to \text{Aut}(T)$ a group homomorphism such that $\displaystyle \forall x, y \in T, \forall i, j \in \mathbb{Z}, x z^i y z^j = x \phi(z^i)(y) z^{i+j}$. If $\displaystyle \phi(z) = \text{id}_T$ then $\displaystyle \forall x \in T, z^x = x^{-1} z x = x^{-1} \phi(z)(x) z = z$, so as $\displaystyle S = \langle z \rangle, S \lhd G$, which is a contradiction. Therefore $\displaystyle \phi(z) \ne \text{id}_T$ and so as $\displaystyle o(\phi(z)) \ne o(z), o(\phi(z)) = 7$. So it follows that $\displaystyle \exists x \in T$ such that $\displaystyle x,\phi(z)(x),\dots,\phi(z^6)(x)$ are distinct and have the same order. So $\displaystyle T$ has at least 7 elements of a specific order.

    $\displaystyle
    \begin{array}{c|ccc}
    & \text{Order} \, 2 & \text{Order} \, 4 & \text{Order} \, 8 \\
    \hline
    C_8 & 1 & 2 & 4 \\
    C_4 \times C_2 & 3 & 4 & 0 \\
    C_2^3 & 7 & 0 & 0 \\
    D_8 & 5 & 2 & 0 \\
    Q & 1 & 6 & 0
    \end{array}$

    Therefore $\displaystyle T \cong C_2^3$.

    Thanks for the help!
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